# The Unapologetic Mathematician

## The First Isomorphism Theorem

Today I want to walk through what’s called the “First Isomorphism Theorem” for groups. There are two more, but the first is really more interesting in my view. I’ll start with a high-level sketch: kernels of homomorphisms are normal subgroups, images are quotient groups, and every homomorphism is a quotient followed by an isomorphism.

First I’m going to need a couple homomorphisms. If we’ve got a group $G$ and a normal subgroup $N$, there’s immediately a homomorphism $\pi_{(G,N)}:G\rightarrow G/N$. Just send each $g$ to its coset $gN$. It should be clear that every coset gets hit at least once, so this is an epimorphism, and that its kernel is exactly $N$. We call $\pi_{(G,N)}$ the “canonical projection” or the “canonical epimorphism” from $G$ to $G/N$.

On the other hand, if $G$ is a group and $H$ is any subgroup, we have a homomorphism $\iota_{(G,H)}:H\rightarrow G$ given by just sending every element of $H$ to itself inside $G$. This is such a natural identification to make that it feels a little weird to think of it as a homomorphism at all, but it actually turns out to be quite useful. The kernel of $\iota_{(G,H)}$ is trivial, making it a monomorphism. We call it the “canonical injection” or “canonical monomorphism” from $H$ to $G$.

Now consider any homomorphism $f:G\rightarrow H$. If $k$ is in the kernel of $f$ $f(k)$ is the identity $e_H$ of $H$ — and $g$ is any element of $G$, we calculate $f(gkg^{-1}) = f(g)f(k)f(g^{-1}) = f(g)f(g)^{-1} = e_H$

so $gkg^{-1}$ is in the kernel as well. Thus the kernel is a normal subgroup.

So every kernel is a normal subgroup, and the canonical projection shows that every normal subgroup shows up as the kernel of some homomorphism.

Now we can write any homomorphism $f$ as a composition $G\rightarrow^{\pi_{(G,{\rm Ker}(f))}}G/{\rm Ker}(f)\rightarrow^{f'} {\rm Im}(f)\rightarrow^{\iota_{(H,{\rm Im}(f)}}H$

where I’ve written the name of each composition next to its arrow. That is, we first project onto the quotient of the domain by the kernel of $f$, then we send that to the image of $f$ by a homomorphism we call $f'$, and finally we inject the image into the codomain. As a bonus, $f'$ is an isomorphism!

Okay, so how do we define $f'$? If we write the kernel of $f$ as $N$, we need to figure out what to do with a coset $gN$. If $g$ and $gn$ are two elements of $gN$, then $f(gn) = f(g)f(n) = f(g)$, so $f$ sends every element of $gN$ to the same element of $H$. We define $f'(gN) = f(g)$.

Now let’s say $f'(gN) = e_H$. This means that $f(g)=e_H$, so $g$ is in $N$ already, and $gN$ is the identity of $G/N$. The kernel of $f'$ is trivial, so $f'$ is a monomorphism. On the other hand, every element in ${\rm Im}(f)$ is $f(g)$ for some $g$, so each one is also $f'(gN)$ for some $gN$. That makes $f'$ an epimorphism, and thus an isomorphism. Q.E.D.

Every homomorphism works like this: you divide out some kernel, hit the quotient group with an isomorphism, and include the result into the target group. Since isomorphisms don’t really change anything about a group and the inclusion is pretty simple too, all the really interesting stuff goes on in the first step. The homomorphisms that can come out of $G$ are essentially determined by the normal subgroups of $G$. Because of this, we call a group with no nontrivial normal subgroups “simple”. The kernel of an homomorphism from a simple group is either trivial or the whole group.

What we’re starting to see here is the tip of a much deeper approach to algebra. The internal structure of a group is intimately bound up with the external structure of the homomorphisms linking it to other groups. Each one determines, and is determined by the other, and this duality can be a powerful tool for answering questions on one side by turning them into questions on the other side.

February 17, 2007