Show that each permutation in $A_4$ has a square root.

Question

Show that $ A_4 = \{ \sigma \in S_4 \mid \sigma = \tau^2 \text{ for some } \tau \in S_4 \} $.

$S_4$ is the permutations of 1,2,3,4. $A_4$ is the alternating group of $S_4$.

Let $ B = \{ \sigma \in S_4 \mid \sigma = \tau^2 \text{ for some } \tau \in S_4 \}$. It is easy to see that $B\subseteq A_4$.

I am having difficulty showing that the cycles of length 3 (e.g. (1,2,3)) have a square root $\tau$.

Below is my work for the other cycles.

Now we note that $A_4 = \{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3), (1,2,3), (1,3,2), (1,3,4), (1,4,3), \\(1,2,4), (1,4,2), (2,3,4), (2,4,3) \}$. Thus $ A_4 $ has three different types of permutations $ e, $ those of length 3, and products of disjoint transpositions. First we note that $ () = e \in B$ since it can be expressed as $ (1,2)(1,2) $.\ Next, for all the permutations consisting of a product of disjoint cycles $ (a_1,a_2)(a_3,a_4) $ we can express them as $ (a_1,a_3,a_2,a_4)^2 $. It remains to be shown that the length-3 cycles have square roots in $S_4$.

Answer 1

So here is 3-cycles as square of three cycles $(abc)=(acb)(acb)$. That's cool.

Answer 2

A $3$-cycle $\sigma$ is order $3$, hence $(\sigma^2)^2=\sigma^4=\sigma^{1}.$ Thus, the square root of $\sigma$ is $\sigma^2$.