Let $|G|=735$. If the number of Sylow $7$-subgroups are more than $1$, then show that there exists a normal Sylow $5$-subgroup.

Let $|G|=735$. If the number of Sylow $7$-subgroups are more than $1$, then show that there exists a normal Sylow $5$-subgroup.

To solve this problem we need to show that if there are more than 1 sylow 7-subgroup then there is only one sylow-5 subgroup.

I tried by assuming that G has more than 1 sylow 7-subgroup and sylow 5-subgroup and I hoped that by assuming this I will get no. of elements will be more than 735 , it didn't happen.

Solution 1:

By Sylow's theorem 3, if there is more than one Sylow $7$-subgroup, then there must be $15$, as it is the only factor of $15$ which is congruent to $1$ modulo $7$, other than $1$.

$G$ acts transitively on the Sylow $7$-subgroups by conjugation (from Sylow's theorem 2). By the orbit-stabiliser theorem, the normalizer of a Sylow $7$-subgroup has size $735/15=49$, so it is just the Sylow $7$-subgroup itself. As groups of order $49$ are abelian, we can conclude that the Sylow $7$-subgroup is central in its normalizer.

We may then apply Burnside's normal p-complement theorem, to conclude that $G$ contains a normal subgroup of order $15$.

Again using Sylow's theorem 3 we know that this group of order $15$ has a normal subgroup of order $3$ and a normal subgroup of order $5$, so it is cyclic. In particular its subgroup of order $5$ is characteristic, hence normal in $G$.

Solution 2:

I hope that tkf will undelete their answer in due time, for the argument in there looks more generalizable than anything I can offer. Promoting my comment to an answer anyway.

Assuming that there is more than one Sylow $7$-subgroup then there will be exactly fifteen of them.

Case 1. There are two Sylow $7$-subgroups $P_1$ and $P_2$ such that they intersect non-trivially.

In this case $H=P_1\cap P_2$ is cyclic of order seven. It follows that in this case $H$ is contained in the center $Z(G)$. To see that let's consider the centralizer $C_G(H)$. The groups $P_1$ and $P_2$ are abelian by virtue of having order $p^2, p=7$. Therefore $P_1,P_2\le C_G(H)$. This implies that $7^2\mid |C_G(H)|$. So $P_1$ and $P_2$ are both Sylow $7$-subgroups of $C_G(H)$. Applying the initial reasoning to $C_G(H)$ instead of $G$ shows that $C_G(H)$ must also have $15$ Sylow $7$-subgroups. Hence $15\mid |C_G(H)|$ and we can conclude that $C_G(H)=G$.

So $G/H$ is a group of order $105$ that has $15$ Sylow $7$-subgroups (correspondence). This is actually impossible. But alternatively you can use a simple counting argument to conclude that $G/H$ must have a unique Sylow $5$-subgroup $Q$. It corresponds to a normal subgroup $N$ of order $35$ of $G$. Such a group is easily seen to be cyclic, hence its subgroup of order five is characteristic in $N$, hence normal in $G$.

Case 2. All the Sylow $7$-subgroups intersect trivially.

In this case the union of Sylow $7$-subgroups has $48\cdot15=720$ non-identity elements, leaving only $15$ elements of orders coprime to $7$. I'm sure you can manage this case.

We actually did not need to prove in the first case that $H$ is central. For the rest of the argument to work as described, it would suffice to show that $H\unlhd G$, and we could have used $N_G(H)$ instead of $C_G(H)$.