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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

Let's use the subgroup test to show that the center of a group $G$, defined by $Z(G)=\{x\in G\mid xg=gx \text{ for all } g\in G\}$, is a subgroup of $G$. First, note that $Z(G)$ is nonempty because the identity $e_G$ of $G$ is in $Z(G)$ as we know $e_Gg=ge_G$ for every $g\in G$ from the definition of an identity. Clearly $Z(G)$ is a subset of $G$ by definition. Let $a,b\in Z(G)$. Then we know $ag=ga$ and $bg=gb$ for every $g\in G$. We will show that $(ab)g=g(ab)$ and $(a^{-1})g=g(a^{-1})$ for every $g\in G$, as this shows that $Z(G)$ is closed under the operation, and closed under taking inverses. Let $g\in G$. We then compute $$(ab)g=a(bg)=a(gb)=(ag)b=(ga)b=g(ab)$$ which shows $ab\in Z(G)$. Since $ag=ga$, left and right multipling by $a^{-1}$ gives $$a^{-1}aga^{-1}=a^{-1}gaa^{-1},$$ which simplies to $ga^{-1}=a^{-1}g$ and hence $a^{-1}\in Z(G)$. Since we've show that $Z(G)$ a nonempty subset of $G$ that is closed under the operation and taking inverses, the subgroup test proves that $Z(G)\leq G$.

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