0

Consider the identity $|Z(G)|=|G|-\sum_{j=1}^m |C_{x_j}|$ where $C_1,...C_m$ are the conjugation classes in which $G\backslash Z(G)$ is partitioned into and $G$ is a non-abelian group with $p^m$ elements where $p$ is prime and $m \in \mathbb{N}$.

My goal is to prove $|Z(G)|>1$. Suppose it has only one element. How can I derive a contradiction then? In this case $|G|=\sum_{j=1}^m |C_{x_j}| + 1=p^m$ for some $p,m$. Since every $C_{x_j}$ has at least two elements the left hand side is bounded by $2m+1$ which is larger than $2^m$ in the case of $p=2$.

Is it enough to derive a contradiction for only one of the $p$'s?

Shaun
  • 47,747
Xaver Wallenstein
  • 31
  • No. You have to prove that your logic works for all primes $p.$ However, you can take motivation from what you trying to do for $p=2.$ – MathRookie2204 Jun 04 '24 at 15:28
  • Use the Orbit-Stabilizer Theorem and the fact that each $C_i$ is nontrivial. – Deif Jun 04 '24 at 15:30
  • The cardinality of $C_{x_i}$s are not arbitrary. It seems you do not know the Orbit-Stabilizer Theorem. Then try to prove $p| |C_{x_i}|$ by proving: (1) For each $x\in G-Z(G)$, $C(x)={a\in G: ax=xa}$ is a subgroup of $G$ unequal to $G$ itself. (2) The map on the set of left cosets of $C(x)$ to the conjugate class of $x$ given by $aC(x)\mapsto axa^{-1}$ is well-defined and bijective. Therefore $|\text{ conjugate class of $x$ }|=[G:C(x)]$ is a multiple of $p$. – Asigan Jun 04 '24 at 15:51

1 Answers1

0

If $Z(G)$ were trivial, then the noncentral/nontrivial elements of $G$ would have centralizers of order $p^n$, for some $n<m$ (depending on the element): in fact, if $g\notin Z(G)$, we have $C_G(g)<G$, strictly. Therefore, the class equation would yield: $$p^m=1+\sum_i p^{m-n_i}$$ Now, $p$ divides the LHS and the sum in the RHS, but $p\nmid 1$: contradiction.

Kan't
  • 4,819