$ \scriptstyle\phi $ is a homomorphism from $ \scriptstyle G $ to $ \scriptstyle H $. Thus, we know that $ \scriptstyle\phi(ab) = \phi(a)\phi(b) | a,b\in G $. $ \scriptstyle\sigma $ is a homomorphism from $ \scriptstyle H $ to $ \scriptstyle K $. Thus, we know that $ \scriptstyle\sigma(ab) = \sigma(a)\sigma(b) | a,b\in H $.
Then, $ \scriptstyle\sigma\phi(ab) = \sigma(\phi(ab)) = \sigma(\phi(a)\phi(b)) = \sigma(\phi(a))\sigma(\phi(b)) = \sigma\phi(a)\sigma\phi(b) $. We've shown that the group operation is preserved in a mapping $ \scriptstyle\sigma\phi $ from $ \scriptstyle G $ to $ \scriptstyle K $, so $ \scriptstyle\sigma\phi $ is therefore a homomorphism from $ \scriptstyle G $ to $ \scriptstyle K $. $ \scriptstyle\Box $
- --Nick Rupley 21:02, 1 October 2008 (UTC)
