UA-AUG-AUMAT229 Pre-read

Section 16.1 Pre-read

Recall that an isomorphism between groups

G_{1}, G_{2}

is an operation-preserving bijective correspondence between the two groups.

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What if we relax the bijective correspondence bit?

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Definition 16.1.1. Homomorphism.

An operation-preserving mapping $φ : G_{1} \to G_{2}$ from one group to another.

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You will be asked to look at a number of examples of homomorphisms in the Discovery guide following this Pre-read section.

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Warning 16.1.2.

Since a homomorphism is not necessarily a bijective correspondence, we must now keep the following in mind.

It may be possible for two (or more!) elements in $G_{1}$ to correspond to the same element in $G_{2} .$ That is, we might find $φ (x) = φ (y)$ to be true for some $x \neq y$ in $G_{1} .$
There may exist elements in $G_{2}$ that have no corresponding element in $G_{1} .$ That is, we may find examples of an element $z$ in $G_{2}$ for which $φ (x) = z$ never occurs.

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Remark 16.1.3.

With this definition of homomorphism now in hand, an isomorphism becomes a special type of homomorphism. In particular, an isomorphism is a bijective homomorphism.

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Definition 16.1.4. Image of a homomorphism.

The collection of all possible outputs of $φ$ in $G_{2} .$ That is, the collection of all elements $z$ in $G_{2}$ for which it is possible to find at least one example of a corresponding element $x$ in $G_{1}$ with $φ (x) = z .$

We write $im φ$ or $φ (G)$ to denote the image of $φ .$

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Definition 16.1.5. Kernel of a homomorphism.

The collection of all inputs in $G_{1}$ that are mapped to the identity element in $G_{2} .$ That is, the collection of all elements $x$ in $G_{1}$ so that $φ (x) = e_{2},$ where $e_{2}$ represents the identity element in $G_{1} .$

We write $\ker φ$ to denote the kernel of $φ .$

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Warning 16.1.6. Keep it straight!

The image $im φ$ is a subcollection of $G_{2},$ while the kernel $\ker φ$ is a subcollection of $G_{1} .$

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Remark 16.1.7.

There was no need to define either concept of image or kernel when we studied isomorphisms, because, as an isomorphism is always a bijective correspondence, the image of an isomorphism $G_{1} \to G_{2}$ is always the full group $G_{2},$ and the kernel always consists of just $e_{1},$ the identity element in $G_{1} .$