Let $G=S_4\times Q_8$. Then show that Sylow 2 subgroup is normal.

My Attempt: $|G|=192=2^6\cdot 3$

Order of Sylow 3 sugroup is 3. Number of elements of order 3 in $S_4=8$.

so there atleast 4 syolw 3 subgroups are there, Hence sylow 3 subgroup is not unique and not normal. Now order of sylow 2 subgroup is $2^6$. Number of sylow 2 subgroup is of the form $1+2k,k=1,2,3,...$ that should divide order of G. number of sylow 2 subgroup is either 1 or 3. Now if there are 3 sylow 2 subgroup then $64\times 3-3$ will be there in Sylow 2 subgroups. then sylow 3 will be unique. which is not possible. but i am not sure that intersection of 2 sylow 2 subgroups is only identity. Please help me . Thanks in advance.


This is plainly false, isn't? If it were true, then the same would be true in any quotient group of $G$. But $S_{4}$, which is isomorphic to the quotient of $G$ by $\{ 1 \} \times Q_{8}$, has three $2$-Sylow subgroups


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