An injective group homomorphism means the group is isomorphic to its image

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This is a corollary to the first group isomorphism theorem
This page is also First-year friendly, which means it is especially verbose.


Suppose [ilmath]\varphi:A\rightarrow B[/ilmath] is any injective group homomorphism, then[1]:

  • [ilmath]A\cong\text{Im}(\varphi)[/ilmath] - there exists a group isomorphism, [ilmath]\theta:A\rightarrow\text{Im}(\varphi)[/ilmath]


As this page is first year friendly, we first include an outline.


We know already by the first group isomorphism theorem that [ilmath]A/\text{Ker}(\varphi)\cong\text{Im}(\varphi)[/ilmath]. Recall that a isomorphism of groups is an equivalence relation, and equivalence relations have a transitive property.

If we show that [ilmath]A\cong A/\text{Ker}(\varphi)[/ilmath] and combine this with what we already know, [ilmath]A/\text{Ker}(\varphi)\cong\text{Im}(\varphi)[/ilmath] we see:

  • [ilmath]A\cong\text{Im}(\varphi)[/ilmath]

Exactly what we are looking for.

Note additionally, that we know [ilmath]A\cong A/\text{Ker}(\varphi)[/ilmath] and we want [ilmath]A\cong\text{Im}(\varphi)[/ilmath]. If these are true then it implies that:

  • [ilmath]A\cong A/\text{Ker}(\varphi)[/ilmath]

This suggests looking to prove [ilmath]A\cong A/\text{Ker}(\varphi)[/ilmath] is the way to go, as if this is false, and we cannot prove it, then the statement is nonsense (as the statement implies we'd have this), this is worth mentioning as if we find proving this difficult, or were to reach a contradiction, we'd know that the statement is nonsense. (This page is first year friendly, so proof remarks are included).

Details of proof

As discussed, we will show:

We have a surjective group homomorphism already, [ilmath]\pi:A\rightarrow A/\text{Ker}(\varphi)[/ilmath] (see: the canonical projection of the quotient group for details about {{M|\pi}), we only need to show it is injective, then it is bijective and a bijective group homomorphism is a group isomorphism.

Proof that [ilmath]\pi[/ilmath] is injective

We wish to show that [ilmath]\forall a,b\in A[\pi(a)=\pi(b)\implies a=b][/ilmath]

  • Let [ilmath]a,b\in A[/ilmath] be given
    1. Suppose [ilmath]\pi(a)\ne\pi(b)[/ilmath] - by the definition of implies we do not care about the truth or falsity of [ilmath]a=b[/ilmath], either way the implication is true, so we are done.
    2. Suppose [ilmath]\pi(a)=\pi(b)[/ilmath] - then by the definition of implies we require that [ilmath]a=b[/ilmath] for the statement to be true.
      • In the proof of the group factorisation theorem we require that [ilmath]\pi(a)=\pi(b)\implies\varphi(a)=\varphi(b)[/ilmath] in order to be able to factor. We can use that again here.
        • We see [ilmath]\pi(a)=\pi(b)\implies\varphi(a)=\varphi(b)[/ilmath], but by hypothesis, [ilmath]\varphi:A\rightarrow B[/ilmath] is injective, so:
          • [ilmath]\forall a,b\in A[\varphi(a)=\varphi(b)\implies a=b][/ilmath] (definition of being injective)
        • Thus [ilmath]\pi(a)=\pi(b)\implies\varphi(a)=\varphi(b)\implies a=b[/ilmath]
      • Or simply: [ilmath]\pi(a)=\pi(b)\implies a=b[/ilmath] as required.

Now [ilmath]\pi:A\rightarrow A/\text{Ker}(\varphi)[/ilmath] is a group isomorphism, thus [ilmath]A\cong A/\text{Ker}(\varphi)[/ilmath]

See also


  1. Abstract Algebra - Pierre Antoine Grillet