If $G$ is a group such that any two commutators commute, $G$ is solvable

Question

I need to prove that if $G$ is a group such that any two commutators of elements of $G$ commute, then $G$ is solvable.

This is the idea that I had: The subgroup of all commutators of $G$, $G^{\prime}$ is a normal subgroup of $G$. Therefore, $G^{\prime}$ is abelian. Hence, $G/G^{\prime}$ is abelian, which in turn implies that $G$ is abelian, and thus solvable.

It doesn't seem right to me, however - it just seems too easy this way. I'll admit, I find cosets and subnormal series intimidating, and I'm not sure I understand them completely. So, I'm ready to be schooled by someone who can explain them well to me if that is the only way this problem can be solved.

In any case, please let me know if what I did was correct. If not, a full solution would be preferred - one with subnormal series explained thoroughly and clearly would be the best case scenario, so I can work backwards from your solution and hopefully learn something.

Answer 1

Be careful, a normal subgroup is not necessarily abelian.

Here, go back to the definition of solvability : $G$ is solvable if $G^{(n)} = {e}$, where $G^0 = G$ and $G^{(n)} = (G^{(n-1)})'$.

If any two commutators of $G$ commute, that means that $G'$, the subgroup formed by the commutators of $G$, is abelian.

Therefore $G'' = (G')'$ is trivial. Indeed, let $z_{1}, z_{2}$ be commutators of $G$, then $z_{1}z_{2}z_{1}^{-1}z_{2}^{-1} = e$ since $z_{1}z_{2} = z_{2}z_{1}$.

So $G$ is solvable.

Answer 2

A group is solvable if and only if its derived series, the descending series of normal subgroups $G>G^{(1)}>G^{(2)}>...$ eventually reaches the trivial subgroup $\{e\}$, where $G^{(1)}=[G,G],\ G^{(2)}=[G^{(1)},G^{(1)}]$ etc.

In your case, $[G,G]$ is already abelian, so $G^{(2)}=\{e\}$.

Answer 3

Why should $G/G'$ abelian imply that $G$ is abelian? An easy counterexample to your claim is to note that $S_3$ is not abelian, but $A_3=S_3'$ is abelian (being of order $3$), and so is the quotient $S_3/A_3=C_2$. This gives indeed the classical solvable series $1\lhd A_3\lhd S_3$. Thus a group may admit a normal abelian subgroup $N$ such that $G/N$ is also abelian without being abelian itself. It is however, an extension of abelian groups, and say if $N$ and $G/N$ are cyclic of prime orders $p<q$ such that $p\not\mid q-1$ then in fact $G$ is abelian.

If $[G,G]=G'$ is abelian then you have a solvable series $$1\lhd G'\lhd G$$ since $G'/1$ is abelian and $G/G'$ is abelian too, being the abelianization of $G$. This works, of course, for any $k$ such that $G^{(k)}=1$.

Answer 4

More generally, define a sequence of subsets of $G$ by $S^0 = G$ and $S^k = \{ ghg^{-1} h^{-1} : g,h \in S^{k-1} \}$. I claim that, if $S^k = \{ e \}$ for some $k$, then $G$ is solvable. (This question is the case $k=2$.) I'm writing this up because I came to math.SE to ask this question, and had the usual rubber duck experience of working it out as I asked it.

Key Lemma: Let $H$ be a group and let $X$ be a subset of $H$ such that (1) $X$ generates $H$ and (2) $hXh^{-1} = X$ for all $h \in H$. Set $X' = \{ x y x^{-1} y^{-1} : x,y \in X \}$ and let $H'$ be the commutator subgroup of $H$. Then $X'$ generates $H'$.

Proof: As explained here, $H'$ is generated by $X'$ and the set of all conjugates of $X'$. But $h (xyx^{-1} y^{-1}) h^{-1} = (h x h^{-1}) (h y h^{-1}) (h x h^{-1})^{-1} (h y h^{-1})^{-1}$, so any conjugate of an element of $X'$ is also in $X'$.

Proof of main theorem: Let $G^k$, as usual, by defined by $G^0=G$ and $G^k$ is the commutator subgroup of $G$. Note that $S^k$ is clearly in $G^k$. Also, $S^k$ is clearly closed under conjugation by $G$, and thus by $G^k$. The lemma thus shows that, if $S^k$ generates $G^k$ then $S^{k+1}$ generates $G^{k+1}$. So, by induction, $S^{k}$ generates $G^{k}$ for all $k$. In particular, if $S^k = \{ e \}$ then $G^k = \{ e \}$, which is the usual definition of solvability.