Element of order 6 in A4 How do I show that A4 cannot have an element of order 6?

I'm not sure how to go about this. So, we would have to have the element either as (6 cycle)(1 cycle), (1 cycle)(6 cycle), (3 cycle)(2 cycle), and (2 cycle)(3 cycle). The first two are not possible because we only have 4 elements to per-mutate. The other two would lead it to be an odd permutation correct? How do I express this in a more formal way?

Take $\sigma \in \Sigma_4$ to be an element of order $6$. Decompose $\sigma$ into a product of cycles, remembering that the order of $\sigma$ is the l.c.m. of the order of its component cycles. Consider the possibilities and show that none of these gives rise to an element in $A_4$.
