Show that $H=A_n$

Question

For $n>1$ ;

Let $H$ be the set of all permutations in $S_n$ that can be expressed as a product of a multiple of $4$ transpositions .Show that $H=A_n$

ATTEMPT:

We know that $A_n$ is generated by $3-$ cycles .

Now $H\subset A_4$ since every element of $H$ is a product of even number of transpositions.

Now take any $\alpha\in A_n$ then $\alpha$ is a $3$-cycle or a product of $3$ cycles.

If $\alpha$ is a product of $3$ cycles of the form $\alpha=(x_1x_2x_3)\ldots (y_1y_2y_3)$ then $\alpha=(x_1x_3)(x_2x_3)\ldots (y_1y_3)(y_1y_2)\in H$.

Problem:

If $\alpha$ is a $3-$ cycle then $\alpha=(x_1x_2x_3)$

If $\alpha\in A_3$ and if we suppose $\alpha=(123)$ in $A_3$ then how can we write $\alpha$ as a product of $4$ $2-$ cycles?

Please help.

Answer 1

Since $n=2, 3$ are allowed, I presume there is no requirement that the transpositions should be distinct?

For large enough $n$, if distinctness is required within each four, and for any $n$ if it is not, when you multiply $(a b)(c d)(e f)(g h)$ by $(g h)(e f)(c d)(x y)$ you get $(a b)(x y)$ - an arbitrary product of two transpositions.

Without distinctness you can also write $(a b)(x y)=(a b)(c d)(c d)(x y)$ for an arbitrary $(c d)$