$H$ prime order implies $G/H$ cyclic?

Question

I am revising some concepts of group theory, but I must admit that it's been a while and I have forgotten quite a bit.

I want to show that if $H$ is a normal subgroup of $G$, where $|H| = p$ ($p$ prime), we have that the factor group $G/H$ cyclic.

In my situation, I have that $|G| = p^2$. In this case I understand that $|G/H| = |G|/|H| = p^2/p = p$. However, I don't understand how to prove that $G/H$ is cyclic.

I feel like the result might be trivial, as I am having trouble with finding a good explanation on the internet. Is there anyone who could explain this result to me? Thanks in advance!

Answer 1

Every group of prime order is cyclic, in fact it is generated by any of its non-neutral elements (what else should that element generate if not the whole group?).

The claim in your title does however does not hold in general (i.e., when $|G|\ne p^2$).

Answer 2

To brush up your knowledge a bit : if for the normal subgroup $H$ of $G$ you would have $|H|=p$ is the smallest prime dividing $|G|$ and $G/H$ is cyclic, then $G$ would be abelian! Why? Since $G/C_G(H)$ embeds homomorphically into $Aut(H) \cong C_{p-1}$, it follows that $G=C_G(H)$, which is equivalent to $H \subseteq Z(G)$. Since $G/H$ is cyclic, it follows that $G/Z(G) \cong (G/H)/(Z(G)/H)$ is cyclic and this implies $G$ being abelian.

Answer 3

This is not true. There are lots of examples where a normal subgroup of order $p$ doesn't yield a cyclic quotient. Just to pick one out of the air, how about $D_8/Z(D_8)\cong V_4$.