# Understanding the definition of the center $Z(G)$ of a group $G$.

I'm having trouble understanding the definition of "center" in group theory. My textbook says:

"The center, $$Z(G)$$, of a group is the subset of elements in $$G$$ that commute with every element of $$G$$. In symbols, $$Z(G)= \{a \in G :\, ax= xa,\forall x\in G\}$$."

What does it mean to commute with every element? Does this just mean it's an Abelian group or is it something entirely different?

This sounds pretty basic but I still don't understand.

Help!

• it means what you have written in your notation, by definition if for every $x\in G$ your candidate $a$ has the property that $ax=xa$ then $a$ is said to be in the center of the group. Feb 5, 2017 at 22:39
• I think that in the groups $GL(n)$ the subgroup of diagonal matrices is a nice example of a center. Feb 6, 2017 at 21:15

If the group $G$ is abelian, then we would have that $Z(G) = G$. Normally in a group which is not necessarily abelian, there are elements which do not commute with every other element. What is true, is that $$Z(G) \unlhd G$$ As an example that in general $Z(G) \neq G$, we have that $$Z(S_n) = \{\operatorname{id}\}$$ whenever $n \geq 3$ (a proof can be found here).
I would use the quaternion group $$Q_8=\{\pm1,\pm i,\pm j,\pm k\}$$ as example.
The center of $Q_8$, $Z(Q_8)=\{\pm 1\}$.
This means that $1x=x1$ and $-1x=x(-1)$ is true when $x$ is any element of $Q_8$.
So $1$ and $-1$ commute with every element in $Q_8$.
In particular, $1$ and $-1$ commute with each other in $Z(Q_8)$.
This means that $Z(Q_8)$ is abelian.