Originally Posted by

**craig** Thanks for the reply.

With it being cyclic, you have the following.

$\displaystyle \{ e \alpha, \alpha^2, \alpha^3 ... \alpha^{11} \}$

$\displaystyle \{ e, \alpha^2, \alpha^4 ... \alpha^{10} \}$

... etc

$\displaystyle \{ e = \alpha^{12} \}$

So the only element of $\displaystyle H$ would be $\displaystyle \alpha^{12} = \langle e \rangle$, which has order 1?

For part 3, I considered $\displaystyle D_{12}$, the symmetries of a regular hexagon.

For this would I just list the 6 symmetries and the 6 reflections in $\displaystyle D_{12}$, and to show that it is not a subgroup explain how it's not commutative?

Thanks again for the pointers.