Nonabelian group of order 95

Question

I have to either find a nonabelian group of order $95$, or show none exists. $95 = 19 \times 5$. I am trying to use the class equation to show that $\vert G\vert = \vert Z(G)\vert$, but I am not sure how to show that there is no subgroup which is not in the center. Can someone gear me in the right direction?

Answer 1

By Cauchy's Theorem, a group $G$ of order $95$ must have two cyclic subgroups $H$ and $K$, isomorphic to $\mathbb{Z}_5$ and $\mathbb{Z}_{19}$.

The next step would be to prove that $G \cong \mathbb{Z}_5 \times \mathbb{Z}_{19}$, which can be done using the fact that if $H \cap K = \{ e \}$, $HK = G$ and both subgroups are normal, then $G \cong H \times K$.

The trivial intersection is quite easy to prove using Lagrange, while for the normality i suggest to use the Sylow Third Theorem to show that $H$ and $K$ are the only two subgroups of order $5$ and $19$, and thus they are normal

[edited, considering Jabin Du remark]. In the end to prove that $G = HK$ just observe that $HK$ is a subgrop since $H$ and $K$ are normal. Thus the order of $HK$ must be divisible by the order of $H$ and the order of $K$, and still have to be less than $95$, so $|HK| = 95$ and then $HK = G$.