Any solvable group of order sixty has a normal subgroup of order five I want to show that any solvable group of order sixty has a normal subgroup of order five. 
Recall that a solvable group is a group that has a normal series in which every factor is abelian.
I think that I can prove this by looking at the Hall subgroups of the group. Let $G$ be a solvable group of order sixty. We know that $60=2^2\cdot 3\cdot 5$. A Hall divisor $d$ of $G$ is a divisor of $60$ such that $\gcd(d, 60/d)=1$, and $H\leq G$ is a Hall subgroup if the order of $H$ is a Hall divisor. In particular, I know that $5$ is a Hall divisor of $G$.
Now since $5$ is a prime divisor of $G$, I know that $G$ has a $5$-Sylow subgroup $S$, and I also know that its order has to be $5$, by Lagrange's Theorem. If I can show that $S$ is the only $5$-Sylow subgroup, then I know it is normal, and I am done. I don't know if this is the case, and I don't know how the fact that $G$ is solvable helps.
Any help is appreciated, thank you.
 A: Let $K$ be a minimal normal subgroup of $G$; $|K|$ is either $2$, $3$, $4$, or $5$.


*

*If $K$ has order $3$, then $G/K$ has order $20$, and has a normal Sylow 5-subgroup (by Sylow's theorems). The preimage is a normal subgroup of order $15$ in $G$.  Again, by Sylow's theorems, the Sylow 5-subgroup is normal (in the subgroup, and then the whole group).

*If $K$ has order $4$, $G/K$ has order $15$, and the arguments above are basically reversed. Namely, $G/K$ has a normal Sylow 5-subgroup, we take the preimage, and get a normal subgroup of order $5$ again.

*If $K$ has order $5$, we're done.


What remains is the case $K$ has order $2$, and thus $G/K$ has order $30$.  If the Sylow 5-subgroup of $G/K$ is normal, then just like above, $G$ has a normal subgroup of order $5$ (since the Sylow 5-subgroup of a group of order $10$ is always normal).
That means all that is left is to prove a group $H$ of order $30$ always has a normal Sylow 5-subgroup.  We can proceed the same way we did above: find a minimal normal subgroup, look at the quotient, etc. The same arguments show $H$ has a normal Sylow 5-subgroup. [Thanks to @DerekHolt for mentioning this approach, which is pedagogically cleaner.]
[Original Version: This can be done by considering the action of $H$ on itself, by right multiplication.  An element of order $2$ will act as an odd permutation, which shows $H$ has a normal subgroup of order $15$.  Once more, we finish by noting that a group of order $15$ always has a normal Sylow 5-subgroup.]
Here's a second approach, which is more case-by-case, but also more elementary:
Assume $G$ has no normal Sylow 5-subgroup.  Then it has $6$ such subgroups, and a case-by-case analysis shows it has $10$ Sylow 3-subgroups.  Counting elements then shows $G$ has $5$ Sylow 2-subgroups.  The conjugation action on these Sylow 2-subgroups embeds $G$ in $S_5$, which means $G\cong A_5$, contradicting solvability.
