First group isomorphism theorem
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- Note:
- Overview of the group isomorphism theorems - all 3 theorems in one place
- Overview of the isomorphism theorems - the first, second and third are pretty much the same just differing by what objects they apply to
First isomorphism theorem | |
isomorphism. | Where [ilmath]\theta[/ilmath] is an|
Properties | |
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something |
Statement
Let [ilmath](G,*)[/ilmath] and [ilmath](H,*)[/ilmath] be groups. Let [ilmath]\varphi:G\rightarrow H[/ilmath] be a group homomorphism, then[1]:
- [ilmath]G/\text{Ker}(\varphi)\cong\text{Im}(\varphi)[/ilmath]
- Explicitly we may state this as: there exists a group isomorphism between [ilmath]G/\text{Ker}(\varphi)[/ilmath] and [ilmath]\text{Im}(\varphi)[/ilmath].
Note: the special case of [ilmath]\varphi[/ilmath] being surjective, then [ilmath]\text{Im}(\varphi)=H[/ilmath], so we see [ilmath]G/\text{Ker}(\varphi)\cong H[/ilmath]
Useful corollaries
- An injective group homomorphism means the group is isomorphic to its image
- If [ilmath]\varphi:A\rightarrow B[/ilmath] is an injective group homomorphism then [ilmath]A\cong \text{Im}(\varphi)[/ilmath]
- A surjective group homomorphism means the target is isomorphic to the quotient of the domain and the kernel
- If [ilmath]\varphi:A\rightarrow B[/ilmath] is a surjective group homomorphism then [ilmath]A/\text{Ker}(\varphi)\cong B[/ilmath]
Proof
Notes
References
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