Question. Suppose $n>2$. Let $$H:=\{f\in\mathcal{S}_n: f\cdot 1=1\ \text{and}\ f\cdot2=2\}.$$ Is $H$ a subgroup of $\mathcal{S}_n$?
Attempt. $H$ describes all the permutations that leave $1$ and $2$ fixed. The order of this group is $(n-2)!$. Since this is a divisor of the order of $\mathcal{S}_n$; by Lagrange, we conclude that $H$ is a subgroup.
Satisfied with this reasoning?
Please correct me if I'm wrong, but this is my best shot.
The solution, I believe, is simple. Consider $a, b \in H$. Since both fix 1 and 2, then $ab$ fixes 1 and 2. (Closure.) Since $a$ fixes 1 and 2, $a^{-1}$ must fix 1 and 2, as well. (Inverses.) The identity fixes 1 and 2. (Identity.) That proves it's a subgroup.
Your reasoning, while more sophisticated, mistakenly reverses the statement made by Lagrange's theorem. Lagrange states that a subgroup's order must divide the order of the group. It doesn't say that there exists a subgroup of each divisor of the group's order.