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I have the group given by the presentation $G= \langle a,b\mid a^2,b^2\rangle$

How can I in general find $G',G/G',G''$ ?

thanks for any hints.

Shaun
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  • have a look at this post for an example: http://math.stackexchange.com/questions/416617/the-abelianization-of-langle-x-a-mid-a2x-xa-rangle/416703#416703 – citedcorpse Jun 11 '13 at 17:38
  • also, i think finding a presentation even for just the commutator subgroup itself is not possible in general – citedcorpse Jun 11 '13 at 17:40

2 Answers2

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In general, $G/G^{\prime}$ is the group attained by making the generators pairwise commute, and is called the abelianisation of $G$ (*). As we only have two generators, here $G/G^{\prime}$ is the group $\langle a, b; a^2, b^2, [a, b]\rangle$. I will leave you to work out what this is isomorphic to.

To find the derived subgroup $G^{\prime}$, you can (in general) use something called "Reidemeister-Schreier" (if you want to know more, I once wrote out an example - note that this will only spit out a finite presentation if $G$ is given by a finite presentation and $G/G^{\prime}$ is finite). However, you needn't do anything quite so fancy here! Indeed, $$G^{\prime}=\langle (ab)^2\rangle.$$ Why? Well, $(ab)^2=abab=a^{-1}b^{-1}ab=[a, b]$, while $\langle (ab)^2\rangle$ is normal in $G$ (why?). Can you see why this is sufficient?

Now, $G^{\prime}$ is cyclic. What does this mean for $G^{\prime\prime}$?

(*) A constructive example of the abelianisation would be something like $\langle a, b, c; aba^{-1}c^{-1}, a^2, b^3\rangle$, then your abelinisation is $\langle a, b, c; aba^{-1}c^{-1}, a^2, c^3, [a, b], [a, c], [b,c]\rangle$...but...as $a$ and $b$ now commute the first relation means that $b=c$, so your group becomes $\langle a, b; a^2, b^3, [a, b]\rangle$. Note that $c^3$ has become $b^3$. Can you see why this is?

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user1729
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    There is one potential restriction on the method in this answer: $G/G'$ should be finite for Reidemeister-Schreier to give a finite presentation. (+1) for easy methods and second example – Jack Schmidt Jun 12 '13 at 14:19
  • @JackSchmidt Good point - I have added this in. However, because finite presentability has been mention, anyone who is reading this should note that one can still use Reidemeister-Schreier even if $G/G^{\prime}$ is infinite. You just need to bring your thinking-cap with you! (Although this is impractical unless $G/G^{\prime}$ is infinite cyclic, or has a similar, very nice, Cayley graph.) – user1729 Jun 12 '13 at 15:29
  • (And it should also be mentioned that if $G/G^{\prime}$ is infinite then $G^{\prime}$ is not necessarily finitely presented, or even finitely generated! See Seirios' answer.) – user1729 Jun 12 '13 at 15:30
  • Thanks! Yes, I had to remove my gloom and doom statements from my comment. A computer cannot do much of anything when $G/G'$ (or in my actual examples $G'/G''$) is infinite, but humans are pretty clever and can often recognize free groups at least, even if R-S is not obviously (to a computer) giving the trivial presentation. – Jack Schmidt Jun 12 '13 at 15:34
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Another way to find $G'$ is to notice that $G$ is the free product $\mathbb{Z}_2 \ast \mathbb{Z}_2$. Then $G'$ is the kernel of the canonical projection $\mathbb{Z}_2 \ast \mathbb{Z}_2 \to \mathbb{Z}_2 \times \mathbb{Z}_2$, and

Lemma: Let $H$ and $K$ be two groups. The kernel of $H \ast K \to H \times K$ is free over the set $\{[h,k] \mid h \in H \backslash \{1\}, k \in K \backslash \{1\}\}$.

For example, you can see this answer.

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Seirios
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    This then implies that unless $H$ and $K$ are both cyclic, $G^{\prime\prime}$ is free on countably many generators, as it is the commutator subgroup of a non-abelian free group. – user1729 Jun 12 '13 at 10:06