# A Group and Its Center, Intuitively

/Last week we took an intuitive peek into the First Isomorphism Theorem as one example in our ongoing discussion on quotient groups. Today we'll explore another quotient that you've likely come across, namely that of a group by its center.

## Example #2: A group and its center

*center*$Z(G)=\{g\in G:gx=xg \text{ for all $x\in G$}\}$ is the subgroup consisting of those elements of $G$ that commute with

*everyone else*in $G$. In line with the the intuition laid out in this mini-series, we'd like to be able to think of (the substantial part of) $G/Z(G)$ as consisting of those elements of $G$ that

*don't*commute with everyone else in $G$.

*or constructing*, such a device?

Here's a BIG hint....

Not too long ago we chatted about the most obvious secret in mathematics: if you want to study (

*or detect!*) properties (

*like the failure to be abelian!**) of an object (

*like a group!*), it's really helpful to look at maps (

*like homomorphisms!*) to/from that object to another (

*like the group itself!*).

So since the secret is out, let's put it to use!

Let's pick an arbitrary element $g\in G$. Our goal is to test $g$ for commutativity: does it commute with every element in $G$? That is, $g$ in $Z(G)$? Or is it not? To find out, let's *define* a map $\phi_g:G\to G$ that takes an element $x$ and sends it to $gxg^{-1}$. It's not too hard to check that this map is acutally a homomorphism. Moreover, we gain two key observations:

- If $g$
*is*in $Z(G)$, then $gxg^{-1}=x$ for*all*$x\in G$ and so $\phi_g$ was really the identity map, i.e. $\phi_g(x)= x$ for all $x\in G.$ - If $g$ is
*not*in $Z(G)$, then there is at least one $x\in G$ so that $gxg^{-1}\neq x$ and so $\phi_g$ has no chance of being the identity map.

*is*the identity if and only if $g \in Z(G)$. Or equivalently, it's

*not*the identity if and only if $g\not\in Z(G)$.

So there we have it! Our commutativity detector! For each element $g\in G$, we simply need to look at the corresponding homomorphism $\phi_g$ and ask, "Is it the identity map?"

The assignment $g\mapsto\phi_g$ turns out to be a group homomorphism in its own right. It maps from $G$ to a group of special homomorphisms $G\to G$ called the *inner automorphisms of $G$*, denoted by $\text{Inn}(G).$ These are precisely the isomorphisms from $G$ to itself that are of the form $x\mapsto gxg^{-1}$ for $g\in G$.

*some*of them do (namely those in the center!) while some of them don't (those

*not*in the center). What's neat is that the size of the quotient $G/Z(G)$ measures just

*how*abelian $G$ is!

For starters, we know that $1\leq |G/Z(G)| \leq |G|$. In fact, $|G/Z(G)|=1$ if and only if $G=Z(G)$ if and only if $G$ is abelian. On the other hand, $|G/Z(G)|=|G|$ if and only if $Z(G)=\{e\}$ if and only if *no non-identity* elements of $G$ commute with any other non-identity elements, i.e. $G$ is as non-abelian as possible. So the abelian-ness of $G$ is inversely proportional to $|G/Z(G)|$ as it ranges from 1 to $|G|$.