The normal subgroup of $\langle a, b\mid a^4, a^2=b^2=(ab)^2\rangle$ generated by $a^2$.

Question

This is Exercise 2.1.5(a) of "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations," by Magnus et al.

The Question:

For each of the following groups $G$, let $H$ be the subgroup generated by the given elements. Show that $H$ is normal and find the order of $G/H$.

(a) $G=\langle a, b\mid a^4, a^2=b^2=(ab)^2\rangle$; $a^2$.

My Attempt:

In $G$, we have $a=bab$ and $b=aba$, so it's elements are of the form $$a^nb^\varepsilon$$ for $n\in \{0,1,2,3\}$, $\varepsilon\in\{0, \pm 1\}$ - is this right? - and the elements of $H$ are of the form $a^{2m}$ for $m\in\Bbb Z$, so consider elements of $G$ of the form $$a^nb^\varepsilon a^{2m} (a^nb^\varepsilon).$$ They should elements of $H$ but I don't know how to show this.

As for the order of $G/H$, we have that $$G/H=\langle a, b\mid a^2=b^2=(ab)^2=e\rangle,$$ so that $a=a^{-1}, b=b^{-1},$ and $abab=e$ give $ab=ba$, which implies $$G/H\cong C_2\times C_2,$$ so $\lvert G/H\rvert=4$.

Please help :)

Answer 1

OK, the normality of $\langle a^2 \rangle$ follows from the relation $a^2=b^2$, which implies that both $a$ and $b$ commute with $a^2$. So $a^2 \in Z(G)$.

Answer 2

For a completely different point of view, you can use the Todd-Coxeter algorithm to construct the Schreier graph of the cosets of $H$. See The Todd-Coxeter procedure. (Cayley graphs are Schreier graphs of the trivial subgroup.)

If the resulting Schreier graph has automorphisms which can carry any vertex to any other vertex, then $H$ is normal.

It is not hard to carry out the algorithm by hand, but there is an online implementation where you can enter a^4;a^2=b^2=(ab)^2 for the "G relations" and a^2 for the "H generators". The result is that $[G:H]=4$ and $H$ is normal. $G/H$ appears to be $C_2\times C_2$.