Check:etermine the number of Sylow $2$-subgroups and Sylow $3$-subgroups that $G$ can have. Let $G$ be a group of order $48$. By the $1$st Sylow theorem $G$ has a Sylow $2$-subgroup and a Sylow $3$-subgroup. Suppose none of these are normal. Determine the number of Sylow $2$-subgroups and Sylow $3$-subgroups that $G$ can have. Justify.
Let $G$ be a group of order $48=3 \cdot 2^4$. The number of Sylow $2$-subgroups $n_2$ divides $24$ and has the form $n_2=2k+1$ by the Sylow Theorems. Therefore $n_2=1,3$. However since we want non-normal subgroups so $n_2=3$.
The number of Sylow $3$-subgroups $n_3$ divides $16$ and has the form $n_3=3k+1$ by the Sylow Theorems. Therefore $n_3=1,4$. However since we want non-normal subgroups so $n_3=4$.

Assume that $n_3 = 16$. If $P_i, P_j \in Syl_3(G)$ and $P_i \neq P_j$, then $P_i \cap P_j = 1$. Each Sylow $3$-subgroup has $3$ elements, the identity and two others. The intersections only contain the identity, and hence from each Sylow $3$-subgroup you can count $2$ new elements.
Now $n_2 = 3$. If $Q_i, Q_j \in Syl_2(G)$ and $Q_i \neq Q_j$, then 
$$
|Q_iQ_j| = \frac{|Q_i||Q_j|}{|Q_i \cap Q_j|} = \frac{16 \cdot 16}{|Q_i \cap Q_j|} \leq 48.
$$
Now $|Q_i \cap Q_j| \ | \ |Q_i| = 16$, thus $|Q_i \cap Q_j| = 8$ or $16$. As $Q_i \neq Q_j$, then $Q_i \cap Q_j \neq Q_i$ and you can deduce that $|Q_i \cap Q_j| = 8$. In the worst case (least amount of distinct elements), you get $3 \cdot 8$ new elements from the $3$ Sylow $2$-subgroups.
In total you get $16 \cdot 2 + 3 \cdot 8 = 56 > 48$ elements, which is a contradiction. Thus $n_3 = 16$ and $n_2 = 3$ is not possible.
