Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Approach:

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?

share|improve this question
    
To be pedantic, you should probably take "$\Rightarrow\phi$" out of your second sentence. –  Bey Oct 16 '12 at 3:47

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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