Let $H = \{\alpha^2 : \alpha \in S_4\}$. Show that $H = A_4$.

Question

Let $H = \{\alpha^2 : \alpha \in S_4\}$. Show that $H = A_4$.

• Suppose $x \in H$. Then $x = \alpha^2$, where $\alpha \in S_4$. Since elements of $S_4$ takes on the form of 2-cycles, 3-cycles, product of two 2-cycles, and 4-cycles, we look at their individual squares. If $\alpha = (a b)$, then $\alpha ^2 = \epsilon \in A_4$ (identity). If $\alpha = (a b c)$, then $\alpha ^2 = (a c b) = (c b )(a b) \in A_4$. If $\alpha = (ab)(cd)$ or $(ac)(bd)$ or $(ad)(bc)$, then $\alpha^2 = \epsilon \in A_4$. If $\alpha = (abcd)$, then $\alpha ^2 = (ac)(bc) \in A_4$. Hence, $ H \subseteq A_4$.

Is this direction correct? Also, how can I prove the other inclusion? Do I look at each element $y \in A_4$ and try to find some element $x \in S_4$ such that $x^2 = y$? Is there another way to do this?

Answer 1

Your proof is correct. For the reverse inclusion, I will show that for each $y\in A_4$, there exists $x\in S_4$ such that $y=x^2$. This can be done similarly by considering the cycle structure of elements in $A_4$.

(i) If $y=\epsilon$, then $y=(12)^2$.
(ii) If $y=(ab)(cd)$, then $y=(acbd)^2$.
(iii) If $y=(abc)$, then $y=(acb)^2$.