# Prove that $\langle (1,2)(3,4), (1,2,3,4,5)\rangle = A_5$ [duplicate]

Let $G = A_5.$ Prove that $\langle (1,2)(3,4), (1,2,3,4,5)\rangle = G$ . State clearly any properties of $A_5$ you use.

So I know $A_5 = \{\sigma | \sigma \in S_5$ and $\sigma$ even permutation$\}$.

I'm trying to expand $\langle (1,2)(3,4), (1,2,3,4,5)\rangle$ to show that it equals $A_5$ but I'm not getting the answer.

$(1,2)(3,4)(1,2)(3,4) = (1)$

$(1,2)(3,4)(1,2,3,4,5) = (1,3,5)$ But this is odd, so it wouldn't be in $A_5$.

How do I approach the question correctly?

• If you want to expand it out, it would simplify the work if you consider the orders of each generator. Also, your second equality is incorrect. Jan 21, 2016 at 17:09
• @Maburo The correctness of the second equality depends on the exact definition of the product of cycles. If you think of them as acting from the right it is correct. However, while $(1,3,5)$ might have odd length it is an even permutation. Jan 21, 2016 at 17:11
• @Jorik Oh yeah, thank you! Because $(1,3,5) = (1,5)(5,3)$ Jan 21, 2016 at 17:14
• Let $H = \langle (1,2)(3,4), (1,2,3,4,5) \rangle$. You have found elements of order $2,3$ and $5$ in $H$, so you know that $|H|$ is at least $30$. But you still need to prove that $|H|=30$ is impossible. Jan 21, 2016 at 17:21
• @DerekHolt I understand that $|H|$ must be a multiple of $30$ so $|H| = 30$ or $60.$ I just don't understand how to show $|H| = 30$ is impossible. Jan 21, 2016 at 17:54

Show directly that each combination of two neighbor-transpositions can be made by multiplying together conjugates of $(12)(34)$ by powers of $(12345)$.
For example, $(12)(34)\,(34)(51)=(12)(51)$ and $(12)(51)\,(51)(23)=(12)(23)$, and so forth.
And now you'll have shown that $(12)(34)$ and $(123\cdots n)$ generate $A_n$ for every odd $n\ge 5$.