Show that every element in the commutator subgroup is a commutator when G/A is cyclic.

Question

Let $A \unlhd G$ be abelian.
1) If $x \in G$ is fixed, show that the map $\Phi: A \to A$ given by $\Phi_x(a)=a^{-1}x^{-1}ax = [a,x]$ is a group homomorphism.
2) Suppose that $G/A$ is cyclic and generated by $xA$. Show that every element in the commutator subgroup $G'=[G,G]=< [x,y] \ | \ x,y \in G>$ is a commutator. (Hint: First show that the image of the map in (1) is a normal subgroup of G.)

For (1):

Let $x \in G$ be fixed. Consider $\Phi_x(ab) = (ab)^{-1}x^{-1}(ab)x$.

Since $A \unlhd G$ we have $x^{-1}(ab)x=ab \in A.$ Thus, $(ab)^{-1}x^{-1}(ab)x = (ab)^{-1}(ab)=1$.

Likewise, $\Phi_x(a) \cdot \Phi_x(b) = (a^{-1}x^{-1}ax)(b^{-1}x^{-1}bx)$.

Since $A \unlhd G$ and $a,b \in A$ we have $x^{-1}ax = a$ and $x^{-1}bx = b$. Hence, $(a^{-1}x^{-1}ax)(b^{-1}x^{-1}bx)=a^{-1}(a)b^{-1}(b)=1$. Therefore $\Phi_x(ab) = \Phi_x(a) \cdot \Phi_x(b)$ as required.

For (2):

Now assume $G/A$ is cyclic and generated by $xA$.

First, we will show the image of the map above is a normal subgroup of G.

Notice $\Phi_x (A) = A^{-1}x^{-1}Ax = A^{-1}A = 1$. Therefore, $A^{-1}x^{-1}Ax = 1 \Rightarrow x^{-1}Ax = A$, as desired.

Now I'm not sure how to reach my conclusion. Also, let me know if what I have here so far is accurate. Thanks!

Answer 1

part 1:

There's an important mistake in your work: $xax^{-1}$ is not necessarily equal to $a$; we can only say that $xax^{-1} \in A$.

That being said, we can make your proof work nevertheless. Here's an edited version:

Let $x \in G$ be fixed. Note that $\Phi_x(ab) = (ab)^{-1}x^{-1}(ab)x$. On the other hand, $\Phi_x(a) \cdot \Phi_x(b) = (a^{-1}x^{-1}ax)(b^{-1}x^{-1}bx)$. Since $A \unlhd G$ and $a,b \in A$ we have $x^{-1}ax\in A$ and $x^{-1}bx \in A$. Hence, $$ (a^{-1}x^{-1}ax)(b^{-1}x^{-1}bx) = \\ a^{-1}(x^{-1}ax)b^{-1}(x^{-1}bx) = \\ a^{-1}b^{-1}(x^{-1}ax)(x^{-1}bx) = \\ (ab)^{-1}(x^{-1}a(xx^{-1})bx) = \\ (ab)^{-1}x^{-1}(ab) x =\\ \Phi_x(ab) $$ Therefore $\Phi_x(ab) = \Phi_x(a) \cdot \Phi_x(b)$ as required.

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Part 2: We argue that $\Phi_x(G)$ is a normal subgroup as follows. For an arbitrary $g \in G$, there must exist a power $k$ such that $gA = (xA)^k = x^k A$, since $xA$ generates $G/A$. Equivalently, we have $x^{-k}g \in A$.

So for any $a \in A$, we have $$ g^{-1}\Phi_x(a)g = \\ g^{-1}(a^{-1}x^{-1}ax) g = \\ g^{-1}x^k x^{-k} a^{-1}x^{k}x^{-k} x^{-1}ax x^{k} x^{-k} g = \\ (x^{-k}g)^{-1} (x^{-k} a^{-1}x^{k})(x^{-(k+1)} a x^{k+1})(x^{-k} g) = \\ (x^{-k}g)^{-1} (x^{-k} g) (x^{-k} a^{-1}x^{k})(x^{-(k+1)} a x^{k+1})= \\ (x^{-k} a^{-1}x^{k})(x^{-(k+1)} a x^{k+1}) = \\ (x^{-k}ax^k)^{-1}\,x^{-1}\,(x^{-k}ax^k) x = \\ \Phi_x(x^{-k}ax^k) $$ Since $x^{-k}ax^k \in A$, this is an element of the image, so that $g^{-1}\Phi_x(a)g \in \operatorname{im}(\Phi_x)$ as desired.

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On using the hint:

Now because $G/A$ is abelian, we can conclude that $G' \subseteq A$. Also, $\operatorname{im}(\Phi_x) \unlhd G$ and $\operatorname{im}(\Phi_x) \subset G' \subset G$, which means that $\operatorname{im}(\Phi_x) \unlhd G'$ with $G'/\operatorname{im}(\Phi_x) \subset G/\operatorname{im}(\Phi_x)$.

So, we have $\operatorname{im}(\Phi_x) \unlhd G' \unlhd A \unlhd G$

I'm not sure how to use this to get the desired result, though.

Answer 2

This is an old thread, but nonetheless: Continuing from Ben Grossmann's response, I proved the result in the following two steps:

1. Show $[x^k,a]\in\mathrm{im}(\Phi)$ for all $a\in A$ and $k\in\mathbb N$.

This is an easy induction (maybe this result is also trivial ... it's late while I'm writing this, so this is how it's going to be).

2. For $g,h\in G$, show $[g,h]\in\mathrm{im}(\Phi)$.

Write $g=x^ka$ and $h=x^\ell b$ with $a,b\in A$ (such a representation comes from $G/A=\langle xA\rangle$).
Disclaimer: I used the definition $[g,h]=ghg^{-1}h^{-1}$ for commutators, which is not the convention the question uses, but again, it's very late and partially I'm writing this here because I know I'm going to need it in a few weeks and would probably forget otherwise. It should be easy to convert my computations to the other convention.
We have $$[g,h]=ghg^{-1}h^{-1}=x^kax^\ell ba^{-1}x^{-k}b^{-1}x^{-\ell}=x^kx^{\ell}x^{-\ell}ax^\ell ba^{-1}x^{-k}b^{-1}x^kx^{-k}x^{-\ell}$$ Since $A$ is abelian, we have $ba^{-1}=a^{-1}b$, so this is equal to $$x^{k+\ell}(x^{-\ell}ax^\ell a^{-1})(bx^{-k}b^{-1}x^k)x^{-(k+\ell)}=x^{k+\ell}[x^{-\ell},a][b,x^{-k}]x^{-(k+\ell)}=x^{k+\ell}[x^{-\ell},a][b,x^{-k}]x^{-(k+\ell)}\\=x^{k+\ell}[x^{-\ell},a][x^{-k},b]^{-1}x^{-(k+\ell)}$$ Note that the two commutators are contained in $\mathrm{im}(\Phi)$ and that $\mathrm{im}(\Phi)$ is a normal subgroup of $G$, so conjugating this with $x^{k+\ell}$ still gives an element in $\mathrm{im}(\Phi)$.

This is the desired result.