Prove or disprove $A_5$ has a subgroup that isomorphic to $\mathbb{Z}_6$

Question

I need to prove or disprove:

To the group $A_5$ has a subgroup that isomorphic to $\mathbb{Z}_6$

My attempt:

I just wrote all the details that I know:

element in $A_5$ should be in form like $\sigma:=\left( \begin{array}{ccc} 1&2&3&4&5\\?&?&?&?&?\end{array} \right)$

such that sgn$(\sigma)=+1$ ,ord$(A_5)=\frac{5!}{2}=60$

Acording to Lagrangh theorem ord(subgroup of $A_5$) $\Big|_{60}$

and $\mathbb{Z}_6=0,1,2,3,4,5$

Now I need to find $\varphi:A_5\rightarrow \mathbb{Z}_6$ such that $\varphi$ should be isomorphism

Is it correct so far? any hints please? how can I proceed?

the answers in this link: Does A5 have a subgroup of order 6? didn't helped me

Answer 1

If it did then $A_5$ would have an element of order $6$, this element would need to have a $3$-cycle and at least two cycles of even length (at least one so the order is even, but at least two because we are in $A_5$), which is impossible in $A_5$.

Answer 2

@dREaM was faster than me, I shall try to detail a bit his answer. Let $\sigma \in A_5$. Decompose it into a product of disjoint cycles, $\sigma = c_1 c_2 \dots c_n$ (this decomposition is not unique, but this does not bother us). Let the order of cycle $c_i$ (equal to its length, i.e. the number of elements that it does not fix) be $o_i$. Since these cycles are disjoint, they will commute with each other, so the order of $\sigma$ will be $6 = \operatorname{lcm} (o_1, \dots, o_n)$. Since each $o_i$ must divide the $\operatorname{lcm}$, there are only two possibilities: either we have only $2$ cycles with orders $2$ and respectively $3$, or a single one of order $6$. In the first case, since one of our cycles has length $2$, it must be a transposition - but transpositions are odd, so this case is impossible. The other posibility is to have a single cycle of length $6$, but this is again impossible when there are only $5$ objects to permute.