An injective group homomorphism means the group is isomorphic to its image
 This is a corollary to the first group isomorphism theorem
 This page is also Firstyear friendly, which means it is especially verbose.
Contents
Statement
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]
Proof
As this page is first year friendly, we first include an outline.
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:
 [ilmath]A\cong A/\text{Ker}(\varphi)[/ilmath] as then by the transitive property of equivalence relations (and isomorphism of groups being an equivalence relation) we see that:
 [ilmath][A\cong A/\text{Ker}(\varphi)\wedge A/\text{Ker}(\varphi)\cong\text{Im}(\varphi)]\implies A\cong\text{Im}(\varphi)[/ilmath], we get [ilmath]A/\text{Ker}(\varphi)\cong \text{Im}(\varphi)[/ilmath] by application of the first group isomorphism theorem.
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
 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.
 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]
 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:
 Or simply: [ilmath]\pi(a)=\pi(b)\implies a=b[/ilmath] as required.
 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.
Now [ilmath]\pi:A\rightarrow A/\text{Ker}(\varphi)[/ilmath] is a group isomorphism, thus [ilmath]A\cong A/\text{Ker}(\varphi)[/ilmath]
 We also know [ilmath]A/\text{Ker}(\varphi)\cong\text{Im}(\varphi)[/ilmath] by the first group isomorphism theorem
 By the transitive property of equivalence relations (and isomorphism of groups being an equivalence relation) we see:
 [ilmath]A\cong A/\text{Ker}(\varphi)\cong\text{Im}(\varphi)[/ilmath]
 By the transitive property of equivalence relations (and isomorphism of groups being an equivalence relation) we see:
 So [ilmath]A\cong \text{Im}(\varphi)[/ilmath]  as required.
See also
 First group isomorphism theorem
 A surjective group homomorphism means the target is isomorphic to the quotient of the domain and the kernel  another corollary, this page deals with injective [ilmath]\varphi[/ilmath], that page deals with surjective [ilmath]\varphi[/ilmath]
 Group factorisation theorem
 Group isomorphism theorems
References
