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I was thinking about the first isomorphism theorem, and something curious occurred to me. If φ is an injective homomorphism from group G to group H, then the kernel is trivial. But then G/kerφ is isomorphic to H, yes? But if kerφ = {e}, then isn't G/kerφ = G/{e} = G? But then wouldn't that make G isomorphic to H?

That doesn't make sense though because that would imply that every injective homomorphism is an isomorphism. What am I doing wrong?

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    $\begingroup$ The isomorphism theorem says that $G/\ker\phi \cong \mathrm{Im}\ \phi$. Taking $H$ as you have assumes that $\phi$ is surjective. $\endgroup$ – Mathmo123 Dec 17 '14 at 0:31
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You makes some mistake. First isomorphism theorem states if $\phi:G\to H$ is a homomorphism then $$G/\operatorname{ker}\phi\cong\operatorname{Im}\phi$$ so $G/\operatorname{ker}\phi$ is not isomorphic to $H$ unless $\phi$ is epimorphism (i.e. surjective homomorphism).

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The first isomorphism theorem says that if you have $\varphi: G \to H$ a surjective homomorphism, then $$G/\ker \varphi \cong H.$$

If $\varphi$ is injective, then $G/\ker \varphi \cong G \cong H$. And that's right, because $\varphi$ would be the isomorphism (both injective and surjective) between $G$ and $H$ itself.

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