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Problem 56 (The Center Of Group Is A Subgroup)

Prove that the center $Z(G)$ of a group $G$ is a subgroup of $G$. If $G$ is Abelian, then what is $Z(G)$?

Solution

In order to prove that the center $Z(G)$ of a group $G$ is a subgroup of $G$, we will use the subgroup test. Recall that $$Z(G)=\{x\in G\mid gx=xg \text{ for all } g\in G\}.$$ Also recall that the subgroup test says that if $G$ is a group, if $Z(G)$ is a nonempty subset of $G$, if $a,b \in Z(G)$ implies $ab \in Z(G)$, and if $a \in Z(G)$ implies $a^{-1} \in Z(G)$, then $Z(G)$ is a subgroup. By assumption, we know that $G$ is a group. We will now show that $Z(G)$ is a nonempty subgroup of $G$. Let $x \in Z(G)$. By definition of the center of a group, we know that $x \in G$. Thus, we know that $Z(G) \subseteq G$. Additionally, consider the identity in $G$, call it $e_G$. By definition of the identity, we know that $ge_G = e_Gg$ for all $g \in G$. Thus, $e_G \in Z(G)$. Because $e_G \in Z(G)$ and $Z(G) \subseteq G$, we know that $Z(G)$ is a nonempty subset of $G$.

Now we sill show that $Z(G)$ is closed under multiplication. Let $a,b \in Z(G)$. Let $g \in G$. We compute $$ (ab)g = a(bg) = a(gb) = (ag)b = (ga)b = g(ab) $$ Thus, we know that $ab \in Z(G)$.

Finally, we will show that $Z(G)$ is closed under taking inverses. Let $a \in Z(G)$. Consider the element $a^{-1}$, which we know is in $G$. Let $g \in G$. We compute $$ ga^{-1} = ega^{-1} = (a^{-1}a)ga^{-1} = a^{-1}(ag)a^{-1} = a^{-1}(ga)a^{-1} = a^{-1}g(aa^{-1}) = a^{-1}ge = a^{-1}g $$ Thus, we know that $a^{-1} \in Z(G)$. Thus, we know that $Z(G) \le G$.

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