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Suppose that $N$ is a normal subgroup of a finite group $G$, and $H$ is a subgroup of $G$. If $|G/N| = p$ for some prime $p$, then show that $H$ is contained in $N$ or that $NH = G$.

I imagine this is related to the fact that $|NH| = |N||H|/|N \cap H|$, but this is not really helping me. I considered the fact that since $N$ is normal, we get that $NH \leq G$, and I then used Largrange, but I'm stuck, and some help would be nice.

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3 Answers 3

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Consider the homomorphism $\phi:G\rightarrow G/N$ that sends $x$ to $xN$. Since $\phi(H)\leq G/N$, therefore $|\phi(H)|||G/N|$. Hence $|\phi(H)|=1$ or $|\phi(H)|=p=|G/N|$. In the first case we get $\phi(H)=\{N\}$, thus $H\leq N$. In the other case $\phi(H)=G/N$, we deduce that $\forall x\in G\ \exists h\in H[xN=hN]$, it is easy to show that this implies that $NH=G$.

(Note that we don't need G to be finite )

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Recall that when we mod out by a normal subgroup $N$, there is a one-to-one correspondence between subgroups in $G/N$ and subgroups $H$ containing $N$ in $G$. Since the order of $G/N$ is prime, there are no proper subgroups of $G/N$ (by Lagrange's Theorem). This implies that there aren't any proper subgroups of $G$ that properly contain $N$, hence $H\subseteq N$.

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Well, if you've got some $h\notin N$, then $|N\langle h\rangle|>|N|$. This is a group since $N$ is normal. Hence if $[G:N]$ is prime, then by Lagrange $N\langle h\rangle=G$.

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