AAR A fundamental isomorphism theorem for groups, rings, vector spaces

Section 1.4 A fundamental isomorphism theorem for groups, rings, vector spaces

If \(X,Y\) are algebraic objects of the same type (group, ring, vector space) and \(\varphi:X \to Y\) is a homomorphism of the appropriate type, then first and foremost all such \(\varphi\) are group homomorphisms, and that is what sets the stage.

It follows that the kernel of the homomorphism is the kernel of the underlying group homomorphism, so \(\ker \varphi = \{x\in X \mid \varphi(x) = e\}\) where \(e\) is the identity of the underlying group, in particular, \(e=0\) for rings and vector spaces.

We also recall that the kernel of any group homomorphism is a normal subgroup; the kernel of any ring homomorphism is a two-sided ideal, and the kernel of any linear map a vector subspace. So in all cases \(X/\ker \varphi\) is an algebraic object of the same type as \(X,\) but always a group.

Theorem 1.4.1. Fundamental theorem for group homomorphisms.

Let \(G, H\) be groups, and let \(\varphi:G \to H\) be a homomorphism. Let \(K \) be any normal subgroup of \(G\) with \(K \subseteq \ker \varphi\text{,}\) and let \(\pi: G\to G/K\) be the usual projection (\(g \mapsto gK\)). Then there exists a unique group homomorphism \(\varphi_*:G/K \to H\text{,}\) so that for all \(g\in G\text{,}\) \(\varphi(g) = \varphi_*(\pi(g))\text{.}\) In this case we say that \(\varphi\) factors through the quotient \(G/K\).

Moreover, the image of \(\varphi_*\) is the same as the image of \(\varphi,\) and \(\ker \varphi_* = \ker\varphi/K.\)

Figure 1.4.2. Factoring a homomorphism through a quotient

Proof.

The condition that \(\varphi(g) = \varphi_*(\pi(g))\) requires that

\begin{equation*} \varphi_*(gK) = \varphi(\pi(g)) = \varphi(g) \end{equation*}

from which it is immediate that there is only one possible definition for \(\varphi_*\) (hence uniquely defined), and also that the images of \(\varphi\) and \(\varphi_*\) are the same.

Presuming the map \(\varphi_*\) is well-defined, we see that it is a group homomorphism since

\begin{equation*} \varphi_*(gK g^\prime K) = \varphi_*(gg^\prime K) = \varphi(gg^\prime) = \varphi(g) \varphi(g^\prime) = \varphi_*(gK) \varphi_*(g^\prime K)\text{,} \end{equation*}

where we have used the group operation on \(G/K\) and that \(\varphi\) is a group homomorphism.

The most important issue is that the map makes sense, i.e., is well-defined. In this case, we must check that

\begin{equation*} \varphi_*(gK) = \varphi_*(gkK) \text{ for any } k \in K. \end{equation*}

But

\begin{equation*} \varphi_*(gkK) = \varphi(gk) = \varphi(g)\varphi(k) = \varphi(g) = \varphi_*(gK) \end{equation*}

since \(\varphi(k) = e\) because \(K \subseteq \ker\varphi.\)

Finally, we compute

\begin{equation*} \ker \varphi_* = \{ gK \in G/K \mid \varphi_*(gK) = \varphi(g) = e_H\}\text{,} \end{equation*}

but that says

\begin{equation*} \ker \varphi_* = \{ gK \in G/K \mid g\in \ker\varphi\} = \ker \varphi/K. \end{equation*}

An immediate corollary of this theorem is the

Theorem 1.4.3. First Isomorphism Theorem.

Let \(G, H\) be groups, and let \(\varphi:G \to H\) be a homomorphism. Then

\begin{equation*} G/\ker\varphi \cong \Im \varphi. \end{equation*}

Proof.

Let \(K=\ker\varphi\text{.}\) The fundamental theorem gives us a surjective homomorphism \(\varphi_*: G/K \to \Im \varphi\) whose kernel is \(K/K = eK = e_{G/K}\text{,}\) the identity of \(G/K\text{,}\) so the map \(\varphi_*\) is injective.

Remark 1.4.4.

The importance of the fundamental theorem cannot be overstated. The main takeaway is that if ever faced with the job of finding a homomorphism \(\Psi: G/N\to H\) you do your best to recognize a homomorphism \(\psi:G \to H\) whose kernel contains \(N.\) In this way, the map you define is not only more natural, but you never have to check that the map \(\Psi:G/N \to H\) is well-defined. Attempting to define the map directly forces you to always check that fact.

Example 1.4.5. The canonical example?

For a positive integer \(n\text{,}\) we have the group \(\Z_n\) consisting of all the congruence classes of integers modulo \(n.\) The set \(n\Z\) (all the integer multiples of \(n\)) is a (normal) subgroup of \(\Z\text{,}\) so \(\Z/n\Z\) is a group. We want to show that

\begin{equation*} \Z/n\Z \cong \Z_n. \end{equation*}

Solution.

Of course we could write down a map taking \(m+n\Z \mapsto [m]_n\text{,}\) but then we would have to check that it is well-defined, a homomorphism, and eventually an isomorphism.

Instead we invoke the fundamental theorem. There is certainly a natural map \(\varphi:\Z\to \Z_n\) which takes an integer \(m\) to its residue class \([m]_n\) modulo \(n.\) It is a homomorphism by virtue that \(\Z_n\) is a group:

\begin{equation*} m+m^\prime \mapsto [m+m^\prime]_n = [m]_n + [m^\prime]_n \end{equation*}

and it is certainly surjective. What is the kernel of \(\varphi\text{?}\)

\begin{equation*} \ker\varphi = \{m \in \Z\mid [m]_n = [0]_n\} = n\Z. \end{equation*}

By the first isomorphism theorem, the result is achieved.

Now what we want to achieve is a fundamental (and first isomorphism) theorem for rings, vector spaces, etc. The key is that all these maps are group homomorphisms at their core and so most of the theorems are already in place.

Theorem 1.4.6. A meta fundamental theorem for homomorphisms.

Let \(X,Y\) be algebraic objects of the same type and \(\varphi: X\to Y\) an associated homomorphism. Suppose that \(Z \subseteq \ker\varphi\) has enough structure so that \(X/Z\) is again an algebraic object of the same type as \(X\) and \(Y.\) Let \(\pi: X \to X/Z\) be the standard projection. Then there is a unique homomorphism \(\varphi_*: X/Z \to Y\) (of the appropriate type) so that \(\varphi = \varphi_*\circ \pi\) whose image is the same as \(\varphi,\) and whose kernel is \(\ker \varphi_* = \ker\varphi/Z.\)

Proof.

Because all these structures have underlying group structures, that map \(\varphi_*\) is uniquely determined, well-defined and has the correct kernel and image. The only thing missing is to verify that \(\varphi_*\) has the additional properties necessary to be a homomorphism of the correct type (e.g., ring or vector space). But this is easily checked. For example, if the objects were rings, then \(Z\) would necessarily be an ideal. The map \(\varphi_*\) takes \(x+Z\) to \(\varphi(x)\text{.}\) We check (using the ring structure of \(X/Z\) and that \(\varphi\) is a ring homomorphism) that

\begin{align*} \varphi_*((x+Z)(x^\prime+Z))\amp = \varphi_*(xx^\prime +Z) = \varphi(xx^\prime)\\ \amp = \varphi(x) \varphi(x^\prime) = \varphi_*(x+Z) \varphi_*(x^\prime+Z). \end{align*}

Remark 1.4.7.

The above meta theorem now tells us that \(\Z/n\Z \cong \Z_n\) as rings as well as abelian groups.

There are further isomorphism theorems that you learned, but they require some new constructions which we give in the next section.