# Subgroup generated by $S$ is $A_5$

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?

• There is no "correct" way to approach a question like this. The standard way, however, is to first use Lagrange's theorem as much as you can to limit the possible orders (because it's the easest thing to do), and then improvise from there. Jan 20, 2018 at 22:01

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 $S=A_5$.
• How does that imply that $S$ doesn't have a unique Sylow $5$ subgroup? Then how does that imply that $|S|$ isn't equal to $10$ or $20$? Jan 20, 2018 at 22:33
• It implies it because it proves that there are two distinct Sylow $5$-subgroups by constructing them explicitly. And the second question follows from Sylow's theorem. Jan 21, 2018 at 11:43