# Nonabelian group of order 95 [duplicate]

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?

• This is the case $(p,q)=(5,19)$. Since $5\nmid 19-1$, all groups of order $pq=95$ are abelian. There are several questions at MSE proving this; also this one. Apr 22 '18 at 19:31

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 $HK = G$.
• Why is it that $H \cap K = \{e\}$ and $H,K$ are normal in $G$ are sufficient conditions to prove that the direct product is isomorphic? Apr 22 '18 at 14:50
• @blanchey It’s necessary to add a condition : $G=HK$. That is , if $H,K$ are normal subgroups of $G$, $H\cap K=1,G=HK$, then $G\cong H\times K$. This is a basic fact of abstract algebra.Without this extra condition, it’s not true. Say, 1 and $K_4$(Klein four group) are normal subgroups of $A_4$, of course $1\times K_4$ is not isomorphic to $A_4$. But in case of here obviously $G=HK$ Apr 22 '18 at 16:19