## Theorem 26: the first isomorphism theorem

Sorry for the gap; things have been pretty busy here.  The good news, though, is that it’s time for another theorem!

I’ve written previously about groups, when I wrote about Lagrange’s theorem in group theory.  Today I’d like to talk some more about groups, so if you don’t know what one is I suggest you pause for a moment and go back to that post.

I’d like to start with a particular example.  Let’s think about two groups.  One is $\mathbb{Z}$, the group of integers under addition.  The other is the group of integers under addition (mod 7), which I’ll write as $\mathbb{Z}_7$.  How are they related?  They’re very similar; the only real difference is that in $\mathbb{Z}_7$, we treat two numbers as the same if they differ by a multiple of 7, because we’re adding (mod 7).  In fact, that’s a precise recipe for how to get $\mathbb{Z}_7$ from $\mathbb{Z}$, isn’t it?

This leads us to the idea of a quotient group.   Here, we take the multiples of 7 (which we could write as $7\mathbb{Z}$), and think of them as being all the same (and in particular the same as 0).  Then we say that two elements of $\mathbb{Z}$ are the same if they differ by a multiple of 7.  (E.g. we think of 6 and 13 as being the same.)  This gives us 7 classes of elements (all the elements in a single class are the same in this sense).  We pick a representative of each class (conventionally 0, 1, 2, 3, 4, 5, 6).  And it turns out that when we do this, with the same group operation as before (addition), we do get something that’s a group.  We write it as the quotient $\mathbb{Z}/7\mathbb{Z}$.

What happens if we’re a bit more general?  Let’s take a group $G$, and a subgroup $H$.   What does it mean to quotient by $H$, and is the resulting object $G/H$ always a group?

Our subgroup $H$ lies somewhere in $G$.  There are also lots of translates of $H$: the sets of the form $gH = \{ gh : h \in H \}$ where $g$ is an element of $G$.  This gives us a translate of $H$ for each element of $G$.  We call them the (left) cosets of $G$.  In our example above, $H$ was $7\mathbb{Z}$, and the translates are all the sets of the form $n + 7 \mathbb{Z}$ with $n$ an integer.

So we’ve got a bunch of cosets.  But some of them are the same, right?  (For example, if $g$ is an element of $H$, then $gH = H$, because $H$ is a subgroup.)  I talked about this when I wrote about Lagrange’s theorem.  Being formal, we can define an equivalence relation $~$ by saying that $gH \sim g' H$ exactly when $g^{-1} g' \in H$.  Then we get equivalence classes from this equivalence relation, and they partition the set.  That is, either two cosets are the same or they’re disjoint.  (We saw this in our example earlier, right?  The cosets $2 + 7 \mathbb{Z}$ and $23 + 7 \mathbb{Z}$ are the same, whereas the cosets $5 + 7 \mathbb{Z}$ and $13 + 7 \mathbb{Z}$ are disjoint.)

This is how we get the quotient $G/H$.  We could represent it by picking a representative from each equivalence class (just as we chose 0, 1, 2, 3, 4, 5, 6 in the example earlier).  But we should remember that really we’re dealing with cosets rather than single elements.

Do we get a group?  We need to define the operation first, before we can check whether we get a group!  We define the product of $gH$ and $g'H$ to be $(gH)(g'H)$.  Does that give us another left coset of $H$?  It turns out that we need an extra condition on $H$: we need it to be a normal subgroup of $G$.  One way of expressing the definition is to say that we need $ghg^{-1}$ to be in $H$ for any $g \in G$, $h\in H$.  Another is to say that we need the left coset to be the same as the right coset: we need $gH = Hg$ for all $g \in G$.  If we know that $H$ is a normal subgroup, then we find that $(gH)(g'H) = (gg')H$, so the operation does indeed combine two cosets to give another.  Of course, when I write $gg'$ I mean that we’re using the group operation in $G$.  (For example, in our earlier example we had $(a + 7\mathbb{Z}) + (b + 7 \mathbb{Z}) = (a + b) + 7\mathbb{Z}$ — that’s what addition in modular arithmetic comes down to.  You might like to check that $7\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$.)

Do we get a group?  We need to check three things: associativity, the existence of an identity, and the existence of inverses.

Associativity first.  We have

$((g_1 H)(g_2 H))(g_3 H) = ((g_1 g_2)H)(g_3 H) = ((g_1 g_2)g_3)H = (g_1 (g_2 g_3))H = (g_1 H)((g_2 g_3)H) = (g_1 H)((g_2 H)(g_3)H)$,

using the associativity of the group operation in $G$.  So far so good.

Is there an identity element?  I think the obvious guess would be to try $eH$ (where $e$ is the identity in $G$).  Let’s check it.  We have $(eH) (gH) = (eg)H = gH = (ge)H = (gH)(eH)$, so that works nicely.    Great.

What about inverses?  Again, let’s try the obvious guess.  For any $g \in G$, we have

$(gH)(g^{-1}H) = (gg^{-1})H = eH = (g^{-1}g)H = (g^{-1}H)(gH)$.

Job done!

So if $H$ is a normal subgroup of $G$, then the quotient $G/H$ is also a group, called (big surprise here) the quotient group.

The title promises that I’ll tell you about the first isomorphism theorem, and I haven’t got there yet.  Let’s start by thinking about maps between groups.

If we have two groups $G_1$ and $G_2$, we can think about functions between them.  Let’s say we have $f : G_1 \to G_2$.  Of course, there are lots of maps between these two sets.  But we know that these sets have some extra structure, the group structure.  So we might ask ourselves which maps preserve those structures.  For example, we might hope that $f$ maps the identity in $G_1$ to the identity in $G_2$.  And we might hope that if $f$ maps $g$ to $f(g)$, then it maps $g^{-1}$ to $[f(g)]^{-1}$.  If you follow these ideas to their natural conclusion, you end up with the following condition.  We want

$f(g g') = f(g) f(g')$

for all $g$, $g' \in G_1$.  Notice that the group operation on the left happens in $G_1$, whereas the group operation on the right is in $G_2$.  This gives us the definition of a homomorphism: $f : G_1 \to G_2$ is a homomorphism if $f(g g') = f(g) f(g')$ for all $g$, $g' \in G_1$.

We say that $f : G_1 \to G_2$ is an isomorphism if it is a bijective homomorphism.

So if we take some groups $G_1$ and $G_2$, and we have a homomorphism $f$ between them, how close is it to being an isomorphism?  If it’s going to be an isomorphism, then we’d certainly need the image of $f$ (the set of points hit by $f$) to be the whole of $G_2$.  It turns out that the image of $f$, $\textrm{Im}(f)$, is always a subgroup of $G_2$ (checking this would be a good exercise!), but it might not be the whole group.   If it isn’t, let’s just focus our attention on $f : G_1 \to \textrm{Im}(f)$.

Now it still might not be an injection, right?  How might it fail to be an injection?  We might have $g$ and $g' \in G_1$ such that $f(g) = f(g')$ (but $g \neq g'$).  But $f$ is a homomorphism, so if $f(g) = f(g')$ then $e_2 = f(g)[f(g')]^{-1} = f(g) f(g'^{-1}) = f(g g'^{-1})$ (where $e_2$ is the identity in $G_2$).  That is, $g g'^{-1}$ must lie in the kernel of $f$.  (The kernel of $f$, $\ker(f)$, is the set of all $g \in G_1$ such that $f(g) = e_2$.  That is, it’s all the points that get sent to the identity.)  Another good exercise would be to check that $\ker(f)$ is a subgroup of $G_1$, and moreover that it’s a normal subgroup.

So we’ve established that if $f(g) = f(g')$, then $g$ and $g'$ `differ’ by an element of the kernel of $f$, in that $gg'^{-1} \in \ker(f)$.  But to have an injection we’d really like these $g$ and $g'$ to be the same, right?  So why don’t we quotient by the normal subgroup $\ker(f)$?  That would have the effect of meaning that we think of all the elements of $\ker(f)$ as being the identity, or, putting that another way, that if two elements of $G_1$ differ by an element of $\ker(f)$ then we can treat them as the same.  And this is what the theorem tells us (although I haven’t given you a formal proof).

Theorem (The first isomorphism theorem) Let $G_1$ and $G_2$ be groups, and let $f : G_1 \to G_2$ be a homomorphism.  Then $G_1/\ker(f)$ is isomorphic to $\textrm{Im}(f)$.

In the special case that $f$ is surjective (that is, that $\textrm{Im}(f) = G_2$), this says that $G_1/\ker(f)$ is isomorphic to $G_2$, of course.

Why is this a good thing?  Apart from being interesting in its own right, it also gives us a way to prove that two groups are isomorphic.  Let me try to illustrate with a simple example.

We saw the group $\mathbb{Z}/7\mathbb{Z}$ above.  Let’s see how we can show that it’s isomorphic to $\mathbb{Z}_7$ as I defined it above (the group $\{0, 1, 2, 3, 4, 5, 6\}$ under addition (mod 7)).  We can define a map $f : \mathbb{Z} \to \mathbb{Z}_7$ by sending $n$ to $n_7$ where $n_7 \equiv n$ (mod 7) is in $\{0, 1, 2, 3, 4, 5, 6\}$ (the reduction of $n$ (mod 7)).  Then we can check that this is a homomorphism (do it!), and that it’s surjective (do it!).  What’s the kernel?  It’s the set of numbers that get sent to 0, which is precisely $7\mathbb{Z}$, the set of multiples of 7.  So the isomorphism theorem tells us that $\mathbb{Z}/7\mathbb{Z}$ is isomorphic to $\mathbb{Z}_7$.

And here’s a very slightly more interesting example.  We also have groups $\mathbb{Z}/3 \mathbb{Z}$ and $\mathbb{Z}/21 \mathbb{Z}$.  One special case of the Chinese Remainder Theorem says that $\mathbb{Z}/21 \mathbb{Z}$ is isomorphic to the product $(\mathbb{Z}/7\mathbb{Z})\times(\mathbb{Z}/3\mathbb{Z})$.  We could prove that by defining a map from $\mathbb{Z}$ to $(\mathbb{Z}/7\mathbb{Z})\times(\mathbb{Z}/3\mathbb{Z})$.  Let’s send $n$ to the pair $(n + 7\mathbb{Z}, n + 3\mathbb{Z})$.  Then we can check that it’s a homomorphism (do it!), and that it’s a surjection (we can prove this in a standard Chinese remainder theorem sort of way).  What’s the kernel?  It’s all the numbers $n$ so that $n \equiv 0$ (mod 7) and $n \equiv 0$ (mod 3).  That is, $\ker(f) = 21 \mathbb{Z}$, right?  So the isomorphism theorem gives what we want.

Those were both pretty simple examples; there are lots of more interesting ones out there…

This should be in any first course on group theory.  It’s also part of a more general theory, as the (perhaps slightly intimidating) Wikipedia page describes.

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