# Modern Alg. : Subgroups and Cosets

Question: If $f\colon G\to H$ is a group homomorphism, and $|G:\mathrm{Ker}(f)|$ and $|H|$ are coprime, then show that $f(G)=1$

Has anyone experienced the ":" notation before? Does it have something to do with cosets? any hints would be greatly appreciated. Thanks for reading.

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"Modern algebra" was modern some 90 years ago. Nowadays it is just "Algebra" :) –  Mariano Suárez-Alvarez Mar 11 '11 at 1:58

$|G\colon\mathrm{ker}(f)|$ is the index of $\mathrm{ker}(f)$ in $G$; that is, the cardinality of the set of cosets of $\mathrm{ker}(f)$ in $G$. You may have seen it denoted by $[G:\mathrm{ker}(f)]$, or $[G:K]$ where $K$ is an arbitrary subgroup of $G$.

Since you are talking about finite groups here, $|G\colon \mathrm{ker}(f)|$ is just the number of left (equivalently, right) cosets of $\mathrm{ker}(f)$ in $G$.

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