Is there a more elegant way of proving $\langle (1,2)(3,4), (1,2,3,4,5) \rangle = A_5$

I'm trying to show the following

$\langle (1,2)(3,4), (1,2,3,4,5) \rangle = A_5$

I managed to prove this but I think my solution is very inelegant. Here's my argument

let $J = \langle (1,2)(3,4), (1,2,3,4,5) \rangle,$ then $\lvert J \rvert \geq 8$ and $\lvert J \rvert$ divides $\lvert A_5 \rvert = 60,$ so the possibilities are

$\lvert J \rvert = 10, 12, 15, 20, 30, 60$.

Then I sat down and calculated 13 more elements of $J$, so now the possibilities are $\lvert J \rvert = 30$ or $60$. But we can't have $\lvert J \rvert = 30$ because then $J$ would be normal (index 2 theorem) which would contradict $A_5$ being simple, so we must have $J = A_5$.

The calculation part make this proof quite long winded, is there a simpler way of getting the result?

Solution 1:

As a more elementary approach $H=<(1 2)(3 4),(1 2 3 4 5)>$ contains an element of order 2, and element of order 5 and an element of order 3 (the product of the two generators).