The Center of a Group

# The Center of a Group

 Definition: Let $(G, *)$ be a group. Then the Center of $G$ denoted $Z(G)$ is the set of elements that commute with every element in $G$, that is $Z(G) = \{ a \in G : \mathrm{for \: all} \: g \in G, a * g = g * a \}$.

One important fact about the center of a group is that a group $(G, *)$ is abelian if and only if $G$ is equal to its center, i.e., $G = Z(G)$ as we prove in the following theorem.

 Theorem 1: Let $(G, *)$ be a group. Then $(G, *)$ is abelian if and only if $G = Z(G)$.
• Proof: $\Leftarrow$ Suppose that $(G, *)$ is abelian. Let $a \in G$. Since $*$ is commutative we have that for all $g \in G$ that $a * g = g * a$. Therefore $a \in Z(G)$, so $G \subseteq Z(G)$. By definition $G \supseteq Z(G)$. Therefore $G = Z(G)$.
• $\Rightarrow$ Now suppose that $G = Z(G)$. Then for all $a \in G$ we have that $a$ commutes with every $g \in G$, i.e, $a * g = g * a$ for all $a, g \in G$, hence $(G, *)$ is abelian. $\blacksquare$

Another nice property of the center of a group is that $(Z(G), *)$ will always be an abelian subgroup of $(G, *)$ which we prove in the following theorem.

 Theorem 2: Let $(G, *)$ be a group. Then $(Z(G), *)$ is an abelian subgroup of $(G, *)$.
• Proof: We have that $Z(G) \subseteq G$ and so we only need to show that $Z(G)$ is closed under $*$ and that for all $x \in Z(G)$ there exists an $x^{-1} \in Z(G)$ such that $x * x^{-1} = e$ and $x^{-1} * x = e$.
• Let $x, y \in Z(G)$. Then $x$ and $y$ commute with all elements in $G$, that is, for all $g \in G$ we have that $x * g = g * x$ and $y * g = g * y$. Consider the product $x * y$, and let $g \in G$. Then by using the associativity property of $*$ inherited on elements in $Z(G)$ we have that:
(1)
\begin{align} \quad (x * y) * g = x * (y * g) = x * (g * y) = (x * g) * y = (g * x) * y = g * (x * y) \end{align}
• Therefore $Z(G)$ is closed under $*$.
• Now let $x \in Z(G)$ and suppose that $x^{-1} \not \in Z(G)$. Then there exists some $g_0 \in G$ such that $x^{-1} * g_0 \neq g_0 * x^{-1}$. Multiplying both sides of this equation by $x$ and using the inherited associativity of $*$ from $G$ and we see that a contradiction arises:
(2)
\begin{align} \quad x * (x^{-1} * g_0) \neq x * (g_0 * x^{-1}) \\ \quad (x * x^{-1}) * g_0 \neq x * (g_0 * x^{-1}) \\ \quad e * g_0 \neq x * (g_0 * x^{-1}) \\ \quad g_0 \neq (x * g_0) * x^{-1} \\ \quad g_0 \neq \underbrace{(g_0 * x)}_{\mathrm{since} \: x \in Z(G)} * x^{-1} \\ \quad g_0 \neq g_0 * (x * x^{-1}) \\ \quad g_0 \neq g_0 * e \\ \quad g_0 \neq g_0 \end{align}
• Therefore our assumption that $x^{-1} \not \in Z(G)$ was false. So for all $x \in Z(G)$ we have that $x^{-1} \in Z(G)$ is such that $x * x^{-1} = e$ and $x^{-1} * x = e$.
• Hence all of the group axioms are satisfied and the commutativity of $*$ is inherited to the subset $Z(G) \subseteq G$, so $(Z(G), *)$ is an abelian group. $\blacksquare$