The first isomorphism theorem is probably the most important result one will learn in a first course on algebra. We will state and prove the theorem for groups. There is, of course, an analogous result for rings. We will state the theorem for rings (the proof is identical to that for groups) and then look at an example using the theorem.

Theorem. If $\varphi : G \to H$ is a homomorphism of groups, then $\text{ker}(\varphi)$ is a normal subgroup of $G$ and $G/\text{ker}(\varphi)\cong\varphi(G)$.

The proof is not terribly complicated; in fact, it is rather “computational”, meaning, we just have to verify a few definitions. It is just a matter of checking that the kernel satisfies the definition of normal and then defining a function from $G/\text{ker}(\varphi)$ to $\varphi(G)$ and checking that the function satisfies the definition of an isomorphism. The “hardest” part of the proof, I think, is actually defining a function from $G/\text{ker}(\varphi)$ to $\varphi(G)$. However, as we will see, the choice of such a function is quite natural.

Proof. First, we will show that the kernel of $\varphi$ is a normal subgroup of $G$. We check the following:

Nonempty. Since $\varphi$ is a homomorphism, it must map the identity in $G$ to the identity in $H$. Thus, $1_G\in\text{ker}(\varphi)$ which shows that the kernel is nonempty.

Closure. Let $a$ and $b$ be in the kernel of $\varphi$. We want to show that $ab^{-1}$ is in the kernel too. We compute that

$\varphi(ab^{-1})=\varphi(a)\varphi(b)^{-1}=(1_H)(1_H)^{-1}=1_H.$

Thus, $ab^{-1}$ is indeed in the kernel of $\varphi$, which shows that $\text{ker}(\varphi)$ is a subgroup of $G$.

Normal. Let $g$ be an element of $G$ and let $k$ be an element in the kernel of $\varphi$. It suffices to show that $gkg^{-1}$ is also in the kernel. We have that

$\varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}=\varphi(g)1_H\varphi(g)^{-1}=1_H.$

Thus, $gkg^{-1}$ is in the kernel of $\varphi$ as desired. This shows that $\text{ker}(\varphi)$ is a normal subgroup of $G$.

Next, we will show that $G/\text{ker}(\varphi)$ is isomorphic to $\varphi(G)$. Define $\psi : G/\text{ker}(\varphi) \to \varphi(G)$ by $\psi(g\text{ker}(\varphi))=\varphi(g)$. I claim that $\psi$ is a well-defined isomorphism. Indeed, we check the following:

Well-Defined. Suppose that $a\text{ker}(\varphi)=b\text{ker}(\varphi)$. Then $a=bk$ for some element $k$ in the kernel of $\varphi$. We wish to show that $\psi(a\text{ker}(\varphi))=\psi(b\text{ker}(\varphi)).$ Evaluating $\psi(a\text{ker}(\varphi))$ we see,

$\psi(a\text{ker}(\varphi))=\varphi(a)=\varphi(bk)=\varphi(b)\varphi(k)=\varphi(b)1_H=\psi(b\text{ker}(\varphi)).$

Thus, $\psi$ is well defined.

Homomorphism. Let $a\text{ker}(\varphi)$ and $b\text{ker}(\varphi)$ be cosets in the factor group $G/\text{ker}(\varphi)$. By multiplying cosets we observe that $a\text{ker}(\varphi)b\text{ker}(\varphi)=ab\text{ker}(\varphi).$ Consequentially,

$\psi(a\text{ker}(\varphi)b\text{ker}(\varphi))=\varphi(ab) =\varphi(a)\varphi(b) =\psi(a\text{ker}(\varphi))\psi(b\text{ker}(\varphi)).$

Thus, $\psi$ is a homomorphism.

Injective. Suppose that $\psi(a\text{ker}(\varphi))=\psi(b\text{ker}(\varphi))$. We wish to show that $a\text{ker}(\varphi)=b\text{ker}(\varphi)$. Consider the following,

$\varphi(a)=\psi(a\text{ker}(\varphi))=\psi(b\text{ker}(\varphi))=\varphi(b).$

Since $\varphi(b)$ is an element of $H$ and $H$ is a group, $\varphi(b)^{-1}$ exists. Therefore,

$\varphi(ab^{-1})=\varphi(a)\varphi(b)^{-1}=1_H,$

which shows that $ab^{-1}$ is in the kernel of $\varphi$. It follows that $a\text{ker}(\varphi)=b\text{ker}(\varphi)$ and so $\psi$ is injective.

Surjective. Let $x$ be an element in the image of $\varphi$. Then there exists an element $g$ in $G$ such that $\varphi(g)=x$. Thus, $\psi(g\text{ker}(\varphi))=\varphi(g)=x$ which shows that $\psi$ is surjective.

We conclude that $\psi$ is a bijective homomorphism, hence an isomorphism. $\Box$

We now state the analogous result for rings.

Theorem. If $\varphi : R \to S$ is a homomorphism of rings, then $\text{ker}(\varphi)$ is an ideal of $R$ and $R/\text{ker}(\varphi)\cong\varphi(R)$.

Example. Let $R$ be a commutative ring with unity. Prove that $R[x] / \langle x \rangle \cong R.$

Define the function $\varphi:R[x]\to R$ to be the evaluation homomorphism $\varphi(f(x))=f(0)$ for any $f(x)$ in $R[x]$. By the first isomorphism theorem, it suffices to check that $\varphi$ is a surjective homomorphism with $\text{ker}\varphi=\langle x \rangle$. We check the following:

Homomorphism. Let $f(x)$ and $g(x)$ be in $R[x]$. We compute that

$\varphi((f+g)(x))=(f+g)(0)=f(0)+g(0)=\varphi(f(x))+\varphi(g(x))$

$\varphi((fg)(x))=(fg)(0)=f(0)g(0)=\varphi(f(x))\varphi(g(x))$.

Thus, $\varphi$ is indeed a ring homomorphism.

Surjective. Let $r$ be an element in $R$. Then $f(x)=x+r$ is in $R[x]$ and $\varphi(f(x))=f(0)=0+r=r$ which shows that $\varphi$ is surjective.

Kernel. Let $f(x)$ be in the kernel of $\varphi$. Then $f(0)=0$ which means that $x$ is a factor of $f(x)$ and so $f(x)$ is in the ideal $\langle x \rangle$. Therefore, $\text{ker}\varphi\subseteq\langle x \rangle$. Now, let $g(x)$ be in the ideal $\langle x \rangle$. Then $g(x)=xf(x)$ for some $f(x)$ in $R[x]$. Thus, $\varphi(g(x))=g(0)=0f(0)=0$ which shows that $g(x)$ is in the kernel of $\varphi$. Therefore, $\langle x \rangle\subseteq\text{ker}\varphi$. We conclude that $\text{ker}\varphi=\langle x \rangle$.

By the first isomorphism theorem we have that $R[x] / \langle x \rangle \cong R.$ $\Box$

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