# Math Help - Prove homomorphic image

1. ## Prove homomorphic image

Suppose that G, G', and G'' are groups. If G' is a homomorphic image of G, and G'' is a homomorphic image of G', prove that G'' is a homomorphic image of G.

Something tells me that it involve transitive property ...

2. Originally Posted by rainyice
Suppose that G, G', and G'' are groups. If G' is a homomorphic image of G, and G'' is a homomorphic image of G', prove that G'' is a homomorphic image of G.

Something tells me that it involve transitive property ...
Obviously.

So, we have $\theta:G\to G',\theta':G'\to G''$ which are both epimorphisms (surjective homomorphisms). So the question is why is $\theta'\circ\theta:G\to G''$ an epimorphism? But this is easy when put this way since the product of surjective maps is surjective and the product of homomoprhisms is a homomorphism. Prove those facts if you don't know them.

3. Originally Posted by Drexel28
Obviously.

So, we have $\theta:G\to G',\theta':G'\to G''$ which are both epimorphisms (surjective homomorphisms). So the question is why is $\theta'\circ\theta:G\to G''$ an epimorphism? But this is easy when put this way since the product of surjective maps is surjective and the product of homomoprhisms is a homomorphism. Prove those facts if you don't know them.
I get what you are saying, and I know how to prove it in term of function. When my professor did one problem of showing a group is one to one or onto ... I have difficulty understanding his strategy of proving it. Group theory is a brand new concept to me.

4. Originally Posted by rainyice
I get what you are saying, and I know how to prove it in term of function. When my professor did one problem of showing a group is one to one or onto ... I have difficulty understanding his strategy of proving it. Group theory is a brand new concept to me.
Is there a question buried in there?

5. Originally Posted by Drexel28
Is there a question buried in there?
Maybe you could tell me how to construct the prove by steps. It is obvious that G --> G'' is onto for all of their (G,G',G'') property or function is onto. However, how can I write it mathematically?

6. Originally Posted by rainyice
Maybe you could tell me how to construct the prove by steps. It is obvious that G --> G'' is onto for all of their (G,G',G'') property or function is onto. However, how can I write it mathematically?
I mean you need to prove that the composition of surjective maps is surjective but this follows since if $G\overset{\theta}{\longrightarrow}G'\overset{\thet a'}{\longrightarrow}G''$ each being a surjection then $(\theta'\circ\theta)(G)=\theta'(\theta(G))=\theta' (G')=G''$ and so surjectivity follows.

The homomorphism property follows since $\theta'(\theta(gg'))=\theta'(\theta(g)\theta(g'))= \theta'(\theta(g))\theta'(\theta(g'))$

7. Originally Posted by Drexel28
Obviously.

So, we have $\theta:G\to G',\theta':G'\to G''$ which are both epimorphisms (surjective homomorphisms). So the question is why is $\theta'\circ\theta:G\to G''$ an epimorphism? But this is easy when put this way since the product of surjective maps is surjective and the product of homomoprhisms is a homomorphism. Prove those facts if you don't know them.
Another question, how do you know that the homomorphism from G to G' and G' to G'' are epimorphism?

8. Originally Posted by rainyice
Another question, how do you know that the homomorphism from G to G' and G' to G'' are epimorphism?
Because we know that $G'$ is a homomorphic image of $G$