# first isomorphism problem - injective homomorphism

I am learning algebra and I do not understand the first isomorphism theorem correctly.

I have an injective group homomorphism $\phi: G \to H$. Moreover I have given that $im(\phi)\cong L$, with $L$ a group.

By the first isomorphism theorem it holds: $L \cong G/ker(\phi)$. Since, $\phi$ is injective it holds $ker(\phi)=\{e\}$. Does this mean $G \cong L$?

• Yes, this is correct. – Gal Porat Jul 20 '18 at 18:32
• Looks like you understand it correctly to me. – The Count Jul 20 '18 at 18:35

The assumption ${\rm im}(\phi)\cong L$ says that $\phi \colon G\rightarrow L$ is surjective, the other that it is injective. Hence it is an isomorphism of groups, i.e., $G\cong L$. The first isomorphism theorem should not come to a different conclusion. So you are right.
Whenever you have a homomorphism $\phi\colon G\to H$, you can factor it as $\phi=\beta\circ\alpha$, where $\alpha$ is surjective and $\beta$ is injective; the trick is to define $$\alpha\colon G\to\operatorname{im}(\phi), \qquad \alpha(x)=\phi(x)$$ and take $\beta$ as the inclusion map $\operatorname{im}(\phi)\to H$.
In your particular case, $\alpha$ is also injective, so it is an isomorphism. Composing it with an isomorphism $\gamma\colon\operatorname{im}(\phi)\to L$ (existing by assumption), we conclude that $\gamma\circ\alpha\colon G\to L$ is an isomorphism.