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Stream: learning: questions

Topic: natural transformations

John Baez (Jun 19 2021 at 20:35):

This is a puzzle for @Jan Pax. Nobody else should answer it.

There's a 2-category of categories, functors and natural transformations. Any group G gives a category BG with one object whose morphisms are elements of G, with composition in BG being multiplication in G. Let me just summarize this by saying "a group is a special sort of category".

So, what's the 2-category of categories that are groups, functors between these, and natural transformations between those?

The last part is the most interesting.

Morgan Rogers (he/him) (Jun 20 2021 at 07:20):

Once they've answered, I have a potentially interesting observation to make about this 2-category.

Jan Pax (Jun 20 2021 at 11:30):

Let $A,B$ be groups and $f,g:A\to B$ homomorphisms . The n.t. are arrows $\eta_c$ in the unique object $c$ of $B$ hence elements $b$ of $B$.
The composition is multiplication in $B$ hence by comutativity of the naturality diagram we have $b\cdot f(a)=g(a)\cdot b$ where $\cdot$ is the operation of $B$.Any such $b$ is a n.t.

Jan Pax (Jun 20 2021 at 14:33):

Now when I clicked on Mentions nothing appears, Before it listed all comments with @Jan Pax :
111.jpg

John Baez (Jun 20 2021 at 15:04):

I don't know why you're seeing no mentions, @Jan Pax. When I click on Mentions I see all comments mentioning my name.

John Baez (Jun 20 2021 at 15:06):

Please don't say anything yet, @Morgan Rogers (he/him), because while @Jan Pax has correctly answered my question, I want a bit more information from him!

John Baez (Jun 20 2021 at 15:07):

So, @Jan Pax: when two group homomorphisms $f, g: A \to B$ obey

$b f(a) = g(a) b$

for some $b \in B$ and all $a \in A$, what do people say about these homomorphisms?

John Baez (Jun 20 2021 at 15:08):

In other words: what's the usual way of talking about this situation?

Jan Pax (Jun 20 2021 at 15:09):

$f(a)=b^{-1}g(a)b$ and they call it conjugation with $b$.

John Baez (Jun 20 2021 at 15:12):

Right! I think people usually write

$g(a) = b f(a) b^{-1}$

and they say $g$ is $f$ conjugated by $b$.

John Baez (Jun 20 2021 at 15:12):

So conjugation, so important in group theory, becomes visible in the 2-category of groups even though it's not very visible in the category of groups!

Jan Pax (Jun 20 2021 at 15:13):

This time your question on me was easier than the tin problem.

John Baez (Jun 20 2021 at 15:14):

Tin can problem. (I guess the British call a tin can a "tin".)

Here's another related question. Say we have a group $G$, an automorphism $f: G \to G$, and a 2-morphism from $1: G \to G$ to $f$. What do we say about $f$ in this situation?

Jan Pax (Jun 20 2021 at 15:20):

I don't know. This means that $b=1$ ?

John Baez (Jun 20 2021 at 15:20):

Oh good, so my questions are finally becoming interesting! Figure it out! Does this imply $b = 1$ or not?

Jan Pax (Jun 20 2021 at 15:22):

You say 'good' so I'd guess from this that it does. But do not know why exactly.

John Baez (Jun 20 2021 at 15:22):

I say "good" because I'm forcing you to think.

Jan Pax (Jun 20 2021 at 15:25):

This is a sufficient condition but may be not necessary .

Jan Pax (Jun 20 2021 at 15:28):

Say if $G$ is commutative.

John Baez (Jun 20 2021 at 15:31):

Let's not say $G$ is commutative: conjugation is extremely boring when $G$ is commutative.

Jan Pax (Jun 20 2021 at 15:32):

So there is something called a centre ?

John Baez (Jun 20 2021 at 15:33):

I suggest doing this: choose a noncommutative group that you understand well, figure out its automorphisms, and figure out which have 2-morphisms to the identity.

John Baez (Jun 20 2021 at 15:33):

I guess the smallest noncommutative group is $S_3$, so I'd use this one.

John Baez (Jun 20 2021 at 15:34):

But if you like geometry you might choose the group of rotations in 3 dimensions, $\mathrm{SO}(3)$.

John Baez (Jun 20 2021 at 15:34):

Or if you like matrices you could use the group of $2 \times 2$ invertible real matrices, $\mathrm{GL}(2,\mathbb{R})$.

John Baez (Jun 20 2021 at 15:35):

It's good to understand the automorphisms of all these groups...

Jan Pax (Jun 20 2021 at 15:36):

$S_3$ is fine for me. I would say that inner automoprhisms is again $S_3$ but I haven't carried out all the calculations.

John Baez (Jun 20 2021 at 15:39):

What's an inner automorphism?

Jan Pax (Jun 20 2021 at 15:40):

one coming from conjugation with $b$. Not every automorphism is inner, I think they are more special.

John Baez (Jun 20 2021 at 15:41):

Now I think maybe you can answer my question:

John Baez said:

Say we have a group $G$, an automorphism $f: G \to G$, and a 2-morphism from $1: G \to G$ to $f$. What do we say about $f$ in this situation?

Jan Pax (Jun 20 2021 at 15:45):

So whenever conjugation of $f$ with $b$ is 1 ?

Jan Pax (Jun 20 2021 at 15:48):

or when $f$ is inner automorphism

John Baez (Jun 20 2021 at 15:52):

Yes: in this situation we say $f$ is an inner automorphism. That's the answer I wanted.

John Baez (Jun 20 2021 at 15:53):

So you were asking if every automorphism is inner:

Jan Pax said:

I don't know. This means that $b=1$ ?

John Baez (Jun 20 2021 at 15:53):

So now I'll ask you: is every automorphism inner?

Jan Pax (Jun 20 2021 at 15:54):

I have rather guessed my answer (with $f$ being inner) than I would know why it is so.

John Baez (Jun 20 2021 at 15:55):

Okay. So, is every automorphism inner?

Jan Pax (Jun 20 2021 at 15:55):

We may have a non-inner automorphism $f$ otherwise the notion would be uninteresting and would not exist.

Jan Pax (Jun 20 2021 at 15:56):

if $b$ is 1 then $f$ is inner.

John Baez (Jun 20 2021 at 15:56):

So you're claiming not every automorphism of every group is inner. Prove it!

Jan Pax (Jun 20 2021 at 15:57):

just divide by center of $G$. If this is trivial then every is inner.

Morgan Rogers (he/him) (Jun 20 2021 at 15:58):

Prove it!

John Baez (Jun 20 2021 at 15:58):

I'm running this show, Morgan.

John Baez (Jun 20 2021 at 15:58):

I don't think someone can stand having two bossy teachers.

John Baez (Jun 20 2021 at 15:59):

At least, not simultaneously!

Jan Pax (Jun 20 2021 at 15:59):

I'm happy with @Morgan Rogers (he/him) intervention :-)

John Baez (Jun 20 2021 at 15:59):

Okay! But mainly it's because he let you avoid answering my question.

John Baez (Jun 20 2021 at 15:59):

Please prove to me that not every automorphism is inner.

Jan Pax (Jun 20 2021 at 16:00):

we just divide out what may cause automorphism to be non-inner.

John Baez (Jun 20 2021 at 16:01):

That's not really a proof.

You seem to be avoiding concrete examples. Please give me an example of a group and an automorphism of that group that is not inner. This is the best way to answer my question.

Jan Pax (Jun 20 2021 at 16:01):

$S_5$ ?

John Baez (Jun 20 2021 at 16:02):

That's a group, please give me a non-inner automorphism of this group.

John Baez (Jun 20 2021 at 16:02):

By the way, I prefer good answers to quick answers.

John Baez (Jun 20 2021 at 16:03):

I'd be perfectly happy if you spend an hour or a day or a week thinking about groups and come back with an example of a non-inner automorphism.

John Baez (Jun 20 2021 at 16:03):

I'm not interested in speed.

Jan Pax (Jun 20 2021 at 16:05):

$Z$ with $1\to -1$ seems to me to be non-inner, but it is not finite.

John Baez (Jun 20 2021 at 16:06):

Why is that automorphism non-inner? (People say "outer".)

John Baez (Jun 20 2021 at 16:06):

I don't care if the group is finite.

Jan Pax (Jun 20 2021 at 16:08):

no integer $b$ will do for all natural numbers to yield their opposite. Again quick and not good ?

John Baez (Jun 20 2021 at 16:10):

It's okay, not great.

There's something about the group $\mathbb{Z}$ that makes it obvious that the identity automorphism is the only inner automorphism of this group. What is it?

Jan Pax (Jun 20 2021 at 16:11):

torsion-less ?

John Baez (Jun 20 2021 at 16:12):

You're just randomly guessing. I urge you to think until you are sure you have the right answer. I don't care if it takes time. It's good to practice figuring things out and being sure about them.

John Baez (Jun 20 2021 at 16:15):

The great thing about math is that you can think and think until you figure something out, just by thinking. This is not true of most subjects.

Jan Pax (Jun 20 2021 at 16:17):

Thank you for your advice. I'm never sure whatever I do or say, though. Are you sure that ZFC is consistent ?

John Baez (Jun 20 2021 at 16:18):

No, but I'm sure that there's no proof within ZFC that ZFC is consistent, because I took two courses where we went through the proof of Goedel's theorem. The second one, taught by Kripke, was extremely detailed.

John Baez (Jun 20 2021 at 16:19):

We had lots of long homeworks.

Jan Pax (Jun 20 2021 at 16:20):

There might be a proof that ZFC is consistent from ZFC. Can you guess when ?

John Baez (Jun 20 2021 at 16:20):

When it's not.

Jan Pax (Jun 20 2021 at 16:20):

Right!

John Baez (Jun 20 2021 at 16:21):

Anyway, the question I'm asking you now is simpler than the consistency of ZFC.

John Baez (Jun 20 2021 at 16:22):

What property of $\mathbb{Z}$ makes it obvious that the identity automorphism is the only inner automorphism of this group?

John Baez (Jun 20 2021 at 16:23):

Take your time.

John Baez (Jun 20 2021 at 16:23):

If you answer this correctly, you'll know other groups for which the identity automorphism is the only inner automorphism.

John Baez (Jun 20 2021 at 16:23):

And then you can find some more groups that have outer automorphisms.

Jan Pax (Jun 20 2021 at 17:45):

For for an automorphism to be inner we need that $-1 = b+1+(-b)$ for some integer $b$ which is not possible.

John Baez (Jun 20 2021 at 17:47):

You said "an" automorphism.

But I think you mean for the automorphism of $\mathbb{Z}$ sending 1 to -1 (and thus -1 to 1) to be inner we'd need $-1 = b + 1 + (-b)$. That's true.

John Baez (Jun 20 2021 at 17:48):

So, here's what I'm asking you again:

John Baez said:

What property of $\mathbb{Z}$ makes it obvious that the identity automorphism is the only inner automorphism of this group?

Jan Pax (Jun 20 2021 at 17:48):

commutativity

John Baez (Jun 20 2021 at 17:49):

Yes!!!

John Baez (Jun 20 2021 at 17:49):

Great!

So, tell me some finite groups whose only inner automorphism is the identity.

Jan Pax (Jun 20 2021 at 17:49):

$S_3$

John Baez (Jun 20 2021 at 17:50):

Why?

Jan Pax (Jun 20 2021 at 17:50):

it is not commutative ?

John Baez (Jun 20 2021 at 17:52):

I said:

What property of $\mathbb{Z}$ makes it obvious that the identity automorphism is the only inner automorphism of this group?

And you said:

commutativitity

And I said:

So, tell me some finite groups whose only inner automorphism is the identity.

And for some mysterious reason you gave me a noncommutative group.

John Baez (Jun 20 2021 at 17:53):

This is an example of why it pays to think carefully before answering my questions. Don't just type the first answer that comes to mind.

John Baez (Jun 20 2021 at 17:53):

Check to see if it's right.

Jan Pax (Jun 20 2021 at 17:53):

I still got stuck why my answer we would need $-1=b+1+(_b)$ is not correct? We need this for all $x$ so taking $\sigma(1)=-1$ we get something obviously false.

John Baez (Jun 20 2021 at 17:54):

I don't understand this.

John Baez (Jun 20 2021 at 17:55):

I think you're confused about something. I think to straighten out your thinking you should prove this for me:

Theorem. If a group is commutative, every inner automorphism of that group is the identity.

Jan Pax (Jun 20 2021 at 17:58):

because by conjugation with any element $b$ we get $x\mapsto b+(x)+(-b)$ hence the inner automorphism is an identity $1$.

John Baez (Jun 20 2021 at 17:59):

Okay, good. Now back to my earlier question:

So, tell me some finite groups whose only inner automorphism is the identity.

John Baez (Jun 20 2021 at 17:59):

$S_3$ is incorrect, unless I'm very very confused.

Jan Pax (Jun 20 2021 at 18:00):

Not you, me.

John Baez (Jun 20 2021 at 18:01):

Okay. Yes, I think $S_3$ has about 5 inner automorphisms that aren't the identity.

John Baez (Jun 20 2021 at 18:01):

So, tell me some finite groups whose only inner automorphism is the identity.

Jan Pax (Jun 20 2021 at 18:01):

so my guess is $S_2$

John Baez (Jun 20 2021 at 18:03):

That's one example. How many automorphisms does this group have, by the way?

Jan Pax (Jun 20 2021 at 18:03):

2

John Baez (Jun 20 2021 at 18:03):

Name them.

John Baez (Jun 20 2021 at 18:03):

That is, say what they do.

Jan Pax (Jun 20 2021 at 18:04):

transposition and identity

John Baez (Jun 20 2021 at 18:04):

I agree that the identity automorphism is an automorphism of $S_2$. Explain the other automorphism of $S_2$. What does it do, exactly?

Jan Pax (Jun 20 2021 at 18:05):

switches 1 and 2, the 2 elements of $S_2$.

John Baez (Jun 20 2021 at 18:05):

So you can have an automorphism of a group that doesn't map the identity element to itself?

Jan Pax (Jun 20 2021 at 18:06):

yes

John Baez (Jun 20 2021 at 18:06):

Hmm, I thought an automorphism of a group preserved multiplication.

John Baez (Jun 20 2021 at 18:07):

Here's the definition of automorphism of a group: it's an invertible map from a group to itself that preserves multiplication in the group.

John Baez (Jun 20 2021 at 18:08):

Can you see why any automorphism of a group must map the identity element to itself?

Jan Pax (Jun 20 2021 at 18:08):

because it is a homomorphism

Jan Pax (Jun 20 2021 at 18:09):

so the transposition is not an automorphism right ?

John Baez (Jun 20 2021 at 18:12):

I don't know. Make up your mind!

John Baez (Jun 20 2021 at 18:14):

Here's the answer I wanted to my last question:

Proof that any automorphism of a group, say $\alpha: G \to G$ must preserve the identity.

It preserves multiplication by definition: $\alpha(gh) = \alpha(g)\alpha(h)$. So, $\alpha(h) = \alpha(1h) = \alpha(1) \alpha(h)$. But $\alpha$ is onto so $\alpha(h)$ can be any element $x$ of the group. So, $\alpha(1)x = x$ for all $x \in G$. So, $\alpha(1)$ must equal 1, since a group can only have one identity element.

John Baez (Jun 20 2021 at 18:15):

If a group had two identity elements, say $1$ and $1'$, we'd have

$1 = 11' = 1'$

John Baez (Jun 20 2021 at 18:16):

So, tell me truthfully this time, and make 100% sure before you answer: how many automorphisms does the group $S_2$ have?

Jan Pax (Jun 20 2021 at 18:23):

I cannot fulfill your query. My guess is "one" but it is still a guess.

Jan Pax (Jun 20 2021 at 18:28):

But it is quite a strong belief.

Jan Pax (Jun 20 2021 at 18:29):

because $\alpha(switch)$ must be $switch$ for $\alpha$ to be bijective. Hence $\alpha=1$.

Jan Pax (Jun 20 2021 at 19:21):

May I have a question: why didn't you just multiply $\alpha(1)x=x$ as $\alpha(1)=\alpha(1)x x^{-1}=x x^{-1}=1$ for some $x$? I just wonder whether this will be correct as the other way.

John Baez (Jun 20 2021 at 20:56):

I was trying to prove that if $\alpha: G \to G$ is a bijection that preserves multiplication it must also have $\alpha(1) = 1$.

Your calculation

$\alpha(1)=\alpha(1)x x^{-1}=x x^{-1}=1$

seems to assume $\alpha(1) = 1$ when you make the step $\alpha(1)x x^{-1}=x x^{-1}$. How do you know $\alpha(1)x x^{-1}=x x^{-1}$ if you don't know $\alpha(1) = 1$?

John Baez (Jun 20 2021 at 20:58):

Jan Pax said:

I cannot fulfill your query. My guess is "one" but it is still a guess.

You don't need to guess. You can list all the bijections from $S_2$ to $S_2$ - there are not very many - and then determine which ones are automorphisms. Take your time!

Jan Pax (Jun 21 2021 at 12:19):

We have only 2 bijections from ${switch,1}$ to ${switch,1}$: one is identity which is an automorphism and the second one doesn't preserve 1.So there is only one automorphism..

Jan Pax (Jun 21 2021 at 12:21):

Also, why have you denied my saying that $1\mapsto -1$ is not an inner automorphism in Z by this fact: for no integer $b$ we have
$-1=\sigma(1)=b+(1)+(-b)$ ?

Jan Pax (Jun 21 2021 at 12:41):

This is by $\alpha(1)x=x$ which you say holds for all $x$ and multiplied by $x^{-1}$ from the right. You do not need 1=1'.

John Baez (Jun 21 2021 at 16:23):

Jan Pax said:

We have only 2 bijections from ${switch,1}$ to ${switch,1}$: one is identity which is an automorphism and the second one doesn't preserve 1.So there is only one automorphism.

Good! Correct. So the group $S_2$ has just one automorphism, the identity.

Jan Pax said:

Also, why have you denied my saying that $1\mapsto -1$ is not an inner automorphism in Z by this fact: for no integer $b$ we have
$-1=\sigma(1)=b+(1)+(-b)$ ?

I didn't deny it. I already agreed that this is a proof that $x \mapsto -x$ is not an inner automorphism of $\mathbb{Z}$. It's a perfectly fine proof.

My next question was: what property of the group $\mathbb{Z}$ makes it obvious that the only inner automorphism of $\mathbb{Z}$ is the identity?

And you answered:

commutativity

And that's correct.

John Baez (Jun 21 2021 at 16:24):

And then my next question was this:

So, tell me some finite groups whose only inner automorphism is the identity.

John Baez (Jun 21 2021 at 16:27):

And you answered

$S_3$

which is false.

John Baez (Jun 21 2021 at 16:28):

So I'm still waiting for you to give me a lot of finite groups whose only inner automorphism is the identity. Take your time: I'd rather have a correct answer than a quick answer.

John Baez (Jun 21 2021 at 16:29):

And when you've done that, I will ask you to find the smallest finite group that has an outer automorphism - that is, an automorphism that's not inner. That's where I'm going with all this.

Jan Pax (Jun 21 2021 at 16:32):

any commutative group has as only inner automorphism the identity.

John Baez (Jun 21 2021 at 16:34):

Okay, good. Now answer my question: give me a lot of finite groups whose only inner automorphism is the identity. I want specific examples.

John Baez (Jun 21 2021 at 16:35):

You seem to think math is about general concepts, but it's also about getting to know specific examples.

John Baez (Jun 21 2021 at 16:35):

So I want a kind of list of finite groups whose only inner automorphism is the identity.

Jan Pax (Jun 21 2021 at 16:36):

you want me to give you examples of finite abelian gorups, like cyclic ones ?

John Baez (Jun 21 2021 at 16:36):

I think my question was perfectly clear.

Jan Pax (Jun 21 2021 at 16:37):

my answer is $Z_n$ but that doesn't suffice to you ? Abelian and infinite number of them.

Jan Pax (Jun 21 2021 at 16:41):

I have a few more, any direct product of cyclic is abelian.

John Baez (Jun 21 2021 at 16:42):

Excellent, now you're giving me lots of specific examples of finite groups whose only inner automorphism is the identity.

John Baez (Jun 21 2021 at 16:42):

So now you're ready to tackle my next question: what is the smallest finite group with an outer automorphism: an automorphism that's not inner?

Jan Pax (Jun 21 2021 at 16:43):

I'm surprised that we have moved from category theory to group theory.

Jan Pax (Jun 21 2021 at 17:06):

What about alternating groups, if symmetric do not work ?

Jan Pax (Jun 21 2021 at 17:14):

subgroups of permutations in $S_n$ of even parity

John Baez (Jun 21 2021 at 17:19):

Jan Pax said:

I'm surprised that we have moved from category theory to group theory.

Groups are categories. I'm trying to lead up to some questions about 2-categories that can use the 2-category of groups as an example. You've learned that the concept of "inner automorphism" is a 2-categorical concept: an automorphism of a group is inner iff there's a 2-morphism between it and the identity automorphism.

John Baez (Jun 21 2021 at 17:20):

By the way, you seem to be randomly guessing answers to the next question, instead of using all the previous information I taught you. My questions aren't random; each one is supposed to build on the previous ones.

John Baez (Jun 21 2021 at 17:21):

So it should be possible now for you to find the smallest group with an outer automorphism and prove your answer is correct.

John Baez (Jun 21 2021 at 17:21):

(An answer without a proof is not worth much.)

Jan Pax (Jun 21 2021 at 19:08):

Another try. Let $A_3$ be the 3 element alternating group ${1,1\to 2 \to 3\to 1,1\to 3\to 2\to 1}$.Conjugation with an odd parity element of $S_3$ ${1\to 2\to 1,3\to 3}$ gives an outer automorphism which switches the 2 non-identity elements of $A_3$. It is outer, since all conjugations with elements of $A_3$ gives the 1 (identity, no switch).Checked by computation on one A4 paper with the hope of no mistakes.

John Baez (Jun 21 2021 at 21:45):

This is interesting. What's another name for the 3 element alternating group?

John Baez (Jun 21 2021 at 21:46):

(You keep not using the hints from the earlier exercises.)

John Baez (Jun 21 2021 at 21:47):

Hint:

Jan Pax said:

my answer is $\mathbb{Z}_n$ but that doesn't suffice to you ? Abelian and infinite number of them.

Jan Pax (Jun 21 2021 at 21:56):

I'm not aware of any special name for $A_3$, but once said, I'll remember. Also I'm sometimes confused by your 'hints': $Z_n$ are abelian hence the only automorphisms are inner ones but here I'm supposed to come up with non-inner one (outer). What's wrong ?

John Baez (Jun 21 2021 at 22:15):

You're mixed up.

1) You didn't prove that for abelian groups all automorphisms are inner. What you proved was almost the opposite of this. Review what you did.
2) $A_3$ is one of the groups $Z_n$ that we were talking about!

Jan Pax (Jun 22 2021 at 12:25):

I'm afraid of saying anything now. $A_3$ may be $Z_3$ ? For outer some $Z_n$ will do ? For commutative inner automorphisms are trivial but the rest of automorphisms are outer. This type of confusion I've made is a general patern for my reasoning: I often mix up opposite things.

John Baez (Jun 22 2021 at 19:31):

Jan Pax said:

I'm afraid of saying anything now.

Good! Just think, and write things down on paper, until you're fairly sure of your answers. Learning to check your work is an important of learning math, so I'm trying to teach you to check your work before answering.

John Baez (Jun 22 2021 at 19:39):

For commutative inner automorphisms are trivial but the rest of automorphisms are outer.

Right.

This type of confusion I've made is a general pattern for my reasoning: I often mix up opposite things.

Part of getting good at math is learning how you tend to make mistakes, so it's very good that you realize you have this pattern. When you might mix up opposite things, you can spend an extra minute or two checking.

For example, I often mix up left and right, so I can easily mix up left and right adjoints unless I think about them conceptually, as "free" and "underlying" rather than the more arbitrary-sounding "left" and "right". When I'm afraid I'm about to make a mistake like this, I write some stuff down on paper and try to make sure I get things right.

John Baez (Jun 22 2021 at 19:41):

$A_3$ may be $Z_3$?

I think you need to learn your small groups. Here are some questions:

• How many 1-element groups are there, up to isomorphism? What are they?

• How many 2-element groups are there, up to isomorphism? What are they?

• How many 3-element groups are there, up to isomorphism? What are they?

• How many 4-element groups are there, up to isomorphism? What are they?

• How many 5-element groups are there, up to isomorphism? What are they?

• How many 6-element groups are there, up to isomorphism? What are they?

These small groups are all very famous and important.

Jan Pax (Jun 22 2021 at 19:43):

This is good from you that you can help on this level (how do I tend to make mistakes), and don't just give up with me.
I'm used to read pretty abstract difficult papers and I do even find mistakes in them.But at the same time I can claim a total non-sense.

John Baez (Jun 22 2021 at 19:47):

Anyway, you are right that $A_3$ is the smallest group with a nontrivial outer automorphism. But it's easier to see this if you know $A_3 \cong \mathbb{Z}_3$, or just forget about $A_3$ and think about $\mathbb{Z}_3$.

You noticed that $\mathbb{Z}$ has a nontrivial outer automorphism sending $1$ to $-1$. So, I wanted you to notice that $\mathbb{Z}_n$ also has an automorphism sending $1$ to $-1$. If this is nontrivial it must be outer - since you know that for an abelian group all nontrivial automorphisms are outer.

For $\mathbb{Z}_3$ the automorphism sending $1$ to $-1$ is nontrivial. For $\mathbb{Z}_2$ and $\mathbb{Z}_1$ this automorphism is trivial.

John Baez (Jun 22 2021 at 19:49):

Anyway, I think you'll be a better category theorist if you know a few groups. So please answer these. Take your time and answer correctly.

• How many 1-element groups are there, up to isomorphism? What are they?

• How many 2-element groups are there, up to isomorphism? What are they?

• How many 3-element groups are there, up to isomorphism? What are they?

• How many 4-element groups are there, up to isomorphism? What are they?

• How many 5-element groups are there, up to isomorphism? What are they?

• How many 6-element groups are there, up to isomorphism? What are they?

Jan Pax (Jun 22 2021 at 19:49):

It is really easier to read than to think and write.

John Baez (Jun 22 2021 at 19:50):

It's easier to read because nobody notices your mistakes when you read. :upside_down:

Jan Pax (Jun 22 2021 at 19:50):

Unless I accept something wrong.

John Baez (Jun 22 2021 at 19:51):

Yes, you may catch your own misunderstandings.

John Baez (Jun 22 2021 at 19:52):

But you're right: it's really important to make the transition from absorbing the math other people have created, to creating your own.

Jan Pax (Jun 22 2021 at 20:01):

I'm afraid that with this pace of mine I'll be for ever just reading without ever publishing. Not that bad, some will not even read.

John Baez (Jun 22 2021 at 20:03):

Well, I don't know your situation and your goals, but creating your own math doesn't require publishing. I started doing math by just coming up with results and writing them down. I have a lot of friends who do math and blog about it.

John Baez (Jun 22 2021 at 20:04):

An academic career requires either a lot of teaching, or publishing, or both... but I know a lot of mathematicians, including some very good ones, who do a lot of teaching and very little publishing.

Jan Pax (Jun 22 2021 at 20:04):

My situation allows me to do whatever I wish to. I just read math papers and contemplate on them.

John Baez (Jun 22 2021 at 20:06):

Okay, then you're in luck! If you want to start creating your own math, you can do so without feeling under pressure.

John Baez (Jun 22 2021 at 20:06):

Most of my grad students feel they need to "crank out papers" if they want to get a research job.

Jan Pax (Jun 22 2021 at 20:08):

May I try to answer your question on finite groups? For primes 2,3,5 there is just one cyclic group of that order. For 4 there are 2, $Z_2\times Z_2$ and $Z_4$ and for 6 $Z_2\times Z_3$ and $Z_6$.

John Baez (Jun 22 2021 at 20:10):

Okay. You seem to have answered a different question than the one I asked. Read my question, and your answer, and you'll see what I mean.

John Baez (Jun 22 2021 at 20:11):

Did you carefully check to see what my question was before writing down your answer?

Jan Pax (Jun 22 2021 at 20:13):

I have given the finite groups of orders 2,3,4,5 and 6. I'm eager to see what's wrong this time.

Jan Pax (Jun 22 2021 at 20:15):

I can see now

Jan Pax (Jun 22 2021 at 20:15):

number 6 has a problem

John Baez (Jun 22 2021 at 20:17):

I asked how many groups there were. You told me there's one cyclic group of orders 2, 3, and 5, and two of order 4...one of which is not really cyclic.

John Baez (Jun 22 2021 at 20:18):

I just want to know how many finite groups there are of each order, I wasn't asking anything about cyclic groups.

John Baez (Jun 22 2021 at 20:19):

So, it's good to practice answering the question asked, without introducing distractions.

John Baez (Jun 22 2021 at 20:19):

This is just a form of practicing how to think clearly.

John Baez (Jun 22 2021 at 20:20):

Anyway, you manage to list all the finite groups of orders 2,3,4, and 5 (up to isomorphism). But you only listed abelian ones, and at some point there will be nonabelian groups too.

Jan Pax (Jun 22 2021 at 20:20):

I just thought that there is a non abelian group of order 6 but I gave 2 abelian? am I roughly correct with this ?

Jan Pax (Jun 22 2021 at 20:22):

I've said cyclic just because I silently assume that all finite groups with prime orders are cyclic by some theorem? But of course I'll take care of your correction about introducing notions to questions.

John Baez (Jun 22 2021 at 20:23):

Yes, all groups of prime order are cyclic.

Jan Pax (Jun 22 2021 at 20:24):

So the only problem is to find that exotic non abelian 6 order group.

John Baez (Jun 22 2021 at 20:24):

There could be more than one.

John Baez (Jun 22 2021 at 20:25):

People who study groups only up to order 6 call nonabelian groups "exotic". :upside_down:

Jan Pax (Jun 22 2021 at 20:30):

I have one, it is $S_3$ but I expect having said something wrong :-)

John Baez (Jun 22 2021 at 20:33):

Is $S_3$ a 6-element group?

Jan Pax (Jun 22 2021 at 20:34):

I think it is but nothing comes as a surprise to me today.

John Baez (Jun 22 2021 at 20:34):

Yeah, I think there are $n!$ permutations of an $n$-element set and 3! = 6.

John Baez (Jun 22 2021 at 20:35):

Is $S_3$ abelian or not?

Jan Pax (Jun 22 2021 at 20:35):

it is not I'd bet on

John Baez (Jun 22 2021 at 20:35):

Why not?

John Baez (Jun 22 2021 at 20:35):

Prove it.

Jan Pax (Jun 22 2021 at 20:36):

there are two cleverly chosen permutations whose composition is not the same

Jan Pax (Jun 22 2021 at 20:36):

I see you need a proof

Jan Pax (Jun 22 2021 at 20:43):

let $f$ be $1\to 2 \to 1,3\to 3$ and $g$ $1\to 1,2\to 3\to 2$. $f\circ g(1)=2$, $g\circ f(1)=3$.

John Baez (Jun 22 2021 at 20:55):

Great!

John Baez (Jun 22 2021 at 20:55):

So $S_3$ is nonabelian, so it cannot be isomorphic to $\mathbb{Z}_6$ or $\mathbb{Z}_2 \times \mathbb{Z}_3$.

John Baez (Jun 22 2021 at 20:56):

By the way, why are $\mathbb{Z}_6$ and $\mathbb{Z}_2 \times \mathbb{Z}_3$ non-isomorphic?

Jan Pax (Jun 22 2021 at 20:56):

Right, but knowing there is no other which I think it is the case seem hopeless without some theorem.

Jan Pax (Jun 22 2021 at 20:57):

one has 1 generator and the other 2

John Baez (Jun 22 2021 at 20:57):

But in fact both groups have 6 generators: we can take every element of a group to be a generator!

Jan Pax (Jun 22 2021 at 20:58):

what about minimal set of generators ?

John Baez (Jun 22 2021 at 20:58):

That's more interesting. So what's your claim, exactly?

Jan Pax (Jun 22 2021 at 20:59):

every minimal set of generators is 1 element in the first case while 2 in the second. If there were an isomorphism then....

Jan Pax (Jun 22 2021 at 21:00):

these sets would be in bijection

John Baez (Jun 22 2021 at 21:05):

One of the things you just said is false. It's probably something you didn't check.

John Baez (Jun 22 2021 at 21:05):

You are again failing to check your work before speaking.

Jan Pax (Jun 22 2021 at 21:06):

would have the same cardinality not necessarily bijective by the isomorphism?

Jan Pax (Jun 22 2021 at 21:09):

I'm still too used to my intuition which works fine while reading only and in fact this intuition is what allows me to absorb the amount of data we have just exchanged. So I cannot just abandon it.

John Baez (Jun 22 2021 at 21:13):

I don't know what "would have the same cardinality not necessarily bijective by the isomorphism?" means - the grammar is strange. But the mistake was somewhere else, and it's a serious mistake.

John Baez (Jun 22 2021 at 21:14):

I'm not telling you to abandon your intuition. I'm telling you to check claims before stating them to me, so that the fraction of times you correctly answer my questions increases.

John Baez (Jun 22 2021 at 21:15):

Intuition is the starting-point of thinking, but not the end.

Jan Pax (Jun 22 2021 at 21:15):

So I'm still not at the end of my thinking.

John Baez (Jun 22 2021 at 21:16):

Right. But please finish your thinking before you answer my questions. Everything goes much slower if you rush.

John Baez (Jun 22 2021 at 21:17):

Something here is false:

every minimal set of generators is 1 element in the first case while 2 in the second. If there were an isomorphism then these sets would be in bijection

Jan Pax (Jun 22 2021 at 21:18):

Is it the number 2 there ?

John Baez (Jun 22 2021 at 21:24):

I'll wait until you find the mistake and correct it. This is not a guessing game.

Jan Pax (Jun 22 2021 at 21:31):

$(1,1)$ generates $Z_2\times Z_3$.

Jan Pax (Jun 22 2021 at 21:32):

so $Z_2\times Z_3 \cong Z_6$ after all.

John Baez (Jun 22 2021 at 21:50):

Okay, great!

John Baez (Jun 22 2021 at 21:51):

Yeah, if $m$ and $n$ are relatively prime then $Z_m \times Z_n \cong Z_{mn}$.

John Baez (Jun 22 2021 at 21:51):

So what do you know about this question?

• How many 6-element groups are there, up to isomorphism? What are they?

Note, I'm not asking you what you guess. I'm asking you what you know, so far.

Jan Pax (Jun 22 2021 at 21:52):

there are at least 2

John Baez (Jun 22 2021 at 21:52):

Distinguishing between what you know and what you guess is important in math.

John Baez (Jun 22 2021 at 21:53):

Just to be sure: why do you know $S_3 \ncong Z_6$?

Jan Pax (Jun 22 2021 at 21:54):

one is commutative and the other not

John Baez (Jun 22 2021 at 21:54):

Okay, good. Now, do you know about "dihedral groups"?

Jan Pax (Jun 22 2021 at 21:55):

no

John Baez (Jun 22 2021 at 21:55):

Oh, those are some other important finite groups.

John Baez (Jun 22 2021 at 21:56):

The symmetry group of a regular n-gon, counting rotations and reflections, is often called $D_n$. It has 2n elements.

John Baez (Jun 22 2021 at 21:56):

So for example the symmetry group of the square is called $D_4$, and it has 8 elements.

John Baez (Jun 22 2021 at 21:58):

You should be glad I didn't ask you about 8-element groups, since there are quite a few:

• $Z_8$
• $Z_2 \times Z_4$
• $Z_2 \times Z_2 \times Z_2$
• $D_4$

and one more, which is the most interesting of all.

Jan Pax (Jun 22 2021 at 21:59):

quaternions ?

John Baez (Jun 22 2021 at 22:00):

Yes, it's called the quaternion group.

John Baez (Jun 22 2021 at 22:01):

It's strange you've heard of that one but not the easier dihedral groups!

Groups whose order is a power of 2 are extremely numerous. There are almost 50 billion groups of order 1024 = $2^{10}$, and over 99% of groups of order < 1024 have order exactly 1024.

John Baez (Jun 22 2021 at 22:01):

Anyway, here's my next question: is the dihedral group $D_3$ isomorphic to one of the other two 6-element groups we've seen, or not?

Jan Pax (Jun 22 2021 at 22:04):

I cannot answer today.

John Baez (Jun 22 2021 at 22:07):

That's okay! Just to help you a tiny bit: $D_3$ is the symmetry group of an equilateral triangle, including rotations and reflections.

Morgan Rogers (he/him) (Jun 23 2021 at 08:27):

John Baez said:

The symmetry group of a regular n-gon, counting rotations and reflections, is often called $D_n$. It has 2n elements.

I was taught (originally by Julia Goedecke) the convention $D_{2n}$, to maintain the convention that subscripts tell you the order of the groups. Of course, extending this convention to symmetric groups would make the notation cumbersome rather quickly...

Oscar Cunningham (Jun 23 2021 at 09:16):

I suppose you could write '$S_{n!}$'.

Jan Pax (Jun 23 2021 at 12:39):

My guess is that $D_3$ is $S_3$ since it has 6 elements and is non-commutative: there are 2 reflections which do not commute. If there is no other non-commutative group of order 6 besides $S_3$ then this $D_3$ is it.

Jan Pax (Jun 23 2021 at 17:18):

I could write the Cayley tables for $D_3$ and $S_3$ and find the isomorphism but it would be tedious.

Oscar Cunningham (Jun 23 2021 at 19:36):

You should be able to see an isomorphism by looking at their definitions. $D_3$ is the group of symmetries of a triangle, and $S_3$ is the group of permutations of a three element set.

Jan Pax (Jun 23 2021 at 19:49):

That's a nice observation. So label the vertices by the elements of the three element set and assign the symmetry to the permutation.I was wrong. Even for $n=4$ this fails. $D_4$ has 8 elements but $S_4$ 24 If I'm not creating another typo by this then switching adjacent vertices and keeping the other 2 is not a symmetry.

Oscar Cunningham (Jun 23 2021 at 20:06):

Right. In fact $S_4$ does give you the symmetries of a regular tetrahedron, since there you can use symmetries to rearrange the vertices however you like.

Jan Pax (Jun 23 2021 at 20:17):

What for higher $n$'s ?

Oscar Cunningham (Jun 23 2021 at 20:18):

The symmetries of the regular simplex with $n$ vertices in $n-1$ dimensional space are given by $S_n$.

John Baez (Jun 23 2021 at 22:37):

Oscar Cunningham said:

You should be able to see an isomorphism by looking at their definitions. $D_3$ is the group of symmetries of a triangle, and $S_3$ is the group of permutations of a three element set.

Good! $D_3 \cong S_3$ because symmetries of an equilateral triangle give permutations of its set of vertices and vice versa.

John Baez (Jun 23 2021 at 22:38):

So, I was trying to see if I could trick you into believing in the existence of a third group of order 6, but there are just two (up to isomorphism). One is

$\mathbb{Z}_6 \cong \mathbb{Z}_2 \times \mathbb{Z}_3$

while the other is

$S_3 \cong D_3$

John Baez (Jun 23 2021 at 22:39):

I will not force you to prove that there are just two!

John Baez (Jun 23 2021 at 22:40):

There's just one 7-element group, $\mathbb{Z}_7$, since 7 is prime.

John Baez (Jun 23 2021 at 22:41):

I mentioned that there are five 8-element groups:

• $Z_8$
• $Z_2 \times Z_4$
• $Z_2 \times Z_2 \times Z_2$
• $D_4$
• and the quaternion group.

John Baez (Jun 23 2021 at 22:42):

I think there are just two 9-element groups:

• $\mathbb{Z}_9$
• $\mathbb{Z}_3 \times \mathbb{Z}_3$

(these should remind you of the two 4-element groups).

John Baez (Jun 23 2021 at 22:43):

I think there are just two 10-element groups:

• $\mathbb{Z}_{10} \cong \mathbb{Z}_2 \times \mathbb{Z}_5$

(a product of two distinct primes always works this way).

John Baez (Jun 23 2021 at 22:43):

There's just one 11-element group:

• $\mathbb{Z}_{11}$

John Baez (Jun 23 2021 at 22:44):

So, 12-element groups are the first case that makes you really think hard, after the 8-element groups.

John Baez (Jun 23 2021 at 22:45):

But there aren't really many. So the first really hard case is 16-element groups; as I said, powers of 2 are really difficult (and so are powers of 3, etc.).

John Baez (Jun 23 2021 at 22:46):

And if you cheat and look it up, you'll see there are 14 different 16-element groups!

Fawzi Hreiki (Jun 23 2021 at 22:52):

In case you want to cheat more, GroupNames is an amazing resource.

John Baez (Jun 23 2021 at 23:00):

I was linking to GroupProps, which is also an amazing resource. Should they join forces and cooperate, or is it good to have some competition among wikis about finite groups? :thinking:

John Baez (Jun 23 2021 at 23:01):

Okay, now let me pose another question to @Jan Pax.

John Baez (Jun 23 2021 at 23:02):

We have been talking about some 2-categories:

• Cat, which is the 2-category of categories, functors and natural transformations
• Grp, which is the "full and 2-full" sub-2-category of Cat where the objects are categories that are just groups.

John Baez (Jun 23 2021 at 23:03):

By full I mean that we include all the 1-morphisms between the chosen objects (namely groups), and by 2-full I mean that we also include all the 2-morphisms between those 1-morphisms.

John Baez (Jun 23 2021 at 23:04):

We have also implicitly been talking about another 2-category:

• AbGrp, which is the "full and 2-full" sub-2-category of Cat where the objects are categories that are just abelian groups.

John Baez (Jun 23 2021 at 23:04):

So here's my question:

Describe the 2-morphisms in AbGp as precisely and simply as possible.

John Baez (Jun 23 2021 at 23:05):

You may start with the definition, and then think about it and simplify it.

Fawzi Hreiki (Jun 23 2021 at 23:29):

John Baez said:

I was linking to GroupProps, which is also an amazing resource. Should they join forces and cooperate, or is it good to have some competition among wikis about finite groups? :thinking:

Yes I noticed. That's why I linked to GroupNames. GroupProps is pretty good but it's less organised than GroupNames so it may be more difficult to find what you want.

John Baez (Jun 23 2021 at 23:30):

GroupProps has lots of pages on properties of finite groups... I don't know if GroupNames does that too.

Jan Pax (Jun 24 2021 at 14:37):

I have one more basic question on the notation yet,before I answer your question on AbGp : in a group as a one object category, the morphisms between the one object $K$ i.e. the elements of the group are called 0-cells or 1-cells ?

Fawzi Hreiki (Jun 24 2021 at 14:52):

1-cells. The 0-cells are the objects.

Fawzi Hreiki (Jun 24 2021 at 14:52):

Which is just the unique one in this case.

Jan Pax (Jun 24 2021 at 14:57):

So do I understand it well that natural transformations between functors $F$ and $G$ are 3-cells ? Is this called 2-category or 3-category ?

Fawzi Hreiki (Jun 24 2021 at 15:42):

A group is only a 1-category. There are no 2-cells in a group - only in the category of groups which is a sub-2-category of the 2-category of categories.

Jan Pax (Jun 24 2021 at 17:46):

In Ab the natural transformations from $f\to f$ with $f:A\to B$ are all elements in $B$ and for $f\neq g$ there is no natural transformation from $f$ to $g$.

John Baez (Jun 25 2021 at 03:37):

Fawzi's answers are correct. A group is a category with one object with all its morphisms invertible. As such Grp becomes a sub-2-category of Cat, as follows:

John Baez (Jun 25 2021 at 03:37):

John Baez said:

• Cat, which is the 2-category of categories, functors and natural transformations
• Grp, which is the "full and 2-full" sub-2-category of Cat where the objects are categories that are just groups.

Jan Pax (Jun 25 2021 at 14:48):

am I again wrong with my answer ? I ask this because I do not see the connection of @John Baez comment with my answer,is there any? I understand that @Fawzi Hreiki is correct. I meant that $A$ and $B$ are groups-the one object categories, $f,g$ are functors between them and I tried to investigate the natural transformations from $f$ to $g$.

Jan Pax (Jun 25 2021 at 14:54):

If $f\neq g$ on a morphism $a\in A$ then for no n.t. $b$ (i.e. an elmenet $b$ of $B$) is $bf(a)=g(a)b$ by commutativity of $B$.

John Baez (Jun 25 2021 at 18:30):

am I again wrong with my answer ?

I didn't see you try to answer the question until just now.

John Baez (Jun 25 2021 at 18:30):

I just saw your attempts to clarify the question.

John Baez (Jun 25 2021 at 18:31):

Now I see your answer:

Jan Pax said:

In Ab the natural transformations from $f\to f$ with $f:A\to B$ are all elements in $B$ and for $f\neq g$ there is no natural transformation from $f$ to $g$.

Yes, that's right!

John Baez (Jun 25 2021 at 18:32):

Okay, I declare this class finished. :+1:

Jan Pax (Jun 25 2021 at 18:44):

If I've made less mistakes you would continue, right ? So shall we meet in October after several months of rest ? It was of a great value to me for start to think, anyway. Thank you.

John Baez (Jun 25 2021 at 18:51):

I'm sorry, I don't feel like doing this any more - it was fun but exhausting. I hope you practice solving lots of problems, e.g. exercises in textbooks. That's a good way to see how well you understand what you're reading.

John Baez (Jun 25 2021 at 18:53):

For example, Borceux's Handbook of Categorical Algebra is 3 volumes, and it covers a lot of category theory, and it has exercises.

Jan Pax (Jun 26 2021 at 17:06):

Please, how can I print the entire history of this page say, on my printer ? I have the Firefox browser but ctrl+P prints only the first page. Is there something like zulipchat meta for this my question ?

John Baez (Jun 26 2021 at 18:41):

https://categorytheory.zulipchat.com/#narrow/stream/229122-general.3A-meta

Oscar Cunningham (Jun 26 2021 at 20:09):

Morgan Rogers (he/him) said:

Once they've answered, I have a potentially interesting observation to make about this 2-category.

Oscar Cunningham (Jun 26 2021 at 20:09):

I had some thoughts too. Interested to see if we're thinking along the same lines.

Morgan Rogers (he/him) (Jun 27 2021 at 10:32):

Ah, yes, I forgot to follow up on this.

My comment is that from any (discrete) group you can build its category $G\text{-}\mathrm{Set} = [G^{\mathrm{op}},\mathrm{Set}]$ of right $G$-sets (I choose the convention of right actions because it coincides with taking presheaves). The resulting categories are Grothendieck toposes; we can actually characterize them as the atomic toposes having a surjective essential point.

Given a group homomorphism $h:G \to H$, we can construct an essential geometric morphism (adjoint triple of functors) $[G^{\mathrm{op}},\mathrm{Set}] \to [H^{\mathrm{op}},\mathrm{Set}]$ whose inverse image functor (the middle functor in the adjoint triple) is the functor $[H^{\mathrm{op}},\mathrm{Set}] \to [G^{\mathrm{op}},\mathrm{Set}]$ consisting of restriction of an action along $h$. It turns out that all geometric morphisms between these toposes are of this form!

Given a conjugation (natural transformation) $\alpha: h \Rightarrow h'$ between group homomorphisms, we get a natural transformation in the opposite direction between the inverse image functors of the corresponding geometric morphisms.

Altogether, the above mean that, up to reversing the direction of the 2-morphisms, the 2-category of groups is (2-)equivalent to the 2-category of atomic toposes having a surjective essential point.

Morgan Rogers (he/him) (Jun 27 2021 at 10:39):

This 2-equivalence can be put to work in both directions, since at the topos level we can use other toposes to help us understand these ones better, and transfer that understanding down to the groups, and conversely it can make constructions involving these toposes easier, especially combined with the observation that they are coreflective in the 2-category of pointed toposes.

Morgan Rogers (he/him) (Jun 27 2021 at 10:40):

It also means that any property of groups which can be expressed (2-)categorically automatically becomes a property of the corresponding topos, an idea which @Jens Hemelaer and I have only just started exploring in the wider context of monoids (where things become a little more subtle)

Oscar Cunningham (Jun 27 2021 at 18:05):

Neat! This is a very geometric way of looking at things, whereas I was thinking more algebraically.

The forgetful functor $\mathbf{Grp}\to\mathbf{Set}$ is represented by the group $\mathbb{Z}$. So for this $2$-category of groups we could look at the functor $\mathrm{Hom}(B\mathbb{Z},-)$ as a sort of 'underlying groupoid' functor. It sends a group to its groupoid of elements, and conjugations between them. So an isomorphism class of this groupoid is a conjugacy class.

So this gives a weird new perspective on groups, in which the 'elements' of a group are the conjugacy classes. I wondered if this could be made into a genuine algebraic theory. In particular, whether this functor $\mathbf{Grp}_2\to\mathbf{Groupoid}$ was a right adjoint. But I'm pretty sure it isn't, because if it had a left adjoint $F$ and we let $G$ be the groupoid with two inequivalent objects, what would $F(G)$ be? It would have to be a group such that conjugacy classes of maps out of that group correspond to pairs of conjugacy classes of elements of the target. I can't see any such beast.

Oscar Cunningham (Jun 27 2021 at 18:14):

But then it occurred to me that while the functor $\mathrm{Hom}(B\mathbb{Z},-):\mathbf{Grp}_2\to\mathbf{Groupoid}$ isn't a right adjoint, the functor $\mathrm{Hom}(B\mathbb{Z},-)$ really is a right adjoint when seen as a funtor $\mathbf{Groupoid}\to\mathbf{Groupoid}$. It's left adjoint is just $B\mathbb{Z}\times -$. This is the tensor-hom adjunction.

This therefore gives us a way to see groups as Eilenberg-Moore algebras of the monad $\mathrm{Hom}(B\mathbb{Z},B\mathbb{Z}\times -)$. I looked this up and found that over $\mathbf{Set}$ the algebras of $\mathrm{Hom}(A,A\times -)$ are called $A$-mnemoids. Therefore we have a way to view groups as $B\mathbb{Z}$-$2$-mnemoids, which is an excellent piece of technobabble.

But I don't yet understand mnemoids very well in general, or $B\mathbb{Z}$-$2$-mnemoids in particular. So that's about as far as I got.

Morgan Rogers (he/him) (Jun 27 2021 at 18:25):

What if you consider pointed groupoids? That is, groupoids equipped with a functor from the trivial groupoid? Does that fix anything, since you can always forget (or trivialize) the components disjoint from the pointed one to recover a group?

Oscar Cunningham (Jun 27 2021 at 18:44):

Interesting thought. I think it's normally natural when considering pointed groupoids to say that the functors between them have to preserve the point, and the natural transformations between functors have to be the identity at the point. (If you do this then you can define the usual $1$-category of groups as the pointed groupoids with one object.) But here we have to forget the second criterion otherwise we won't get a functor from $\mathbf{Grp}_2$.

Oscar Cunningham (Jun 27 2021 at 18:48):

So I think you're saying we should look at if the functor $\mathrm{Hom}(B\mathbb{Z},-):\mathbf{Grp}_2\to\mathbf{PointedGroupoid}$ is a right adjoint.

Oscar Cunningham (Jun 27 2021 at 18:55):

We could look at the pointed groupoid $G$ with $n$ inequivalent objects and no nontrivial morphisms (where now one of those objects is the point) and ask what $F(G)$ would be. A functor from $G$ to the 'underlying groupoid' of a group, is a list of $n$ conjugacy classes of the group where the one corresponding to the point has to be the conjugacy class of the identity. So when $n=1$ we could take $F(G)=0$, and when $n=2$ we could take $F(G)=\mathbb{Z}$, but I think we get stuck in the same way as we did before when $n=3$.

John Baez (Jun 27 2021 at 19:51):

Morgan Rogers (he/him) said:

Ah, yes, I forgot to follow up on this.

My comment is that from any (discrete) group you can build its category $G\text{-}\mathrm{Set} = [G^{\mathrm{op}},\mathrm{Set}]$ of right $G$-sets (I choose the convention of right actions because it coincides with taking presheaves). The resulting categories are Grothendieck toposes; we can actually characterize them as the atomic toposes having a surjective essential point.

What's a "surjective essential point"? The group $G$ itself being a $G$-set, we get a geometric morphism $\mathrm{Set} \to G\text{-}\mathrm{Set}$, i.e. a "point". And this is an essential geometric morphism; I guess that's why you're calling it an "essential point". If I'm right so far, I just need to know precisely how it's "surjective". And I'll bet this has something to do with how every $G$-set is a colimit of representables, which are just copies of $G$ itself.

John Baez (Jun 27 2021 at 19:59):

(I'm very bad at understanding the different adjectives people stick in front of "geometric morphism", but I want to get better.)

Fawzi Hreiki (Jun 27 2021 at 20:27):

Surjective just means that the inverse image is faithful.

Fawzi Hreiki (Jun 27 2021 at 20:30):

(This is really in the opposite direction since maps of toposes are reversed, like how a sublocale is actually a quotient frame and vice versa)

John Baez (Jun 27 2021 at 21:04):

Thanks. So my guess was wrong; this sort of surjectivity is more about morphisms than objects.

Morgan Rogers (he/him) (Jun 28 2021 at 06:59):

A geometric morphism $f:\mathcal{F} \to \mathcal{E}$ with faithful inverse image is called a surjection for a few reasons. The original one is that if we take the image factorization of a continuous map, this will correspond to the surjection-inclusion factorization of the corresponding geometric morphism. But for me a better justification is that a geometric morphism is a surjection if and only if its inverse image functor is comonadic; this means that the $\mathcal{E}$ can be completely recovered from some data on $\mathcal{F}$, and while that data doesn't take the form of an equivalence relation, it still somehow carries the intuition that $\mathcal{E}$ is "covered" by $\mathcal{F}$.

John Baez (Jun 28 2021 at 15:56):

Thanks! I think I need to start by seeing how properties of maps between topological spaces translate into properties of geometric morphisms between their sheaf topoi.