Order of element of $A_5$

Question

I need to show there is no injective homomorphism $f : D_6 \hookrightarrow A_5$. I know that there is an element of $D_6$ with order 2 (namely $\sigma$, a reflection in some line). If $f$ is a homomorphism, we obtain $f(\sigma)^2 = f(\sigma^2) = f(e_{D_6}) = e_{A_5}$. Hence $f(\sigma)$ is of order 1 or 2. If it is of order 1, $\sigma$ would be an element of ker$(f)$, which would in turn be nontrivial. Therefore, if we demand $f$ to be injective, the order of $f(\sigma)$ must be 2. Now, if I would be able to prove that there is no element of order 2 in $A_5$, I would be done. My gut tells me that is indeed the case, since $A_5$ is generated by the 3-cycles. The proof however, eludes me.

Answer 1

$D_6$ has an element of order $6$, but the only elements $x\in S_5$ of order $6$ have disjoint cycle representation of the form $x=(a\;b\;c)(d\;e)$ which is the product of the even element $(a\;b\;c)$, and the odd element $(d\;e)$, so $x$ is odd, hence $x\notin A_5$.

Note that $|A_5|=60$, hence $A_5$ does have elements of order $2$, since any group of even order has an element of order $2$. Explicitly, if $x,y\in S_5$ are disjoint $2$-cycles, and $z=xy$, then $z$ has order $2$, and $z$ is even, so $z\in A_5$.