# Abstract Algebra - Group of Order 12 with Conjugacy Class of Order 4

1. Nov 29, 2012

### Szichedelic

1. The problem statement, all variables and given/known data

A group G of order 12 contains a conjugacy class C(x) of order 4. Prove that the center of G is trivial.

2. Relevant equations

|G| = |Z(x)| * |C(x)|

(Z(x) is the centralizer of an element x$\in$G, the center of a group will be denoted as Z(G))

3. The attempt at a solution

Let G be a group of order 12 with a conjugacy class C(x) of order 4. Then, by the counting formula, |Z(x)|=3. Further, Z(x) is a cyclic subgroup of G of order 3 and since x$\in$Z(x), the order of x is 3, as well.

Now, because Z(G)$\subseteq$Z(x), we have that |Z(G)| = 1,2, or 3 (by Lagrange's Theorem). However, since |Z(x)|≠12, x$\notin$Z(G) and hence, |Z(G)|=1 or 2.

Suppose that |Z(G)|=2, then Z(G) = Z(x) - {x}. However, since Z(x) is a subgroup, this implies that x$^{-1}$$\in$Z(x), as well. This means that if |Z(G)| = 2, then Z(G) = {1, x$^{-1}$}. This is not possible since Z(G) is a subgroup we have that if x$^{-1}$$\in$Z(G) then x must also be an element of Z(G) as well. Further, in any group, the order of an element and its inverse must be the same. Hence, x$^{-1}$ is of order 3.

Therefore, the only other possibility is that |Z(G)|=1 which implies that a group G of order 12 with a conjugacy class of order 4 must have a trivial center.

(I'm not sure if this is right, and I'm not sure where I'm being redundant... I know that this group is isomorphic to A4, since A4 is the only group of order 12 with a conjugacy class of order 4. However, I didn't want to go the route of just showing the conjugacy classes of every possible group of order 12 and saying... Hey, A4 is the only one with a conjugacy class of order 4 and it has the trivial center!)

Last edited: Nov 29, 2012
2. Nov 29, 2012

### Dick

Seems pretty ok to me. Sure it would be a cop out to just say "I know it's A4". But you could simplify it a little. You know Z(G) is a subgroup of Z(x). Once you know x is not in Z(G) there no subgroup of Z(x) of order 2. Lagrange's theorem again.

3. Nov 29, 2012

### Szichedelic

Oh duh.. That makes things a lot easier.