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Let $$\phi : G \rightarrow \bar{G}$$ be a homomorphism. If $$|\ker\phi | = n $$ then $\phi$ is an $n$-to-1 mapping from $G$ onto $\phi(G)$.


Since $|\ker\phi | = n$, then $\ker\phi = \{g_1,...,g_n\}$ and $\phi(g_i) = e$ Therefore $\phi^{-1}(e) = g_i\ker\phi$ for all $i = 1,\dots,n$. So all the cosets of $\ker\phi$ have the same number of elements. Is this enough to conclude the result? Or do I need to explain more?

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To be pedantic, you should probably take "$\Rightarrow\phi$" out of your second sentence. – Bey Oct 16 '12 at 3:47
up vote 3 down vote accepted

Let $K=ker(\phi)$. Since $\phi(x)=\phi(y)$ if, and only if, $\phi(xy^{-1})=e$ if, and only if, $xy^{-1}\in K$ it follows that $\phi(x)=\phi(y)$ if, and only if, $y=hx$, $h\in K$. That is, the elements that have the same image as $x$ are precisely the elements in the coset $xK$, and there are precisely $|K|$ many such elements.

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