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 $|G|$.

The First Isomorphism Theorem, Intuitively

Welcome back to our little discussion on quotient groups! (If you're just now tuning in, be sure to check out "What's a Quotient Group, Really?" Part 1 and Part 2!) We're wrapping up this mini series by looking at a few examples. I'd like to take my time emphasizing intuition, so I've decided to give each example its own post. Today we'll take an intuitive look at the quotient given in the First Isomorphism Theorem.

Example #1: The First Isomorphism Theorem

Suppose $\phi:G\to H$ is a homomorphism of groups (let's assume it's not the map that sends everything to the identity, otherwise there's nothing interesting to say) and recall that $\ker\phi\subset G$ means "You belong to $\ker\phi$ if and only if you map to the identity $e_H$ in $H$." I'd like to convince you why it's helpful to think of the quotient $G/\ker\phi$ as consisting of all the stuff in $G$ that doesn't map to $e_H$.
 
 
First notice that every element of $G$ is either 1) in $\ker\phi$ or 2) is not. And there's only one way to satisfy 1)---you're simply in the kernel. This is why we have exactly one "trivial" coset, $\ker\phi$. On the other hand, there may be many ways to satisfy 2) and is why there may be many "nontrivial" cosets.

But just how might an element $g\in G$ satisfy 2)? Well, $\phi(g)\neq e_H$, of course! But notice! There could be many elements besides $g$ who also map to the same $\phi(g)$ under $\phi$. (After all, we haven't required that $\phi$ be injective.) In fact, every element of the form $gg'$ where $g'\in\ker\phi$ fits the bill. So we group all those elements together in one pile, one coset, and denote it $g\ker\phi$. The notation for this is quite good: the little $g$ reminds us, "Hey, these are all the folks that map to the value of $\phi$ at that $g$." And multiplying $g$ by $\ker\phi$ on the right is suggestive of what we just observed: we can obtain other elements with the same image $\phi(g)$ by multiplying $g$ on the right by things in $\ker\phi$.

I like to imagine the elements of $G$ as starting off as dots scattered everywhere,

 
before.jpg
 
which we can then organize into little piles according to their image under $\phi$. In fact, let's color-code them:
(Notice $\phi$ isn't necessarily surjective.) Now here's the key observation: we get one such pile for every element in the set $\phi(G)=\{h\in H|\phi(g)=h \text{ for some $g\in G$}\}$. The idea, then, behind forming the quotient $G/\ker\phi$ is that we might as well consider the collection of green dots as a single green dot and call it the coset $\ker\phi$. And we might as well consider the collection of pink dots as a single pink dot and call it the coset $g_1\ker\phi$, and so on. So we get a nice, clean picture like this:
bijection.jpg
Intuitively, then, we should expect a one-to-one correspondence between the cosets of $G/\ker\phi$ and the elements of $\phi(G)$. That's what the image above is indicating. And that's exactly what the First Isomorphism Theorem means when it tells us there is a bijection $$G/\ker \phi \cong \phi(G).$$ (In fact, it's richer than a bijection of sets---it's actually an isomorphism of groups!) Pretty cool, huh? We should also notice that there are exactly $|\phi(G)\smallsetminus\{e_H\}|$ ways to "fail" to be in $\ker\phi$, and exactly $1=|\{e_h\}|$ way to be in $\ker\phi$. Typically, $|\phi(G)\smallsetminus\{e_H\}|$ is bigger than one, and so the interesting or substantial part of the quotient $G/\ker\phi$ lies in its subset of nontrival cosets, $g_1\ker\phi,g_2\ker\phi,\ldots.$ The First Isomorphism Theorem implies that this is the same as viewing the interesting or substantial part of $\phi(G)$ as lying in all the elements of $g$ that don't map to the identity in $H$.

And this is why I like to think of $G/\ker\phi$ as "things in $G$ that don't map to the identity."

A closing remark

At some point, you may have seen the First Isomorphism Theorem stated as follows:
Theorem: Let $\phi:G\to H$ be a group homomorphism and let $\pi:G\to G/\ker\phi$ be the cannonical (surjective) homomorphism $g\mapsto g\ker\phi$. Then there is a unique isomorphism $\psi:G/\ker\phi\to \phi(G)$ so that $\phi=\psi\circ\pi$, i.e. so that the diagram on the right commutes.
On the surface, this sounds fancy, but it's really just the concise version of our discussion above. We can see this by matching up our pictures with the diagram:
 
 
The diagram simply states the obvious: the map $\phi$ from $G$ naturally partitions the elements of $G$ into little (color-coded) piles, according to where they land in $H$. This therefore gives us two ways to get from an arbitrary $g\in G$ to its image $\phi(g)\in H$. We can either send it directly there via $\phi$. This is the diagonal arrow in the diagram. Or we can first send it to its corresponding color/pile/coset, and then realize, "Aha, everyone in that particular color/pile/coset maps to $\phi(g)$, therefore $g$ goes there too." This is the composition of the horizontal and vertical maps, $\pi$ and $\psi$. And the uniqueness of $\psi$ just says, "This is super obvious!"

(By the way, there's nothing really special about groups here. There's a first isomorphism theorem for other algebraic objects, and the same intuition holds.)

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