In $S_{10}$ let us consider the permutation $\sigma = (123)(456)$ and the cyclic subgroup $ G = \langle \sigma \rangle$.
I can fairly easily find that $$\langle \alpha = (1,4,2,5,3,6)(7,8) \rangle \neq G$$ is a cyclic subgroup of the alternating group $A_{10}$ such that $\sigma \in \langle \alpha \rangle$ (and hence contains $\langle \sigma \rangle$). Indeed, $\alpha^2 = \sigma$.
However, I'd like to ask:
How does one find an abelian non-cyclic proper subgroup of $A_{10}$ that contains $\langle \sigma \rangle$?
How does one find a non-abelian (hence, obviously, non-cyclic) proper subgroup of $S_{10}$ that contains $\langle \sigma \rangle$?
For 1, think about following property of cyclic group: for every divisor of order of a cyclic group, there is unique subgroup of that order.
So if you want to find a non-cyclic group in $A_{10}$ containing $\sigma=(123)(456)$, then, take a commuting element with it of order $3$, but not a power of $\sigma$ (so that it gives one more cyclic subgroup of order $3$); a simplest choice is $(123)$ (since disjoint cycles commute). So $(123)$ and $(123)(456)$ for a non-cyclic (but abelian) group in $A_{10}$ since it violates a property of cyclic group mentioned in beginning.
For non-abelian group, we want to find an element non-commuting with $(123)(456)$ in $A_{10}$. We know $(12)$ do not commute with $(123)$ and $(45)$ do not commute with $(456)$. It is then easy to gess an element non-commuting with $(123)(456)$: it is $\tau=(12)(45)$. This is also in $A_{10}$.
Consider subgroup generated by $\sigma$ and $\tau$. We want to ensure that this subgroup is proper. But it is easy: they form the diagonal subgroup of $S_3\times S_3$ where first $S_3$ is permutation group on $\{1,2,3\}$ while second on $\{4,5,6\}$.
[If $G$ is any group, then in $G\times G$, $\{(h,h)\colon h\in G\}$ is a proper subgroup, called diagonal.]