TITLE: Every group of order $150$ has a normal subgroup of order $25$
QUESTION [12 upvotes]: Let $G$ be a group of order $150$.  I must show that it has a normal subgroup of order $25$.  The hint says to show that is has a normal subgroup of order $5$ or $25$.
Now from Sylow, I know that the number $n_5$ of Sylow-$5$ subgroups (which each have $25$ elements) must be either $1$ or $6$, since $n_5$ must also divide $6$.  Now clearly if $n_5=1$ we are done, so I can assume that $n_5=6$.  My problem is I haven't figured out how to use this information (I am going for a contradiction, just based on what the problem asks me to prove).  I know by Cauchy's theorem I can get a subgroup of order $5$, but I don't see immediately if it must be normal. 
Any direction I should try to be moving?

REPLY [2 votes]: Just to state the simplest answer alone so people see it: Since $|G|=2m$ with $m$ odd, $G$ has an index $2$ subgroup $H$ of order $75$.  By Sylow, $H$ has a unique 5-Sylow $P$ of order $25$.  Since $P$ is characteristic in $H$, which is normal in $G$ (because index $2$), $P$ is normal in $G$, so it is the unique $5$-Sylow of $G$.