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.

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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,  

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:        

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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."

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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.

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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.)