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?
Best Answer
As you say, by Sylow's theorem, either $n_5=1$ or $n_5=6$. If $n_5=1$, we have found a normal subgroup of $25$, so we are done. So let's assume that $n_5=6$. We will exhibit a normal subgroup of order $5$ as follows. Let $P$ and $Q$ be distinct Sylow $5$-subgroups. Using the fact that $|PQ| = \frac{|P|\cdot |Q|}{|P\cap Q|}$, it is easy to see that $|P\cap Q| = 1$ is impossible! So we conclude that the subgroup $T=P\cap Q$ has order $5$.
We claim that $T$ is normal in $G$. It is a fact if $H$ is a proper subgroup of a $p$-group $L$, then $H$ is properly contained in its normalizer $N_{L}(H)$ (see here for proof). Since $P$ and $Q$ are $p$-groups, we can apply this result. Hence, $T\subsetneq N_{P}(T)$ and $T\subsetneq N_{Q}( T)$. So $N_{P}(T) = P$ and $N_{Q}(T)=Q$. This immediately gives $P\subseteq N_{G}(T)$ and $Q\subseteq N_{G}(T)$, which implies $PQ \subseteq N_{G}(T)$. Again using the formula above, $PQ=\{p \cdot q: p\in P, q\in Q\}$ (as a subset) has cardinality $125$. By order considerations, we get that $N_{G}(T)=150$, so $N_{G}(T)=G$ and $T$ is a normal subgroup of order $5$.
This proves the hint. Now let's finally show that $G$ has a normal subgroup of order $25$. So let $T$ be the normal subgroup of order $5$ constructed above. Then the quotient group $G/T$ has order $150/5=30$. Now every group of order $30$ contains a normal Sylow $5$-subgroup (see this MSE thread). Let $H$ be a normal subgroup of order $5$ in $G/T$, and consider the canonical map $\pi: G\to G/T$. Then $\pi^{-1}(H)$ is normal subgroup of $G$. By correspondence theorem, $[G: \pi^{-1}(H)]=[G/T: H]=6$, so $\pi^{-1}(H)$ is the desired normal subgroup of index $6$, i.e. desired subgroup of order $25$.
(This shows that the case $n_5=6$ is actually impossible, and there is only one subgroup of order $25$, which is the $5$-Sylow subgroup)