If $G$ is a group of finite order and $p$ is a prime, then $N\triangleleft G,\ H\leq G,\ |G/N|=p\Longrightarrow H\subset N$ or $NH= G$.

Question

If $\displaystyle N$ is a normal subgroup of finite group $\displaystyle G$ (finite group) and $\displaystyle H$ is a subgroup of $\displaystyle G$ such that the quotient group $\displaystyle G/N\ $is of prime order, then prove that either $\displaystyle H\subset N$ or $\displaystyle NH=G$.

Since $\displaystyle G/N$ is of prime order $\displaystyle p$, it must be cyclic. It follows that $\displaystyle \exists g\in G\setminus N$ such that

$$G/N=\langle gN \rangle$$ Suppose that $\displaystyle H\not\subset N$.

It follows that $\displaystyle \exists h\in H\setminus N$ and therefore $\displaystyle hN=g^{i} N$ for some $\displaystyle i\geqslant 1$. I am stuck here and don't know how to show $\displaystyle NH=G$ from here. I'll appreciate any hint in this. Thanks.

Answer 1

Hint: $NH$ is a subgroup since $N$ is normal. In addition $N \subseteq NH \subseteq G$, so $|G:N|=|G:NH| \cdot |NH:N|$ is a prime, so ....