Suppose that $G$ is a group of order $48.$ Show that the intersection of any two distinct Sylow $2$-subgroups of G has order $8.$
Let $H,K$ be two distinct Sylow $2$-subgroups of $G.$ Then $$O(H\cap K)=\dfrac{O(H)O(K)}{O(HK)}=\frac{2^4\times2^4}{O(HK)}$$ where $O(HK)$ simply means number of elements in the set $HK.$
Since any two Sylow $2$-subgroups are conjugate to each other $K=gHg^{-1}$ for some $g\in G.$
Also $H,K$ being distinct subgroups of $G,$ $g\notin H.$ Thus $$O(H\cap K)=\dfrac{O(H)O(K)}{O(HK)}=\frac{16\times16}{O(gHg^{-1}.K)}$$
Since $O(H\cap K)|2^4$ and $H\ne K$ the possible values of $O(H\cap K)$ are $1,2,4,8.$ I don't know how to reject the possibilities of $1,2,4.$
Please help.
Best Answer
I think this outline should work : Assuming that there isn't a unique $2-$Sylow subgroup, there have to be 3. Let $G$ act on the set of $2-$Sylow subgroups by conjugation. This gives a homomorphism $$ \varphi : G\to S_3 $$ It shouldn't be too hard to check that this map is surjective, and so $\ker(\varphi)$ has order 8.
Check that $\ker(\varphi)$ is contained in any $2-$Sylow subgroup (since it is contained in one 2-sylow subgroup).
Hence, $\ker(\varphi)$ is contained in $H\cap K$, and so $|H\cap K| \geq 8$.
By what you have, $|H\cap K| \leq 8$, which proves your result.