Exercise 2.2.2 ($C_G(Z(G)) = G$ and $N_G(Z(G)) = G$)

Question

Prove that $C_G(Z(G)) = G$ and deduce that $N_G(Z(G)) = G$.

Richard Ganaye · Accepted Answer

\begin{proof} 
For all $x \in G$, by definition of $Z(G)$,
$$\forall y \in Z(G),\ xy = yx.$$
So $x$ commutes with every element of $\Z(G)$, thus $x \in C_G(Z(G))$.

This shows the inclusion $G \subseteq C_G(Z(G))$, and since $C_G(Z(G)) \subseteq G$ by definition,
\begin{align*}
C_G(Z(G)) = G.
\end{align*}

We know that the normalizer of any subset $A$ contains the centralizer of $A$. Thus $G = C_G(Z(G)) \subseteq N_G(Z(G)) \subseteq G$, so
\begin{align*}
N_G(Z(G)) = G.
\end{align*}
\end{proof}

Exercise 2.2.2 ($C_G(Z(G)) = G$ and $N_G(Z(G)) = G$)

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Exercise 2.2.2 ($C_G(Z(G)) = G$ and $N_G(Z(G)) = G$)

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