# Conjugacy class and normal abelian subgroup

• Nov 15th 2009, 06:53 PM
apple2009
Conjugacy class and normal abelian subgroup
Consider the group S₄.
a)Prove that {(1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}is an S₄ - conjugacy class.
b)Let V:={(1), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}. Prove that V is a normal abelian subgroup of S₄ isomorphic to the Klein 4-subroup V₄
c)Let H= {(1), (1, 2, 3), (1, 3, 2)}. Prove: VH is a normal subgroup of S₄ of cardinality 12, which is the union of three S₄ - conjugacy classes
• Nov 16th 2009, 01:14 AM
Swlabr
Quote:

Originally Posted by apple2009
Consider the group S₄.
a)Prove that {(1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}is an S₄ - conjugacy class.
b)Let V:={(1), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}. Prove that V is a normal abelian subgroup of S₄ isomorphic to the Klein 4-subroup V₄
c)Let H= {(1), (1, 2, 3), (1, 3, 2)}. Prove: VH is a normal subgroup of S₄ of cardinality 12, which is the union of three S₄ - conjugacy classes

For (a), what have you done already? Do you know what it means for a set of elements to form a conjugacy class?

For (b), you need to prove that these elements form a subgroup (as you will get normality for free because they are closed under conjugation). To prove that it is isomorphic to the Klien 4-group, you just need to show that it is not cyclic (there are only 2 groups of order 4, and if it is not one it must then be the other).

For (c) you need to calculate the set $\displaystyle VH=/{vh: v \in V, h \in H/}$. This is a group because $\displaystyle V \lhd S_4$:

$\displaystyle v_1h_1v_2h_2=v_1h_1v_2h_1^{-1}h_1h_2=v_1v_2'h_1h_2 \in VH$.

Now, as you have calculated the set you know it has order 12, and that it contains at least two conjugacy classes (the identity, and the one in part (a)). The third will just be the rest of the group. You should prove this is a conjugacy class.

Now, neatly, you have that your group is also normal as it is a union of conjugacy classes so it MUST be normal.