Question about 3rd Sylow theorem

Question

Say we have a group of order $12 = 2^2 \cdot 3$. Then by the third Sylow theorem, we know there is either $1$ Sylow $2$-subgroup, or $3$ Sylow $2$-subgroups. If there is $1$ Sylow $2$-subgroup, then it must have order $4$. Since $2^2$ is the order of the Sylow $2$-subgroup. But if we have $3$ Sylow $2$-subgroups, then how come that doesn't mean we have $3$ Sylow $2$-subgroups each of order $4$? If we have $3$ $2$-subgroups of order $2$, then we don't have a Sylow $2$-subgroup because a Sylow $2$-subgroup is the maximal $2$-subgroup, i.e. $2^2$. That means $2$-subgroups of order $2^1 = 2$ are not Sylow $2$-subgroups; they are just regular $2$-subgroups, doesn't it?

Answer 1

You are right; a subgroup of order $2$ in a group of order $12$ is not a Sylow $2$-subgroup, since it is not maximal. If $G$ is a group of order $12$, then its Sylow $2$-subgroups have order $4$, regardless of how many there are. For example, the alternating group $A_4$ has a unique Sylow $2$-subgroup, its derived subgroup. On the other hand, the dihedral group $D_6$ of order $12$ has $3$ Sylow $2$-subgroups, all of order $4$.