Quotient of $G= \left\langle a, b \ \middle|\ a^4, b^2=a^4, aba=b \right\rangle$ by $\langle a^2 \rangle$

Question

Let $G$ be finite group of order $8$ of the form: $G= \left\langle a, b \ \middle|\ a^4, b^2=a^4, aba=b \right\rangle$.

The elements are $\left\lbrace 1, a, a^2, a^3, b, ab, a^2b, a^3b\right\rbrace$.

Question 1: Show that the subgroup $H=\left\langle a^2 \right\rangle$ is a normal subgroup of $G$

$H$ is a normal subgroup if it is a union of conjugacy classes of $G$

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The conjugacy classes of $G$ are:

$\{1\}$ and $\{a^2\}$ since they commute with everything

$bab=bba^3=a^3$ so $\{a, a^3\}$ is another

I also found $\{b, a^2b\}$ and $\{ab, a^3b\}$

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So $\left\langle a^2 \right\rangle$ is a union of of the first two conjuacy clases so is a normal subgroup - is this correct?

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Further I want to be able to determine the following:

Question 2: What is $G/H$ isomorphic to?

This is a quotient of a normal subgroup, and the set of cosets of $N$ in $G$. I am not sure what it could be isomorphic to, but I guess it would be some abelian group $\mathbb{Z}_n$- would that be right?

Thank you for your time

Answer 1

That's correct that if $H$ is a union of conjugacy classes, it's normal. You can also note that, as the center of $G$, it must be normal.

Your group $G$ has order $8$, and the normal subgroup $H$ has order $2$, so there aren't a lot of possibilities for the quotient $G/H$ -- it's either the cyclic group of order $4$, or the direct product of two cyclic groups of order $2$.

However, since $$G/Z(G) \text{ cyclic} \implies G \text{ Abelian},$$

you can narrow things down quite a bit (i.e., completely) if you know that $G$ itself is not Abelian.

Answer 2

You can check from the presentation that the group is $D_4$, with center $\langle a^2\rangle$. We know that the center is a normal subgroup. It has been shown here at MSE that the quotient is isomorphic to $C_2\times C_2$, the Klein $4$-group. But we can also show this directly, by considering the relations "modulo $a^2$".