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If my $G=\langle a,b\ |\ a^4,b^4,a^2=b^2 \rangle$, then how to find presentation for $G'$ (derived subgroup of $G$).

Should I proceed by Reidemeister-Schreier process, that is too cumbersome. Iisn't there any other way. Actually I have to show that $G$ is metabelian.

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  • $\begingroup$ What is $G'$? Is it the abelianization? $\endgroup$
    – msteve
    Mar 20, 2015 at 1:09
  • $\begingroup$ If you like, it's the kernel of the abelianization (aka commutator subgroup). $\endgroup$ Mar 20, 2015 at 1:28

1 Answer 1

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Your group $G$ is presented as the amalgamated product $\mathbb{Z}_4*_{\mathbb{Z}_2}\mathbb{Z}_4$ which is a central extension of order $2$ of the infinite dyhedral group $\mathbb Z_2*\mathbb Z_2$. The latter is isomorphic to $\mathbb Z \rtimes \mathbb Z_2$ which is clearly metabelian. A central extension of a metabelian group is metabelian.

The commutator subgroup of $\mathbb Z_2*\mathbb Z_2$ is cyclic and generated by $ABAB$, where $A$ and $B$ are the generators of the two free factors and are the images of $a$ and $b$ under the quotient of $G$ by $\langle a^2\rangle$. The element $ABAB$ lifts in $G$ to the commutator $[a,b]=[a,b^{-1}]$ which generates the commutator subgroup $G'$ (this is easy to check). Whence $G'$ is infinite cyclic and generated for example by $[a,b]$.

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  • $\begingroup$ Why is latter clearly metabelian. $\endgroup$ Mar 20, 2015 at 2:35
  • $\begingroup$ ok. We have N as group of rotations then G/N is abelian. thanks $\endgroup$ Mar 20, 2015 at 7:26

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