# Section9.1The First Isomorphism Theorem

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.

##### Proof

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

##### Proof

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:

##### Proof

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

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.

##### Example9.1.6

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 $GL(n,\R)/SL(n,\R) \simeq \R^*\text{.}$

##### Example9.1.7

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.

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.

##### Remark9.1.9

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

##### Proof

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