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Given an exact sequence $$ 0\to A\to B\overset\alpha\to C\to\dots $$ is $$ \ker\alpha\cong A $$ true? I feel that the answer is trivially "yes, by definition" but I couldn't find this expression anywhere so now I'm in doubt.

Anyway, I'm interested in the case where these are groups, and ideally the $\cong$ refers to a group isomorphism.

AccidentalFourierTransform
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Of course. Since the map $A \to B$ is monic, its image is isomorphic to $A$. But its image is equal to $\mathrm{ker}(\alpha)$ by exactness.

Randall
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