Prove that if $H$ is a normal subgroup in $S_n$, and $H$ contains a $2$ cycle, then $H = S_n.$
My attempt:
I have proved in a previous part of the question that given any two arbitrary $k$-cycles, there exists some permutation $\tau$ in $S_n$ such that the $k$ cycles are conjugate.
By this, any $2$-cycle in $S_n$ must be conjugate to our given $2$-cycle. Thus $H$ must contain every $2$-cycle in $S_n,$ by the definition of a Normal Subgroup.
But we know that any permutation in $S_n$ can be written as a product of transpositions. Therefore the $2$-cycles generate $S_n.$
As $H$ is a subgroup, it contains all compositions of its elements, as I is closed under the binary operation. It contains all compositions of all $2$-cycles in $S_n,$ thus $H$ must be equal to $S_n,$ as the $2$-cycles generate $S_n.$
Is my reasoning above correct?