I'm reading "Contemporary Abstract Algebra (Eighth Edition)," by Gallian.
This is Exercise 5.80 ibid. Answers that use only tools available in the textbook so far are preferred.
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 four transpositions. Show that $H=A_n$.
This is a proof verification question (plus $\varepsilon>0$).
My Attempt:
Well, my first observation is that $e=(12)(12)(12)(12)\in H$.
For any $\sigma\in H$, by definition, $\sigma$ is a product of a multiple of four transpositions; in particular, each multiple of four is even; so $\sigma\in A_n$. Hence $H\subseteq A_n$.
If $\alpha\in A_n$ is a product of $t$ transpositions with $t\equiv 2\pmod{4}$, then $\alpha=e\alpha=(12)(12)\alpha\in H$.
If $\beta\in A_n$ is a product of a multiple of four transpositions, then there is nothing to prove; $\beta\in H$.
Thus, since any $\tau\in A_n$ is by definition an even product of transpositions and products of a number in either class of even numbers modulo four are in $H$, we have $\tau\in H$. Hence $A_n\subseteq H$.
Hence $H=A_n$. $\square$
Another idea is to prove that $H$, as a subset of $A_n$, is also a subgroup of $A_n$. I'd use the "one step subgroup lemma" since I already have that $H$ is nonempty. (But how?) It would then be a matter of showing $\lvert H\rvert=\lvert A_n\rvert$. (But again, how?)
Please help :)