Let $G$ be a group of order $120$ with a normal subgroup $N$ of order 5. Let $H$ be any subgroup of $G$ of order $15$. Prove $N$ is a subgroup of $H$.
I have a proof (see below), but I am wondering: Is there a simpler proof? In particular, something that doesn't use that $HN \simeq H \times N$.
Proof: Since $N$ is normal, $HN$ is a subgroup $G$. By Lagrange, since $N$ has prime order it must be cyclic and every non-identity element is a generator. By way of contradiction, suppose $H$ does not contain $N$. Then $H \cap N = \{e\}$. From this, it follows that $HN \simeq H \times N$. Thus $|HN| = |H||N| = 15 \cdot 5 = 75$. By Lagrange, $75 \mid 120$. Contradiction.
Let $G$ be a group of order $|G| = 120$ with a normal subgroup $N$ of order $|N| = 5$.
Note that since $\gcd(|G/N|, |N|) = 1$, it follows that $N$ is the only subgroup of order $5$ in $G$.
Now let $H \leq G$ be a subgroup of order $15$. Because $H$ contains a subgroup of order $5$ (for example by Cauchy's theorem, or by Sylow's theorem), conclude that $N \leq H$.