Two 3-cycles generate $A_5$ I want to solve the following exercise, from Dummit & Foote's Abstract Algebra

Let $x$ and $y$ be distinct 3-cycles in $S_5$ with $x \neq y^{-1} $.
  Prove that if $x$ and $y$ do not fix a common element of $\{1,2,3,4,5\}$ then $\langle x,y\rangle =A_5$.

I know that using brute force one can find $31$ distinct permutations in $\langle x,y \rangle$, and then Lagrange's Theorem finishes the proof. 
However, is there a more elegant way to go about this? (Preferably without using the fact that $A_5$ is simple, since it wasn't proven up to this point in the text.)
Thank you!
 A: Let $H = \langle x,y \rangle \leq A_{5}.$ Then $H$ has more than one Sylow $3$-subgroup, so has either $4$ or $10$ Sylow $3$-subgroups. If $H$ has $10$ Sylow $3$-subgroups, then $|H|$ is divisible by $30.$ If $|H| = 60,$ we are done. If $|H| = 30,$ then $H \lhd A_{5},$ and furthermore, since $H$ already contains $20$-elements of order $3,$ it can't have $6$ Sylow $5$-subgroups. Hence $H$ has a unique Sylow $5$-subgroup $S$ so $S^{g} = S$ for each $g \in A_{5}$ (since $S^{g} \leq H).$  But then $S \lhd A_{5},$ and $A_{5}$ has a unique Sylow $5$-subgroup, a contradiction. 
Hence we may now suppose that $H$ has $4$ Sylow $3$-subgroups. Hence $12$ divides $|H|.$ Suppose that $H \neq A_{5}.$ Then $|H| = 12.$ Now $H$ contains $8$ elements of order $3,$ so $H$ has a normal Sylow $2$-subgroup. Now the action of $A_{5}$ on $5$-points may be realised by (or, if you prefer, identified with)  its conjugation action on its five Sylow $2$-subgroups. Now $H$ is the stabilizer of one of these Sylow $2$-subgroups in that action, and $x$ and $y$ have a common fixed point.
A: Hint: $y(1,2,3)y^{-1} = (y(1),y(2),y(3))
$
solution:
You can assume that $x=(1,2,3)$. As $x$ and $y$ do not fix either 4 or 5, you can assume that 
$y=(3,4,5)$.
Now
$$
y(1,2,3)y^{-1} = (y(1),y(2),y(3)) = (1,2,4)
$$
If you replace $y$ with $y^2$, you get that $\langle x,y\rangle$ contains
$(1,2,3),(1,2,4),(1,2,5)$.
Do the same switching $x$ and $y$ you get $(3,4,5),(1,4,5),(2,4,5)$.
To get the remaining 3 cycles:
$$
x(1,2,4)x^ {-1} = (2,3,4)\\
x(2,3,4)x^ {-1} = (3,1,4)\\
x(1,2,5)x^ {-1} = (2,3,5)\\
x(2,3,5)x^ {-1} = (3,1,5)\\
$$
You have proved that 
$\langle x,y\rangle$ contains
$\{(1,2,3),(1,2,4),(1,2,5),(3,4,5),(1,4,5),(2,4,5),
(2,3,4),(3,1,4),(2,3,5),(3,1,5)\}$ and the powers of these elements are all the 3 cycles. Then $\langle x,y\rangle\supseteq A_5$ and eventually $\langle x,y\rangle=A_5$. 

To fit OP's definition of alternate permutations: as
$
(1,2)(2,3) = (1,2,3)$ every product of an odd number of transpositions is a product of 3-cycles.
A: Yet another answer. We can assume $x=(1,2,3)$, $y=(3,4,5)$. Let $H=\langle x,y \rangle$. Now $z := x^y = (1,2,4) \in H$. Clearly $H$ is transitive on $\{1,2,3,4,5\}$, $\langle x,z \rangle \le H_5$ is transitive on $\{1,2,3,4\}$ and $\langle x \rangle \le H_{4,5}$ is transitive on $\{1,2,3\}$, so by the Orbit-Stabilizer Theorem, $|H| \ge 5 \times 4 \times 3 = 60$.
A: Just as an another approach, we can assume that the $3$-cycles are $(123)$ and $(345)$ as in the first answer. $(345)(123)(345)^2 = (124)$. Hence by the previous exercise of the same book, $\langle (123)(124)\rangle$ generates the whole $A_4$. Since $|A_5:A_4| = 5$ and $5$ is a prime, $A_4$ is a maximal subgroup of $A_5$, thus $\langle A_4, (345)\rangle = A_5$.
Maximality follows from the fact that, if $A_4\leq L \leq A_5$, then $|A_5:A_4| = p = |A_5: L|\cdot |L:A_4|$ forcing $L$ to be either $A_4$ or $A_5$.
