3
$\begingroup$

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 $G \cong L$?

$\endgroup$
  • 3
    $\begingroup$ Yes, this is correct. $\endgroup$ – Gal Porat Jul 20 '18 at 18:32
  • $\begingroup$ Looks like you understand it correctly to me. $\endgroup$ – The Count Jul 20 '18 at 18:35
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.