**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.1**

Prove that $C_G(A) = \{ g \in G \ |\ g^{-1}ag = a\ \mathrm{for\ all}\ a \in A \}$.

Solution: By definition, $C_G(A) = \{ g \in G \ |\ gag^{-1} = a\ \mathrm{for\ all}\ a \in A \}$.

($\subseteq$) If $g \in C_G(A)$, then $gag^{-1} = a$ for all $a \in A$. Left multiplying by $g^{-1}$ and right multiplying by $g$, we have that $a = g^{-1}ag$ for all $a \in A$.

($\supseteq$) If $g \in G$ such that $g^{-1}ag = a$ for all $a \in A$, then left multiplying by $g$ and right multiplying by $g^{-1}$ we have that $a = gag^{-1}$ for all $a \in A$.