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 $\displaystyle \theta:G\to G',\theta':G'\to G''$ which are both epimorphisms (surjective homomorphisms). So the question is why is $\displaystyle \theta'\circ\theta:G\to G''$ 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 mean you need to prove that the composition of surjective maps is surjective but this follows since if $\displaystyle G\overset{\theta}{\longrightarrow}G'\overset{\thet a'}{\longrightarrow}G''$ each being a surjection then $\displaystyle (\theta'\circ\theta)(G)=\theta'(\theta(G))=\theta' (G')=G''$ and so surjectivity follows.
The homomorphism property follows since $\displaystyle \theta'(\theta(gg'))=\theta'(\theta(g)\theta(g'))= \theta'(\theta(g))\theta'(\theta(g'))$