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A group G of order 12, with conjugacy class of order 4 has trivial center.

My attempt:

$|C(x)|=4 \implies |Z(x)|=3$. This implies that Z(x) is a cyclic subgroup of order 3. Thus $Z(x)= \{1,x,x^{-1}\}$. We know the center of the group $Z(G) \subset Z(x)$.Therefore, $|Z(G)|= 1,2,$ or $3$.

If $|Z(G)|= 1$, then we are done. If $|Z(G)|= 2$, then consider WLOG $Z(x)-\{x\}$. That means $|Z(G)| =\{1,x^{-1}\}$ which is contradiction as we know $G$ is not abelian and $Z(G)$ is a subgroup of $G$.

Now, consider $|Z(G)|=3$. This implies that $Z(G)=Z(x)$.

I am stuck in this case?? I dont see any contradiction. Any help? Also, out of curiosity are there any alternate proofs to prove the same claim?

user104221
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  • What do you mean by conjugacy class of order 4? Do you mean there are 4 distinct conjugacy classes? Maybe you mean that there is a conjugacy class that has 4 elements? In general different conjugacy classes will not have the same number of elements. – D Wiggles Nov 02 '13 at 22:18
  • @IBWiglin: I mean the group contains a conjugacy class of order 4. (has 4 elements) – user104221 Nov 02 '13 at 22:20

2 Answers2

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Since the conjugacy classes partition $G$ we have the well known Class Equation

$$|G| = |C_1| + \dots + |C_n|$$

where $C_1, \dots , C_n$ are the conjugacy classes. The conjugacy classes of order $1$ correspond to the elements in the center $Z(G)$. Also, all the $|C_i|$ divide $|G| = 12$.

So if the center has order three and we have (at least) one conjugacy class of order $4$, the Class Equation can only look like this:

$$12 = 1 + 1 + 1 + 4 + 3 + 2$$

So we have a class of order $|C(y)|=3$. Then $|Z(y)| = 4$ and since $Z(G)$ is a subgroup of $Z(y)$, $3$ would have to divide $4$ (by Lagrange's Theorem), which it doesn't.

Brusko651
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More simple: you had prove $Z(G) \subset Z(x)$. Since $|Z(x)|=3$ and by assumption $|Z(G)|\ne 1$ then $|Z(G)|=3$. Hence $x\in Z(G)$. But then $|C(x)|=1$.

Boris Novikov
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