# 4.2: Image and Kernel

- Page ID
- 675

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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 |
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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 |
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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 |
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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 |
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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 |
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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)