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 $NH=G$.

  • 1
    $\begingroup$ 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). $\endgroup$ – user1729 Mar 19 at 16:29
  • $\begingroup$ Since $N$ is normal, $NH$ is a subgroup, and $N$ is its (normal) subgroup. Now what can you say about $NH/N$? $\endgroup$ – M. Vinay Mar 19 at 16:36
  • $\begingroup$ $\vert NH/N\vert=\frac{H}{H\cap N}$. $\endgroup$ – AColoredReptile Mar 19 at 16:44
  • $\begingroup$ @AColoredReptile Correct. What else? As a group $NH/N$ is…? A subgroup of something? $\endgroup$ – M. Vinay Mar 19 at 16:45
  • $\begingroup$ It's a subgroup of $G/N$. $\endgroup$ – 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 $NH = G$.


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