Let $G$ be a group, assume $[G:Z(G)]=n$ is finite. Let $A \subseteq G$ be any subset. Prove that the number of conjugates of $A$ is at most $n.$

Question

The title is an exercise in Aluffi's Algebra: Chapter 0.

In the section before the exercise, it states: if $A \subseteq G$ is a subset and $g \in G,$ the conjugate of A is the subset $gAg^{-1}.$

Does this mean that the number of conjugates of $A$ is the cardinality of the following set, $\{gAg^{-1} : g \in G\}?$

I'd like to correctly understand this notion of the number of conjugates of a subset of a group.

Answer 1

Hint: prove that the number of conjugates of $A$ equals the index $|G:N_G(A)|$, where $N_G(A)=\{g \in G: gA=Ag\}$, the normaliser of $A$ in $G$ (this is a subgroup of $G$). Show then that $Z(G) \subseteq N_G(A) \subseteq G$.

Answer 2

If $xAx^{-1}=yAy^{-1}$, then $y^{-1}xA(y^{-1}x)^{-1}=A$ and $y^{-1}x$ normalizes $A$.

This is another way of saying $x$ and $y$ are in the same coset of $G/N$, where $N$ is the normalizer of $A$.

But that the center $Z(G)$ is contained in the normalizer $N$ is trivial. Therefore $\mid G/N\mid\le\mid G/Z(G)\mid=n$.

Answer 3

Note that the cardinal of the inner automorphism group is $|\text{Inn}(G)|=n$, since $G/Z(G)\cong\text{Inn}(G)$. Thus, $$|\{A^g:g\in G\}|=|\{\alpha(A):\alpha\in\text{Inn}(G)\}|\leq |\text{Inn}(G)|=n.$$