Show every subgroup of order $5^3$ of $S_{15} \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$ 
Show that every subgroup of order $5^3$ of the Symmetric Group $S_{15}$ is isomorphic to $\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$.

I thought about the p-Sylow laws first, but didn't find any relation. Might it be true that all members of that subgroups are of order 5? 
Can you please help me to find an approach?
 A: Let $H \leq S_{15}$ such that $|H| = 125$.  For any non-identity $a \in H$, we know that $|a| = 5, 25 \text{ or } 125$ by Lagrange's theorem.
Recall that, for any $\pi \in S_n$, we know $|\pi|$ will be the least common multiple of the cycle lengths when $\pi$ has been decomposed into a composition of disjoint cycles.  Applying this, can you see why $|a| \neq 25 \text{ and } 125$?  So it is indeed true that every non-identity element of $H$ has order $5$.
Now, I claim that $H = \langle a, b, c \rangle$ where $a, b, c \in H$ are disjoint $5$-cycles (which commute with each other).  Suppose for contradiction this were not true.  Then we could find two $5$-cycles $x, y \in H$ that are not disjoint and also do not generate the same subgroup.  
We can then consider the element $xy$.  Once you've shown that $xy$ cannot be written either as a single $5$-cycle or a composition of disjoint $5$-cycles, then notice that $|xy| \neq 5$.  This is a contradiction!
Therefore, $H$ is generated by $3$ disjoint $5$ cycles that commute with each other, so $H$ is abelian.  You can now use the structure theorem to finish up.
A: As noted in the comment, the only elements of order a power of $\;5\;$ in $\;S_3\;$ are those of order $\;5\;$ , and from here that if we take three disjoint $\;5$-cycles here we get an abelian group of order $\;125\;$ as disjoint cycles commute.
