The question reads as follows:
Suppose that $H$ is a subgroup of $A_6$ with index $|A_6: H| \leq 4$. By considering the cosets $H\sigma ^i$, show that any 5-cycle $\sigma$ belongs to $H$. Hence, show that $H = A_6$.
I know that through using Lagrange's theorem the order of $H$ is greater than or equal to 90, and that $A_6$ is a simple group. But I'm not sure how to use the cosets to verify containment.
As you see, two of the cosets $H\sigma^i\ (i=0,\dots,4)$ for a 5-cycle $\sigma$ must coincide since $|A_6\colon H|\leq 4$. Namely $H\sigma^i=H\sigma^j$ for some $i<j<5$ and $H\sigma^{j-i}=H$, $\sigma^{j-i}\in H$, hence $\sigma\in H$ follows.
Thus, $H$ contains all 144 5-cycles in $A_6$ and so $H$ has 36 5-Sylow subgroups. By index assumption, $H$ is of order 360, 180, 120 or 90. The latter two are not multiples of 36 so impossible. The second case is where $H$ is of index two and so normal in $A_6$, thus impossible by simplicity. Therefore $H=A_6$.