Show a quotient group is not cyclic.

Question

Consider $D_{12}$ the dihedral group of order 12. This is the group of symmetries of the regular hexagon. Label the vertices of the hexagon clockwise as 1, 2, 3, 4, 5, 6. Let σ = (1 2 3 4 5 6) and τ = (2 6)(3 5). Regard $D_{12}$ as the group

$D_{12}$ = {e, σ, $σ^2$ , $σ^3$ , $σ^4$ , $σ^5$ , τ, στ, $σ^2$ τ, $σ^3$ τ, $σ^4$ τ, $σ^5$ τ}

Let H = <$σ^2$>. Show that the quotient group $D_{12}$/H is not cyclic. What is this group?

I worked out <$σ^2$> to be the conjugacy class {$σ^2$, $σ^4$}

We are given that the quotient group is made up of 4 distinct cosets, how would I find these cosets? I thought you were supposed to multiply each element of $D_{12}$ by <$σ^2$> but that gives me much more than 4 distinct cosets.

Then, since $D_{12}$/H has order 4, we must show that none of the elements of this are of order 4 to show it is not cyclic. This implies that all the elements have order 1 or 2 and a coset only has order 1 or 2 if $(aH)^2$ = eH, where $a\in D_{12}$ and e is the identity element. How do I go about proving this?

Answer 1

From @Derek Holt 's suggestion you should see that there are four cosets of $\langle\sigma^2\rangle$,

1. $\langle\sigma^2\rangle =\langle\sigma^4\rangle =H$
2. $\sigma\langle\sigma^2\rangle = \sigma^3\langle\sigma^2\rangle = \sigma^5\langle\sigma^2\rangle$
3. $\tau\langle\sigma^2\rangle = \tau\sigma^2\langle\sigma^2\rangle = \tau\sigma^4\langle\sigma^2\rangle$
4. $\tau\sigma\langle\sigma^2\rangle = \tau\sigma^3\langle\sigma^2\rangle = \tau\sigma^5\langle\sigma^2\rangle$

Each of them has three members and they partition $G$ (which I like to call $D_6$).

The square of the coset $xH$ is $x^2H$ for all $x$. From which you should see that $G/H$ is not cyclic. Leaving just one alternative.