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

[Szichedelic]

Homework Statement

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

Homework Equations

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

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

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\inZ(x), the order of x is 3, as well.

Now, because Z(G)\subseteqZ(x), we have that |Z(G)| = 1,2, or 3 (by Lagrange's Theorem). However, since |Z(x)|≠12, x\notinZ(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}\inZ(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}\inZ(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!)

 

[Dick]

Homework Statement

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

Homework Equations

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

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

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\inZ(x), the order of x is 3, as well.

Now, because Z(G)\subseteqZ(x), we have that |Z(G)| = 1,2, or 3 (by Lagrange's Theorem). However, since |Z(x)|≠12, x\notinZ(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}\inZ(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}\inZ(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!)

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.

 

[Szichedelic]

Oh duh.. That makes things a lot easier.