Showing $F_X \cong F_Y\implies |X| = |Y|$

Question

Lately I've been studying free groups, I'm a layman on the subject but I came across a step in the demonstration that I couldn't move forward. I know the question seems to be good but: If $F_X \cong F_Y$ implies $|X| = |Y|$.

In the proof the author constructs a quotient group $F_X/N_1$ where $N_1 = \langle x^2 : x \in F_X \rangle$ and $F_Y/N_2$ where $N_2= \langle y^2 : y \in F_Y \rangle$.

My question is: If $F_X \cong F_Y$ and $N_1 \cong N_2$, then $F_X/N_1 \cong F_Y/N_2$? If so, how did you naturally perceive $N_1 \cong N_2$?

It's not generally true that if $G_1\cong G_2$ and $H_1,H_2$ are normal subgroups of $G_1,G_2$ with $H_1\cong H_2,$ then $G_1/H_1\cong G_2/H_2.$ They key here is that the subgroups are defined independent of $X,Y,$ but simply in terms in terms of the entire group. — Thomas Andrews, Sep 14 '22 at 14:27
If $G$ is a group, we can define $H_G=\langle g^2\mid g\in G\rangle,$ prove $H_G$ is normal. Then if $G_1\cong G_2,$ then $H_{G_1}\cong H_{G_2}$ and $G_1/H_{G_1}\cong G_2/H_{G_2}.$ This is because nothing about the definition of $H_G$ breaks the isomorphism. The elements of $H_{G_1}$ correspond to elements of $H_{G_2}$ in obvious ways. — Thomas Andrews, Sep 14 '22 at 14:29
OK, I think it would be obvious. But I thought that the isomorphism between the quotients would come from some property of free groups. Is there anything that actually guarantees this isomorphism? — kots, Sep 14 '22 at 14:39
No specific name, and the notation is non-standard. I just wanted to express the general case absent anything about free groups. If you replaced $2$ in the exponent with $k$ it would still be true: $N_{G,k}=\langle g^k: g\in G\rangle.$ Then $N_{G,k}$ is normal, and if $G_1\cong G_2$ then $N_{G_1,k}\cong N_{G_2,k}$ and $G_1/N_{G_1,k}\cong G_2/N_{G_2,k}.$ — Thomas Andrews, Sep 14 '22 at 14:45
This is because the definition of $G/N_{G,k}$ is defined just in terms of the elements of $G,$ not the "names". of the elements, and $G_1\cong G_2$ means that $G_2$ is just a renaming of the elements. You can go about a formal proof of this fairly directly, but it can be fairly messy. — Thomas Andrews, Sep 14 '22 at 14:46
I understand, unfortunately this "formal" private I can't do :(. Before even asking this question here I had already tried, but unfortunately without success. — kots, Sep 14 '22 at 14:49
Unfortunately not. I would just like to understand the step-by-step of claim 3. — kots, Sep 14 '22 at 17:51

score 1 · Answer 1 · answered Sep 14 '22 at 15:13

1

This has nothing to do with free groups.

So the key isn't just that the $N_1\cong N_2,$ but that the isomorphism $\phi:F_X\to F_Y$ restricts to an isomorphism $\phi_{|N_1}:N_1\to N_2.$

Definition: If $G$ is a group, define $N(G)=\langle g^2\mid g\in G\rangle.$

Claim 1: $N(G)$ is a normal subgroup.

Proof: The follows since $gg_1^2g^{-1}=\left(gg_1g^{-1}\right)$.

If $h\in N(G)$ then $h=g_1^2g_2^2\cdots g_k^2$ for some sequence $g_1,\dots,g_k\in G.$ If $g\in G,$ then $$ghg^{-1}=\left(gg_1g^{-1}\right)^2\left(gg_2g^{-1}\right)^2\cdots \left(gg_kg^{-1}\right)^2\in N(G).$$

Claim 2: If $\phi:G_1\to G_2$ is an isomorphism, then $\phi(N(G_1))= N(G_2).$

Proof: Let $\phi:G_1\to G_2$ be an isomorphism, so $\phi^{-1}$ is also an isomorphism. If $h\in N(G_1),$ then $h=g_1^2g_2^2\cdots g_k^2$ and $\phi(h)=\phi(g_1)^2\phi(g_2)^2\cdots \phi(g_k)^2\in N(G_2).$ Likewise, if $h\in N(G_2),$ $\phi^{-1}(h)\in N(G_1).$ So $\phi,$ restricted to $N(G_1),$ is an isomorphism between $N(G_1)$ and $N(G_2).$

Claim 3: If $\phi$ is an isomorphism between $G_1$ and $G_2,$ and $N_1$ is a normal subgroup of $G_1,$ then $\phi(N_1)$ is a normal subgroup of $G_2,$ isomorphic to $N_1,$ and $G_1/N_1\cong G_2/\phi(N_1).$

I'll leave this last claim to you.

At heart this is all due to the fact that an isomorphism can be thought of as a renaming of the elements of a group. Since the definition of $N(G)$ doesn't care about the names of the elements, the process of computing $N(G)$ and $G/N(G)$ do not depend on the names.

answered Sep 14 '22 at 15:13

Thomas Andrews

186,215

Thanks, Thomas! – kots Sep 14 '22 at 15:17
In claim 2, why is $\phi $restricted to N(G_1) an isomorphism between N(G_1) and N(G_2)? Since $\phi$ restricted to N(G_1) is the map $\alpha: N(G_1) \rightarrow G_2$? – kots Sep 14 '22 at 16:06
But the image $\phi(N(G_1))=N(G_2),$ so $\phi$ is a bijective homomorphism between $N(G_1)$ and $N(G_2).$ @kots – Thomas Andrews Sep 14 '22 at 16:12
Got it, the last claim seems impossible to me, I'm not getting it (claim 3) – kots Sep 14 '22 at 16:14
It takes some time, but it is worth doing yourself, to get used to what it really means to be an isomorphism. At heart, if $gN_1$ is a coset of $N_1,$ then you show that $\phi(gN_1)=\phi(g)\phi(N_1)=\phi(g)N_2.$ Recall that $G/N$ is defined in terms of cosets. – Thomas Andrews Sep 14 '22 at 16:20
Thomas, but $\phi$ is already an isomorphism. Given that, in my head we have the result. Does this last equality guarantee isomorphism? – kots Sep 14 '22 at 16:31
You mean $\phi(gN_1)=\phi(g)N_2?$ Once you prove that, it is a few steps to write the isomorphism $\rho:G_1/N_1$ and $G_2/N_2$ and prove $\rho$ is an isomorphism. It isn't just "$\ phi$ is an isomorphism, so..." but the argument isn't much more. – Thomas Andrews Sep 14 '22 at 16:39
But why should I show that $\phi(gN) = \phi(g)N_2$? Just to show that $\rho$ is well defined? And what guarantees me that $gN_1$ is a set of $N_1$? ps: thank you so much for helping me – kots Sep 14 '22 at 16:46
Do I actually use this $\phi(gN_1) = \phi(g)N_2$ to show that $rho$ is a homomorphism? – kots Sep 14 '22 at 16:50
You can use it to prove $\rho$ is well-defined - that it is a function $G_1/N_1\to G_2/N_2.$ If you just defined $\rho(gN_1)=\phi(g)N_2,$ you'd have to show that if $g_1N_1=g_2N_1,$ then $\phi(g_1)N_2=\phi(g_2)N_2.$ But the fact that $\rho(g_1N_1)$ is the image of $g_1N_1$ under $\phi$ means that it is independent of choice of $g_1$ or $g_2.$ – Thomas Andrews Sep 14 '22 at 16:56
So is it ok to set $\rho$ from $\phi$? – kots Sep 14 '22 at 17:02
No idea what you mean by that. I realize it is hard to ask questions in comments, but I've had to struggle reading your questions to me here because they often have me guessing what you mean. – Thomas Andrews Sep 14 '22 at 17:04
I mean setting $\rho(gN)=\phi(gN)$ is a good way to go? – kots Sep 14 '22 at 17:05
Yes, but it is important to realize that $\rho(gN)$ is a function applied to single elements $gN$ of $G/N,$ while $\phi(gN)$ is the image of the set of elements $gN\subset G,$ under the function $\phi.$ The two notations can look the same, but $\rho(gN)=\phi(gN)$ risks making it look like $\rho=\phi,$ which is not the case. Possibly better to use some notation, that $\phi_{im}(X)$ for the image of a set $X$ under $\phi,$ just for clarity. – Thomas Andrews Sep 14 '22 at 17:10
Got it, I've been trying to do it here, but claim 3 is not coming out at all. I'm going to study, maybe one day I'll understand. – kots Sep 14 '22 at 17:14
It is one of those things that takes some time to understand, but once you do, it makes total sense. An isomorphism ultimately means the groups are essentially the same, so proving this sort of thing is abstract nonsense. The subgroup $N_1$ of $G_1$ corresponds to the subgroup $N_2$ of $G_2,$ and thus the quotients are the same. – Thomas Andrews Sep 14 '22 at 17:21
Claim 3 isn't true in general. If $G_1\cong G_2$ and $N_1,N_2$ are subgroups of $G_1,G_2$ with $N_1\cong N_2,$ it is not always true that $G/N_1\cong G/N_2.$ You need $\phi(N_1)=N_2$ in some isomorphism $\phi:G_1\to G_2.$ This condition is, intuitively, just that "$N_2$ is a renaming of $N_1$ under the isomorphism $\phi.$" – Thomas Andrews Sep 14 '22 at 17:24
Yes, it is in this "abstract" that I get lost. I would just like to understand that claim 3. If I understand, I'm at peace. This has bothered me a lot – kots Sep 14 '22 at 17:25
It's demonstrating claim 3 which is confusing me, you explained it very well but I don't have enough skills to demonstrate such a thing – kots Sep 14 '22 at 17:27

Showing $F_X \cong F_Y\implies |X| = |Y|$

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