# Suppose $N\trianglelefteq G$ and $H\leqslant G$. If $\vert G/N \vert$ is prime prove $H\subseteq N$ or $NH=G$

Suppose $$N\trianglelefteq G$$ and $$H\leqslant G$$. If $$\vert G/N \vert$$ is prime prove $$H\subseteq N$$ or $$NH=G$$

I believe I want to make use of this fact that if $$H,K\leqslant G$$ that $$HK=H \iff K\subseteq H$$.

So here I would proceed by cases either $$H\subseteq N$$ or it is not. Case 1 there is nothing to prove.

For $$H\not\subseteq N$$ then $$NH\not = H$$ but I'm not sure how to proceed from here. I'm thinking something about $$N$$ being normal should give me a reason that $$NH=G$$.

• It is helpful to state the question in the body of the post (the title is just a title, and it gets confusing if you do not state the question properly). – user1729 Mar 19 at 16:29
• Since $N$ is normal, $NH$ is a subgroup, and $N$ is its (normal) subgroup. Now what can you say about $NH/N$? – M. Vinay Mar 19 at 16:36
• $\vert NH/N\vert=\frac{H}{H\cap N}$. – AColoredReptile Mar 19 at 16:44
• @AColoredReptile Correct. What else? As a group $NH/N$ is…? A subgroup of something? – M. Vinay Mar 19 at 16:45
• It's a subgroup of $G/N$. – AColoredReptile Mar 19 at 16:46

$$N \unlhd G$$ and $$|G/N| = p$$ is prime. For any $$H \le G$$, $$H \subseteq N$$ or $$NH = G$$.
Since $$N$$ is normal, $$NH = HN \implies NH \le G$$, and further, $$N \unlhd NH$$, so that $$NH/N \le G/N$$. But since $$G/N$$ is of order $$p$$, a prime, by Lagrange's theorem, $$|NH/N| = 1$$ or $$p$$.
If $$|NH/N| = 1$$, then $$NH = N$$, and therefore, $$H \subseteq N$$.
(Or: $$|NH/N| = |H / (H \cap N)| = 1 \implies |H| = |H \cap N| \implies H \subseteq N$$).
If $$|NH/N| = p = |G/N|$$, then $$NH = G$$.