Prove that the conjugacy classes in $\displaystyle A_5$ have sizes 1,12 ,12, 15 and 20..

so to prove the conjugacy class of size 1, its trivial, its just the identity.

to prove the conjugacy class of size 20, i used a lemma saying... All 3 cycles in $\displaystyle A_5$ are conjugate to each other,

$\displaystyle x=(1 2 3) \in A_5

$

$\displaystyle g \in G_x \implies gxg^{-1}=x \implies gx=xg$

$\displaystyle (1),(1 2 3),(1 3 2) \implies |G_x| = 3$

$\displaystyle |G| = 60$

$\displaystyle

|C_g(x)| = 60/3 = 20$

to prove the conjugacy class of 15 i used

$\displaystyle x =(12)(34) \in A_5

$

there are $\displaystyle \binom{5}{2}$ ways to do the first cycle, of x and there are $\displaystyle \binom{3}{2}$ to do the second cycle in x, and to eliminate similar cycle structures we divide by 2.

$\displaystyle \frac{\binom{5}{2}\binom{3}{2}}{2} = \frac{30}{2}=15$

Not really sure if this actually proves the 15, and i know there are 24 elements left of order 5... (process of elimination, how do i prove they have to be 12 + 12...)??