About a generating set and orders of elements in $A_5$ It is well known that $A_5$ has presentation $$\langle a,b: a^2=1,b^3=1,(ab)^5=1\rangle.$$

Suppose in $A_5$, $x$ is an element of order $2$ and $y$ an element of order $3$ such that $\langle x,y\rangle=A_5$. Is it necessary that order of $xy$ is $5$? 

I feel it should happen, but I am unable to justify it.
 A: You can prove this by brute force by just looking at what permutations $x$ and $y$ can be.  Since $y$ must be a $3$-cycle, we may assume without loss of generality that $y=(1\ 2\ 3)$.  Now $x$ must be a pair of $2$-cycles.  If $x$ does not involve $4$ (or $5$), then every element of the subgroup generated by $x$ and $y$ fixes $4$ (or $5$).  So $x$ must involve $4$ or $5$, and furthermore $(4\ 5)$ cannot be one of its cycles since then the set $\{4,5\}$ would be invariant under the subgroup generated by $x$ and $y$.  So $x$ has two $2$-cycles, each of which contains one of $4$ and $5$.  Without loss of generality $x=(1\ 4)(2\ 5)$ (any other possibility can be obtained by cyclically permuting $\{1,2,3\}$ and/or swapping $4$ and $5$).  Now you can just compute that $xy$ has order $5$ directly.
(In fact, this argument shows that any two pairs of such elements $(x,y)$ are conjugate by an element of $S_5$.)
A: The only other possible orders for $xy$ are $2$ and $3$, but the groups defined by the presentations $\langle x,y \mid x^2=y^3=(xy)^n=1 \rangle$ with $n=2$ and $3$ have order $6$ and $12$ (the groups are $D_6$ and $A_4$), so $x$ and $y$ could not generate $G$.
