Let $G$ be a group of order $120$ with no normal subgroup of order $5$. Then $G$ has a normal subgroup of order $2$ or index $2$.
My attempt: Let $n_5$ denote the number of Sylow $5$-subgroups of $G$. By Sylow's theorem, we obtain $n_5=6$ and $G$ acts on the set of Sylow $5$-subgroups by conjugation. Therefore there exists a homomorphism $\phi: G\to S_6$. We now consider two cases:
Case 1: G is simple. Then $\ker \phi = 1$ or $G$. If $\ker \phi=1$, then $G=S_6$ is not simple, a contradiction. If $\ker \phi=G$, then the action of $G$ on $X$ is trivial, which contradicts that the Sylow $5$-subgroups are not normal. So, this case cannot hold.
Case 2: G is not simple. Then $G$ has a proper non-trivial normal subgroup, say $H$.
How to proceed?
Best Answer
Let me correct and finish the OP's attempt:
Case 1: $\phi$ is injective. Since $A_6$ is simple, it cannot have a subgroup of index $3$ (otherwise there would be a non-trivial homomorphismus $A_6\to S_3$). Hence, $\phi(G)$ is not contained in $A_6$. Consequently, $S_6=A_6\phi(G)$ and $|\phi(G):\phi(G)\cap A_6|=|S_6:A_6|=2$. Since $G\cong\phi(G)$, we are done.
Case 2: $\phi$ is not injective. Since $G$ acts transitively on the set of Sylow $5$-subgroups, $|\phi(G)|$ is divisible by $6$. Therefore, $|\ker\phi|$ divides $20$. If $\ker\phi$ contains a Sylow $5$-subgroup $P$, then $P$ would be characteristic in $\ker\phi$ (by Sylow's theorem) and normal in $G$. This was excluded. Thus, $|\ker\phi|$ divides $4$. We may assume that $|\ker\phi|=4$. Then $|G/\ker\phi|=30$. It is easy to show that every group of order $30$ has a normal Sylow $5$-subgroup (https://math.stackexchange.com/a/569260/960602). Taking the preimage, we obtain a normal subgroup of $G$ of order $20$, which was already excluded.
There are probably shorter arguments.
Addendum: The statement is true without the hypothesis on the Sylow $5$-subgroups: Every group of order 120 possesses a normal subgroup of order 2 or index 2.