Limitations on the order of a subgroup of $S_5$ The goal is to show $S_5$ has no subgroup of order $15$ or $30$, but does have one of order $20$. I think I have part of the first part of it:
There are only two groups of order $15$, $\mathbb{Z}_3\times \mathbb{Z}_5$ and $\mathbb{Z}_{15}$ (Of course, these two groups are isomorphic). While $S_5$ contains copies of $\mathbb{Z}_3$ and $\mathbb{Z}_5$, neither of them are normal, and hence their direct product is not a subgroup. Alternatively, I'm sure $S_5$ doesn't have an element of order $15$, but I forget how to show it, which would lead to $S_5$ not containing $\mathbb{Z}_{15}$, therefore no subgroup of order 15.
I know that if a subgroup of $S_5$ has index $k\le n$, then the only possibilities for $k$ are $1,2,5$. Thus, as an order $30$ subgroup would have index $4$, such a subgroup doesn't exist.
I'm then having trouble showing there's a group of order $20$. I mean, I've seen on the groupprops wiki what it's generated by, but I'm not sure on how I'd come up with elements to test (I suppose since $4\cdot 5=20$, but that logic doesn't work with $3\cdot 5=15$). Any guidance would be appreciated.
 A: Consider the Sylow $5$-subgroups. It's easy to verify that there are six of these. In general, if $n_p$ denotes the number of Sylow $p$-subgroups and $P$ is any of these subgroups, then $n_p = |G:N_G(P)|,$ where $N_G(P)$ is the normalizer of $P$.
So in this example, if $P$ is any Sylow $5$-subgroup, then we have $|G:N_G(P)| = 6$, hence $|N_G(P)| = |G|/6 = 120/6 = 20$.

Suppose that we want to find an explicit subgroup of order $20$. By Cauchy's theorem, such a subgroup must contain an element of order $5$, and in $S_5$ the only such elements are $5$-cycles, so let's start with some arbitrary $5$-cycle $(abcde)$.
Let $H$ denote the subgroup generated by $(abcde)$, hence $H$ has order $5$. 
Let's try to find some element which is not in $H$ but which normalizes $H$. For example, let's find some $k \in G = S_5$ such that $k(abcde)k^{-1} = (abcde)^2 = (acebd)$. It's easy to check that $k = (bced)$ satisfies this requirement. Since conjugation by $k$ sends $(abcde)$ to a power of $(abcde)$ and both of these generate $H$, it follows that $k$ normalizes $H$.
Now let $K$ be the subgroup generated by $k$, hence $K$ has order $4$.
Let $N = N_G(H)$ denote the normalizer of $H$ in $G$. Then $H \lhd N$ by definition. Also, we just showed that $k \in N$, hence $K = \langle k \rangle \leq N$. It follows that $HK$ is a subgroup of $N$, hence also a subgroup of $G$. The order of $HK$ is
$$|HK| = |H||K|/|H \cap K| = |H||K| = 5\cdot 4 = 20$$
since $|H \cap K| = 1$ (because $|H|=5$ and $|K|=4$ are relatively prime). Since $HK$ is a subgroup with the desired order, we're done. Note that this is a semidirect product since $H \lhd HK$ but $K \not\lhd HK$.
