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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.

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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?

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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