##### Theorem9.1.1

Let \(G\) and \(G'\) be groups, let \(\phi\) be a homomorphism from \(G\) to \(G'\text{,}\) and let \(K=\Ker \phi\text{.}\) Then for \(a,b\in G\text{,}\) \(aK=bK\) if and only if \(\phi(a)=\phi(b)\text{.}\)

A very powerful theorem, called the *First Isomorphism Theorem*, lets us in many cases identify factor groups (up to isomorphism) in a very slick way. Kernels will play an extremely important role in this. We therefore first provide some theorems relating to kernels.

Let \(G\) and \(G'\) be groups, let \(\phi\) be a homomorphism from \(G\) to \(G'\text{,}\) and let \(K=\Ker \phi\text{.}\) Then for \(a,b\in G\text{,}\) \(aK=bK\) if and only if \(\phi(a)=\phi(b)\text{.}\)

Let \(\phi\) be a homomorphism from group \(G\) to group \(G'\text{.}\) Then \(\phi\) is one-to-one (hence a monomorphism) if and only if \(\Ker \phi=\{e_G\}\text{.}\)

We now prove a theorem that provides the meat and potatoes of the First Isomorphism Theorem.

Let \(G\) and \(G'\) be groups, let \(\phi\) be a homomorphism from \(G\) to \(G'\text{,}\) let \(K=\Ker \phi\text{,}\) let \(N\) be a normal subgroup of \(G\) with \(N\subseteq K\text{,}\) and let \(\Psi\) be the canonical epimorphism from \(G\) to \(G/N\text{.}\) Then the map \(\phibar: G/N \to G'\) defined by \(\phibar(aN)=\phi(a)\) is a well-defined homomorphism, with \(\phibar \circ \Psi=\phi\text{.}\)

We summarize this in the following picture:

Note that the above theorem does *not* state that \(\phibar\) is a monomorphism or an epimorphism. This is because in general it may be neither! We do have the following theorem:

Let \(G\text{,}\) \(G'\text{,}\) \(\phi\text{,}\) \(N\text{,}\) and \(\phibar\) be as defined in the Factorization Theorem. Then

\(\phibar\) is onto (an epimorphism) if and only if \(\phi\) is onto (an epimorphism); and

\(\phibar\) is one-to-one (a monomorphism) if and only if \(N=\Ker \phi\text{.}\)

We are now ready to state the all-important First Isomorphism Theorem, which follows directly from the Factorization Theorem and Theorem 9.1.4.

Let \(G\) and \(G'\) be groups, with homomorphism \(\phi:G \rightarrow G'\text{.}\) Let \(K=\Ker \phi\text{.}\) Then \(G/K \simeq \phi(G)\text{.}\) In particular, if \(\phi\) is onto, then \(G/K\simeq G'\text{.}\)

So to prove that a factor group \(G/N\) is isomorphic to a group \(G'\text{,}\) it suffices to show there exists an epimorphism from \(G\) to \(G'\) that has \(N\) as its kernel.

Letting \(n\in \Z^+\text{,}\) let's identify a familiar group to which \(GL(n,\R)/SL(n,\R)\) is isomorphic. As in Example 8.2.9, the map \(\phi:GL(n,\R)\to \R^*\) defined by \(\phi(A)\) is a homomorphism with kernel \(SL(n,\R)\text{.}\) Moreover, \(Phi\) clearly maps onto \(\R^*\text{:}\) indeed, given \(\lambda \in \R^*\text{,}\) the diagonal matrix having \(\lambda\) in the uppermost left position and 1's elsewhere down the diagonal gets sent to \(\lambda\) by \(\phi\text{.}\) So by the First Isomorphism Theorem, we have \(GL(n,\R)/SL(n,\R) \simeq \R^*\text{.}\)

Let \(G=S_3\times \Z_{52}\) and let \(N=S_3 \times \{0\}\subseteq G\text{.}\) It is straightforward to show that \(N\) is normal in \(G\text{.}\) What is the structure of \(G/N\text{?}\) Well, define \(\phi:G\to \Z_{52}\) by \(\phi((\sigma, a))=a\text{.}\) Then \(\phi\) is clearly an epimorphism and \(\Ker \phi=\{(\sigma,a)\in G\,:a=0\}=N\text{.}\) So \(G/N\) is isomorphic to \(\Z_{52}\text{.}\)

Generalizing the above example, we have the following theorem, whose proof we leave to the reader.

Let \(G=G_1\times G_2 \times \cdots \times G_k\) (where \(k\in \Z^+\)) and let \(N_i\) be a normal subgroup of \(G_i\) for each \(i=1,2,\ldots, k\text{.}\) Then \(N=N_1 \times N_2 \times \cdots \times N_k\) is a normal subgroup of \(G\text{,}\) with \(G/N \simeq G_1/N_1 \times G_2/N_2 \cdots \times G_k/N_k.\)

We provide one more cool example of using the First Isomorphism Theorem. Clearly, since \(\R\) is abelian, \(\Z\) is a normal subgroup of \(\R\text{.}\) What is the structure of \(\R/\Z\text{?}\) Well, in modding \(\R\) out by \(\Z\) we have essentially identified together all real numbers that are an integer distance apart. So we can think of the canonical epimorphism from \(\R\) to \(\R/\Z\) as wrapping up \(\R\) like a garden hose! Thus, one might guess that \(\R/\Z\) has some circle-like structure—but if we want to think of it as a group, we have to figure out what the group structure on such a “circle” would be!

We leave, for a moment, our group \(\R/\Z\text{,}\) and look at how we can consider a circle to be a group.

Recall that for every \(\theta \in \R\text{,}\) \(e^{i\theta}\) is defined to be \(\cos \theta + i\sin \theta\text{.}\) It is clear then that the set \(S^1=\{e^{i\theta} \,:\, \theta\in \R\}\) is the unit circle in the complex plane.

Note that if \(\theta_1, \theta_2\in \R\text{,}\) then \(e^{i\theta_1}=e^{i\theta_2}\) if and only if \(\theta_1-\theta_2 \in 2\pi \Z\text{.}\)

\(S^1\) is a group under the multiplication \(e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)}\text{.}\)

Beautifully, it turns out that our group \(\R/\Z\) is isomorphic to \(S^1\text{.}\)

\(\R/\Z \simeq S^1\text{.}\)