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I found this question in an algebra qualifying exam and I'm interested if anyone has a way to organize the way I should be trying to solve it or if there's a better way to approach it.

Question: Let $G$ be a group generated by two elements $a,b$ where $a^2=b^2=1$. Show that the commutator subgroup $G'$ is cyclic.

So far I have reasoned that we can write any element of $G$ as $(ab)^n$ or $(ab)^nb$ or $(ba)^n$ or $(ba)^nb$ for some $n\geq0$. Then we have identities such as $(ab)^{-1}=ba$ and $(ab)^nb=b(ba)^n$ so we can check that every commutator of combinations of the four options above (for instance $[(ab)^n,(ba)^mb]$) can be written as $(ab)^{2l}$ for some integer $l$. Thus $G'$ is generated by $abab=[a,b]$.

I'd rather not do all that work so is there a way to avoid it that might give more understanding?

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Two involutions $a,b$ always generate a dihedral group of order $2n$ where $n$ is the order of $ab$. This is easy to show: $G$ turns easily out to be the semidirect product $\langle ab \rangle \rtimes \langle a \rangle$. – Martino Sep 27 '12 at 17:00
@MartinoGaronzi The element $ab$ does not have finite order in this (free) group. Maybe this comment requires a deeper step that I didn't take? – rschwieb Sep 27 '12 at 17:06
Well, it can't be a free group as there are non-trivial elements with finite order, but from what we know it could be the free product $\,C_2*C_2\,$ – DonAntonio Sep 27 '12 at 17:12
@DonAntonio Yeah that was silly of me to say: what I intended to say is exactly that without "free". "free" crept in from a previous thought that does not fit in with this comment. Thanks Don. – rschwieb Sep 27 '12 at 17:24
@rschwieb: right! But if $ab$ has infinite order we just get the infinite dihedral group $D_{\infty} = \mathbb{Z} \rtimes C_2$. – Martino Sep 27 '12 at 17:25
up vote 1 down vote accepted

Try this: You know $G$ consists of all manner of finite strings of $a$ and $b$. You also know that $G/G'$ has to be Abelian, and that $G'$ is the smallest such normal subgroup to do that.

Examining $G$, you can see that if you make $a$ commute with $b$, (i.e., require $ab=ba$, aka $aba^{-1}b^{-1}=1$) by modding out the normal (edit) subgroup generated by $[a,b]$, then the quotient is abelian.

So, it makes sense that $G/\langle[a,b]\rangle$ would be abelian.

That gives you that $\langle[a,b]\rangle\supseteq G'$, and the reverse containment is trivial.

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Modding out by the normal subgroup generated by $[a,b]$. It is a short but easy calculation that $\langle [a,b] \rangle$ is normal (but this is a special feature of order 2). In general your argument applied to any 2-generated group, but the simple group of order 60 is 2-generated and its commutator subgroup is not cyclic. – Jack Schmidt Sep 27 '12 at 18:02
@JackSchmidt Ah yeah that omission slipped in: thank you. – rschwieb Sep 27 '12 at 18:13

As rschweib mentions, the commutator subgroup of $G=\langle a,b\rangle$ is the smallest normal subgroup containing the commutator $[a,b]$.

For a general group generated by two elements $\langle a,b\rangle$, the commutator subgroup certainly need not be cyclic. For instance if $a=(1,3,2)$ and $b=(2,5)(3,4)$, then $\langle [a,b] \rangle = \langle (1,2,3,4,5) \rangle$ has order 5 and is not a normal subgroup of $\langle a,b \rangle$. Indeed the commutator subgroup of $\langle a,b \rangle$ is the entire $\langle a,b \rangle$ in this case and one certainly does not have that $\langle [a,b] \rangle \supseteq G'$.

Hence one needs to show that $\langle [a,b] \rangle$ is normal, which makes essential use of the fact that $a$ and $b$ have order 2.

When $a,b$ have order 2, one gets the very simple formula $x^a = axa$ and $[a,b] = abab$, so that $[a,b]^a = aababa = baba = [b,a] = [a,b]^{-1}$ and $[a,b]^b = bababb = baba = [b,a] =[a,b]^{-1}$. In particular, $\langle [a,b] \rangle$ is normal, and both $a$ and $b$ act as the automorphism $x \mapsto x^{-1}$.

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