$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 4.2: Image and Kernel

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Definition 4.2.0

The image of a homomorphism $$\rho:G\rightarrow H$$ is the set $$\{\rho(g) \mid g\in G\}\subset H$$, written $$\rho(G)$$. The kernel of $$\rho$$ is the set $$\{g \mid g\in G, \rho(g)=1 \}$$, written $$\rho^{-1}(1)$$, where $$1$$ is the identity of $$H$$.

Let's try an example. Recall the homomorphism $$\phi:\mathbb{Z}\rightarrow \mathbb{Z}$$, defined by $$\phi(n)=2n$$ for any $$n\in \mathbb{Z}$$. The image of $$\phi$$ is the set of all even integers. Notice that the set of all even integers is a subgroup of $$\mathbb{Z}$$. The kernel of $$\phi$$ is just $$0$$.

Here's another example. Consider the map $$\phi: \mathbb{Z}_3\rightarrow \mathbb{Z}_6$$ given by $$\phi(n)=2n$$. So $$\phi(0)=0$$, $$\phi(1)=2$$, and $$\phi(2)=4$$. This is actually a homomorphism (of additive groups): $$\phi(a+b) = 2(a+b) = 2a+2b =\phi(a)+\phi(b)$$. The image is the set $$\{0, 2, 4\}$$, and, again, the kernel is just $$0$$.

And another example. There's a homomorphism $$\rho: \mathbb{Z}_6\rightarrow \mathbb{Z}_3$$ given by $$\rho(a)=a%3$$ (divide by 3 and keep the remainder). Then $$\rho(0)=0$$, $$\rho(1)=1$$, $$\rho(2)=2$$, $$\rho(3)=0$$, $$\rho(4)=1$$ and finally $$\rho(5)=2$$. You can check that this is actually a homomorphism, whose image is all of $$\mathbb{Z}_3$$ and whose kernel is $$\{0, 3\}$$.

So the image is the set of everything in $$H$$ which has something in $$G$$ which maps to it. The kernel is the set of elements of $$G$$ which map to the identity of $$H$$. The kernel is a subset of $$G$$, while the kernel is a subset of $$H$$. In fact, both are subgroups!

Proposition 4.2.1

The image $$\rho(G)$$ is a subgroup of $$H$$. The kernel $$\rho^{-1}(1)$$ is a subgroup of $$G$$.

To see that the kernel is a subgroup, we need to show that for any $$g$$ and $$h$$ in the kernel, $$gh$$ is also in the kernel; in other words, we need to show that $$\rho(gh)=1$$. But that follows from the definition of a homomorphism: $$\rho(gh)=\rho(g)\rho(h)=1\cdot 1=1$$. We leave it to the reader to find the proof that the image is a subgroup of $$H$$.

Exercise 4.2.2

Show that for any homomorphism $$\rho: G\rightarrow H$$, $$\rho(G)$$ is a subgroup of $$H$$.

We can use the kernel and image to discern important properties of $$\rho$$ as a function.

Proposition 4.2.3

Let $$\rho:G\rightarrow H$$ be a homomorphism. Then $$\rho$$ is injective (one-to-one) if and only if the kernel $$\rho^{-1}(1)=\{1\}$$.

Proof 4.2.4

If we assume $$\rho$$ is injective, then we know (from the exercise in the last section) that $$\rho^{-1}(1)=\{1\}$$. For the reverse direction, suppose $$\rho^{-1}(1)=\{1\}$$, and assume (for contradiction) that $$\rho$$ is not injective. Then there exist $$x\neq y$$ with $$\rho(x)=\rho(y)$$. But then $$\rho(x)\rho(y)^{-1}=\rho(xy^{-1})=1$$. Since $$x\neq y$$, $$xy^{-1}\neq 1$$, giving a contradiction.

The kernel is actually a very special kind of subgroup.

Proposition 4.2.5

Let $$\rho: G\rightarrow H$$ be a homomorphism, and let $$K$$ be the kernel of $$\rho$$. Then for any $$k\in K$$ and $$x\in G$$, we have $$xkx^{-1}\in K$$.

Proof 4.2.6

The proof is a simple computation: $$\rho(xkx^{-1}) = \rho(x)\rho(k)\rho(x^{-1}) = \rho(x)1\rho(x^{-1}) = 1$$. Therefore, $$xkx^{-1}$$ is in the kernel of $$\rho$$.

## Contributors

• Tom Denton (Fields Institute/York University in Toronto)