A problem about the alternating group $A_4$ How can I prove that $A_4=\{\sigma^2:\sigma\in S_4\}$?
My approach is the following: If $T:S_4\to S_4$ is defined by $T(\sigma)=\sigma^2$, then we have to prove that $T(S_4)=A_4$.
Since $sgn(\sigma^2)=(sgn (\sigma))^2 $, we have that $T(S_4)\subset A_4$. 
On the other hand $A_4=\langle(12)(34),(123)\rangle$. Is it enough to show that these generators are in $T(S_4)$? Do I have to prove first that $T(S_4)$ is a subgroup of $S_4$ if I want to use this approach?
Is there any simpler or intuitive approach to solve this problem? 
Is it true that   $sgn(\sigma^2)=(sgn (\sigma))^2 $ for every $\sigma$ in $S_n$?
Can we generalize this result to S_n for any $n$? Is true that $A_n=\{\sigma^2:\sigma\in S_n\}$?
 A: All elements  of $A_4$ are of the form $(a\,b)(c\,d)=(a\,c\,b\,d)^2$ or $(a\,b\,c)=(a\,c\,b)^2$ or $()=()^2$.
A: The answer to the generalised question is no. For instance, $(1234)(56) \in \mathfrak A(6)$ is not a square.
To see this, you have to notice that when you square a permutation which is written as a product of disjoint cycles $\sigma = c_1 \ldots c_p$, the square is simply $\sigma^2 = c_1^2 \ldots c_p^2$. Now, it is not hard to figure out what the square of a $q$-cycle is (a $q$-cycle if $q$ is odd, the product of two disjoint $q/2$-cycles if $q$ is even), so you can check that no square of an element of $\mathfrak S(6)$ has the shape $(*\,*\,*\,*)(*\,*)$.
Basically, this discussion gives a criterion to decide if an element $\sigma$ of $\mathfrak S(n)$ is a square: it is necessary and sufficient that for all even numbers $q$, the number of $q$-cycles in the decomposition in disjoint cycles of $\sigma$ is itself even. It clearly implies that it's an even permutation (we already knew that...) but as soon as $n \geq 6$, it's a stronger condition.
A: Yes, $sgn(\sigma^2) = (sgn\, \sigma)^2$ for all $\sigma\in S_n$. The right-hand side is always $1$, so all we have to do is show that the LHS is also $1$. But if we write $\sigma$ as a product of transpositions, then $\sigma^2$ is that product concatenated with itself; clearly there are an even number of transpositions there, so $\sigma^2$ is an even permutation.
