Let $G$ be a group of order $56$ and let $P$ and $Q$ be subgroups of $G$ of order $8$. Prove that $P \cap Q \neq \{e\}$.
Here is my thinking thus far:
We begin by using the Sylow Theorems. Let $|G| = 56 = 2^3 \cdot 7$. Denote $n_2, n_7$ by the number of $2$-Sylow subgroups and the number of $7$-Sylow subgroups, respectively. Then we have $n_2 | 7$ and $n_2 \equiv 1$ (mod $2$) and $n_7|8$ and $n_7 \equiv 1$ (mod $7$) $\Rightarrow n_2 = 1$ or $7$ and $n_7 = 1$ or $8$. We note that each of the $2$-Sylow subgroups has order $8$.
By a simple counting argument of the number of elements in $|G|$, it's easy to show that $n_2 = 7$ and $n_7 = 8$ is not possible. That leaves us with three possibilities — $n_2 = 1$ and $n_7 = 1$, $n_2 = 1$ and $n_7 = 8$, or $n_2 = 7$ and $n_7 = 1$.
By nature of the question, though, since $P$ and $Q$ are subgroups of $G$ of order $8$, they are telling us that $n_2 = 1$ is not desired. Thus, we are left with the case $n_2 = 7$ and $n_7 = 1$.
I'm struggling now with showing that $n_2 = 7$ and $n_7 = 1$ gives that, for two $2$-Sylow subgroups $P$ and $Q$ of $G$, $P \cap Q \neq \{e\}$. Suppose $P \cap Q = \{e\}$. Then, by a counting argument, we get $7(7) = 49$ non-identity elements in $G$ coming from the $2$-Sylow subgroups, and $6(1) = 6$ nonidentity elements in $G$ coming from the unique $7$-Sylow subgroup. This gives $49 + 6 = 55$ nonidentity elements in $G$. Plus the identity element in $G$ this gives exactly 56 elements. So, by a counting argument, I don't see the problem with $P \cap Q = \{e\}$ occuring.
How can I reach the desired contradiction?
Thanks!
Best Answer
The Sylow $7$-subgroup $N$ is normal. Therefore $G$ is a semi-direct product of $N$ with any Sylow $2$-subgroup $P$. Your counting argument also gives that all elements of $G$ have orders $1$, $7$ or $2^k$. In particular, $G$ has no element of order $14$. But $P$ acts on $N$ by conjugation. The order of $\text{Aut}(N)$ is $6$ and that of $P$ is $8$. So the kernel of this action (which is a homomorphism $P\to\text{Aut}(N)$) is nontrivial, so there's an element $g$ of order $2$ acting trivially on $N$, which means that it commutes with the generator $h$ of $N$. Then $gh$ has order $14$.