Let $G$ be a finite group, $H\le G$, and $N\unlhd G$. If $|H|=5$ and $|G/N|=7$, prove $H\le N$.

Question

Let $G$ be a finite group and $H$ is a subgroup and $N$ is a normal subgroup of $G$. If $|H|=5$ and $|G/N|=7$. Prove that $H$ is a subgroup of $N$.

By Lagrange's theorem, $|N|$ is a multiple of 5. But will that imply $H\subseteq N$?

Answer 1

Consider $p:G\rightarrow G/N$ the quotient map, $p(H)$ is a subgroup of $G/N$, thus its order divides $7$ (Lagrange), but the order of $p(H)$ divides $5$ since $|H|=|\ker(p_{\mid H})||p(H)|$, this implies that the order of $p(H)$ is $1$ and $p(H)$ is the neutral element of $G/N$, we deduce that for every $x\in H, p(x)=e$ and $x\in N$.