# 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

If $G$ is a group, its *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 $G$.

But how can we see this? Wouldn't it be nice if we had a "commutativity detector"? I think so! But how would we go about finding, *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.

Well this is great! The commutativity (or lack thereof) of $g$ is entirely reflected in whether or not the correponding map $\phi_g$ is the identity! That is, $\phi_g$ *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 $g\in G$.

Notice that several elements of $G$ may give rise to the same inner automorphism. In fact, $\phi_g=\phi_{gh}$ whenever $h\in Z(G)$. (Check this: if $x\in G$, then $\phi_{gh}(x)=(gh)x(gh)^{-1}=g(hxh^{-1})g^{-1}=gxg^{-1}=\phi_g(x)$.) So we might as well lump those elements together in a single pile, or coset, and call it $gZ(G)$. We might expect, then, a one-to-one correspondence between the cosets $gZ(G)$ in $G/Z(G)$ and the inner automorphisms $G\to G$. And that's exactly what we get! The map $G\to\text{Inn}(G)$ given by $g\mapsto\phi_g$ is surjective, and its kernel---all the elements of $G$ whose corresponding $\phi_g$ is the identity map---is exactly $Z(G)$. So by the First Isomorphism Theorem, $$G/Z(G)\cong\text{Inn(G)}.$$

Of course, there's only one way for an element $g\in G$ to induce the identity map, and that's if $g\in Z(G)$. (This is why there's only one "trivial" coset, namely $Z(G)$, in $G/Z(G)$.) But there may be lots of non-trivial cosets, i.e. lots of elements $g\not\in Z(G)$ that induce different, non-identity inner automorphisms of $G$. But the latter comprise the substantial or interesting part of the quotient. And that is why I think it's helpful to view $G/Z(G)$ as being made up of those elements of $G$ that don't commute!

*"Abelian" is a property that a group may or may not possess: either the elements commute with each other or they don't. Or maybe *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 $|G|$.