Any group of order $12$ must contain a normal Sylow subgroup This is part of a question from Hungerford's section on Sylow theorems, which is to show that any group with order 12, 28, 56, or 200 has a normal Sylow subgroup. I am just trying the case for $|G| = 12$ first.
I have read already that one can't conclude in general that $G$ will have a normal Sylow 2-subgroup or a normal Sylow 3-subgroup, so I am a bit confused on how to prove this. Here is my start, but it's not that far. Let $n_2, n_3$ denote the number of Sylow $2$- and $3$-subgroups, respectively. Then:
$n_2 \mid 3$ and $n_2 \equiv 1 (\operatorname{mod } 2)$, so $n_2 = 1$ or $3$. Similarly, $n_3 \mid 4$ and $n_3 \equiv 1 (\operatorname{mod } 3)$, so $n_3 = 1$ or $4$.
I was thinking to assume that $n_2 \ne 1$, and prove that in this case, $n_3$ must equal $1$. But I don't see how to do this. One could also do it the other way: if $n_3 \ne 1$, then show $n_2 = 1$.

Hint: If $n_3 = 4$, then how many elements of order 3 are there in the group? How many elements does that leave for your groups of order 4?
