# conjugacy classes

• Nov 5th 2009, 07:16 AM
ux0
conjugacy classes
Prove that the conjugacy classes in $A_5$ 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 $A_5$ are conjugate to each other,

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

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

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

$|G| = 60$

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

to prove the conjugacy class of 15 i used

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

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

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

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...)??
• Nov 16th 2009, 11:25 AM
johnt4335
They are 12 + 12 because . . .