If $G$ is a group of order $48$, show that the intersection of any two distinct Sylow $2$-subgroups has order $8$

Question

All I know is that we have $3$ Sylow-$2$ subgroups of order $16$. $$o(H \cap K)= o(H)o(K)/o(HK)$$

How to proceed further?

Answer 1

I will elaborate the hint given and the solution indicated by @Derek Holt

Denote by $o(M)$ the cardinality (order) of the subset $M$. For two subgroups $H$ and $K$ define $HK$ to be set of all the products $h \cdot k$ where $h \in H$ and $k \in K$. We have the surjective map $$H \times K \to HK\\ (h,k) \mapsto h\cdot k $$ Two pairs $(h,k)$ and $(h',k')$ have the same product if and only if $(h',k') = (h\cdot l, l^{-1} \cdot k)$ for some $l$. That $l$ must be in both $H$ and $K$ so in $H\cap K$. Therefore the fibers of the map $(h,k) \mapsto h\cdot k$ have cardinality $o(H\cap K)$. We conclude that $$o(H)\times o( K) = o(H\cap K) \cdot o(H\cdot K)$$ analogous to the formula for the dimensions of sums and intersection of vector spaces ($\dim(\cdot) \simeq \log o(\cdot)\ $ ) and so $$o (H\cap K ) = \frac{o(H)\cdot o(K)}{o(H \cdot K)} \ge \frac{o(H)\cdot o(K)}{o(G)}$$

In our situation $o(G) = 48 = 2^4 \cdot 3$, $\ o(H)= o(K) = 2^4$ so $o(H\cap K) \ge 2^4/3> 2^4/4=2^2$. However, by Lagrance $\ o(H\cap K) \, |\, 2^4$. Therefore $\ o(H\cap K)= 2^3$ or $2^4$. Since the subgroups $H$, $K$ are distinct $\ o(H\cap K) < o(H) = o(K)$ so $\ o(H\cap K)= 2^3=8$.