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 $G$ and $G'$ be groups and $\varphi : G \rightarrow G'$ a homomorphism. Suppose that $N$ is a normal subgroup of $G$ with the property that $N$ contained in $\text{ker}(\varphi)$. Define $F : G/N \rightarrow G'$ as $F(Ng) = \varphi(g)$.

a. Show that $F$ is a well-defined function.

b. Show that $F$ is a homomorphism.

For part b. I have $F(Ngh) = \varphi(gh) = \varphi(g)\varphi(h)$ so $F$ is a homomorphism.

For part a. I know that I need to somehow show that for all $Ng=Nh$ are elements of G/N, then $F(Ng) = F(Nh)$

So if $Ng = Nh$ then $gh^{-1}$ is an element of $N \subseteq \text{ker}(\varphi)$ so

$\varphi(gh^{-1}) = e'$

$\varphi(g)\varphi(h)^{-1} = e'$

$\varphi(g)\varphi(h)$

so $F(Ng) = F(Nh)$

share|improve this question
    
Do you know what it means for a function to be well-defined? What have you already tried? –  Daniel Hast Dec 5 '13 at 22:02
    
I have edited my question. –  user111230 Dec 5 '13 at 22:16
    
I don't understand your statement of what you need to show for part (a). Just to clarify, for part (a) you must show that if $Ng = Nh$, then $F(Ng) = F(Nh)$, i.e. $\varphi(g) = \varphi(h)$. Now you can use the fact that $g \in Nh$. –  dc631 Dec 6 '13 at 0:58
1  
@DanielHast: I think this is a tricky question. When we say that a function is well-defined we don't really mean that a function is well-defined (every function is well-defined, a least in the naive sense), but rather that a formula, construction or something defines unambiguously an object which is, in fact, a function. –  tomasz Dec 6 '13 at 3:05
    
I just edited my question and added my work to showing part a. Feedback would be greatly appreciated. –  user111230 Dec 6 '13 at 3:10

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.