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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 $\langle ab \rangle \rtimes \langle a \rangle$.2012-09-27
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    @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?2012-09-27
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    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\,$2012-09-27
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    @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.2012-09-27
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    @rschwieb: right! But if $ab$ has infinite order we just get the infinite dihedral group $D_{\infty} = \mathbb{Z} \rtimes C_2$.2012-09-27
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    @MartinoGaronzi I'm curious also what you mean by "dihedral". For me it means "symmetries of a regular 2-D" polygon, but are you using it for higher dimensional symmetries or something? Sorry I misinterpreted your use of $n$ in the first comment.2012-09-27
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    Martino's notation is standard. The infinite dihedral group is AGL(1,Z), the affine symmetries of the integers, generated by the "rotation" $x\mapsto x+1$ and the reflection $x\mapsto -x$. It has presentation $\langle a,b : a^2=b^2=1\rangle$. If you view an $\infty$-gon in 2D as a very obtuse angled set of edges (so obtuse that we just have to draw in the corners at the integers) then the standard symmetries still apply and give the infinite dihedral group. If you view a $\infty$-gon as a circle, this is a subgroup of its symmetries, generated by any two reflections at an irrational angle.2012-09-27
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    @JackSchmidt Now I know... thanks!2012-09-27

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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.2012-09-27
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    @JackSchmidt Ah yeah that omission slipped in: thank you.2012-09-27
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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 $x \mapsto x^{-1}$.