Let $p$ be a prime number. I've to figure out for which $p$ $$G_p = \{\sigma \in S_{12}: \text{ord}(\sigma) \text{ divides } p \}$$ is a subgroup of $S_{12}$.
I realize that we have to consider $p \le 11$ and I can find an example to show that it is not the case for $p = 2$: in fact $(1,2)\circ(2,3) \not\in G_2$. But how does one solve such a problem?
What happens if we consider $G_n$ if $n$ is non-prime?
Suppose $n$ is an integer at least $2$. If $G_n$ is a subgroup of $S_{12}$, then it is a normal subgroup. In particular, it must be either $A_{12}$ or $S_{12}$.
We cannot have $G_n=A_{12}$, because $n$ would have to be even and thus $G_n$ would contain a transposition. On the other hand, we have $G_n=S_{12}$ if and only if $n$ is a multiple of the exponent of $S_{12}$, namely $27720=2^3\cdot 3^2\cdot 5\cdot 7\cdot 11$.