The Center of a Group G, Z(G)

# The Center of a Group G, Z(G)

Definition: Let $(G, \cdot)$ 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 $Z(G) = \{ a \in G : \mathrm{for \: all} \: g \in G, a \cdot g = g \cdot a \}$. |

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

Proposition 1: Let $(G, \cdot)$ be a group. Then $(G, \cdot)$ is abelian if and only if $G = Z(G)$. |

**Proof:**$\Leftarrow$ Suppose that $(G, \cdot)$ is abelian. Let $a \in G$. Since $\cdot$ is commutative we have that for all $g \in G$ that $a \cdot g = g \cdot a$. 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, $a \cdot g = g \cdot a$ for all $a, g \in G$, hence $(G, \cdot)$ is abelian. $\blacksquare$

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

Proposition 2: Let $(G, \cdot)$ be a group. Then $(Z(G), \cdot)$ is an abelian subgroup of $(G, \cdot)$. |

**Proof:**We have that $Z(G) \subseteq G$ and so we only need to show that $Z(G)$ is closed under $\cdot$ and that for all $x \in Z(G)$ there exists an $x^{-1} \in Z(G)$ such that $x \cdot x^{-1} = e$ and $x^{-1} \cdot x = e$.

- 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 $x \cdot g = g \cdot x$ and $y \cdot g = g \cdot y$. Consider the product $x \cdot y$, and let $g \in G$. Then by using the associativity property of $\cdot$ inherited on elements in $Z(G)$ we have that:

\begin{align} \quad (x \cdot y) \cdot g = x \cdot (y \cdot g) = x \cdot (g \cdot y) = (x \cdot g) \cdot y = (g \cdot x) \cdot y = g \cdot (x \cdot y) \end{align}

- Therefore $Z(G)$ is closed under $\cdot$.

- 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 $x^{-1} \cdot g_0 \neq g_0 \cdot x^{-1}$. Multiplying both sides of this equation by $x$ and using the inherited associativity of $\cdot$ from $G$ and we see that a contradiction arises:

\begin{align} \quad x \cdot (x^{-1} \cdot g_0) &\neq x \cdot (g_0 \cdot x^{-1}) \\ \quad (x \cdot x^{-1}) \cdot g_0 &\neq x \cdot (g_0 \cdot x^{-1}) \\ \quad e \cdot g_0 &\neq x \cdot (g_0 \cdot x^{-1}) \\ \quad g_0 &\neq (x \cdot g_0) \cdot x^{-1} \\ \quad g_0 &\neq \underbrace{(g_0 \cdot x)}_{\mathrm{since} \: x \in Z(G)} \cdot x^{-1} \\ \quad g_0 &\neq g_0 \cdot (x \cdot x^{-1}) \\ \quad g_0 &\neq g_0 \cdot 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 $x \cdot x^{-1} = e$ and $x^{-1} \cdot x = e$.

- Hence all of the group axioms are satisfied and the commutativity of $\cdot$ is inherited to the subset $Z(G) \subseteq G$, so $(Z(G), \cdot)$ is an abelian group. $\blacksquare$