In $A_4 \times S_4$, is it possible to have three $2$-Sylow subgroups?
We know that $|A_4 \times S_4|=2^5 \cdot 3^2$ ; |$2$-Sylow| $= 32$ and |$3$-Sylow| $= 9$.
I found that the number of $3$-Sylow subgroups is $16$, but the number of $2$-Sylow subgroups is $3$ ; makes it $(2^5) - 1=31$ & $31 \cdot 3=93$ but $93+(9×16=144) =237$ which isn't $288$!…