Let $G = A_5$. I want to show that $ S = \langle(1,2)(3,4), (1,2,3,4,5)\rangle = A_5$, where $A_5$ is the alternating group.
My thoughts are that $A_5$ is simple, and if I can show that $S$ is a normal subgroup of $A_5$ then I am done. I have shown that $S$ is a subgroup of $A_5$, but I can't show it's normal.
Is this the correct way of tackling the question or is there a better way?
Clearly $|S|$ is a mutliple of $10$, so it must be $10,20,30$ or $60$.
Conjugating the second generator by the first does not give a power of the second generator, so $S$ does not have a unique Sylow $5$-subgroup, so $|S|$ is not equal to $10$ or $20$. Since $A_5$ is simple, it has no subgroup of order $30$, so $|S|=60$, and $S=A_5$.