# How to prove that a mapping is homomorphic

Let $f:(A, \cdot) \to (B, \ast)$ and $g:(B,\ast) \to (C,\times)$ be Operation preserving maps. Then I must prove that $g \circ f$ is an operation preserving map too. This is what I have so far: Since $f$ is a homomorphism $(A, \cdot)$ and $(B, \ast)$ are groups and $f(x \cdot y)=f(x)\ast f(y)$ Since $(C,\times)$ is a group so $g(f(x)\ast (f(y))=g(f(x)) \times g(f(y))$. Hence $g\circ f$ is homomorphic.

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What is 'operation preserving maps'? Do you want to prove that composition of homomorphisms is homomorphism? – Sigur Aug 17 '12 at 0:45
An operation preserving map I think is another way of saying that a function is homomorphic. My textbook uses the weirdest terms – math101 Aug 17 '12 at 0:49

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You should probably write out the full line: $$(g\circ f)(x\cdot y)=g(f(x\cdot y))=g(f(x)*f(y))=g(f(x))\times g(f(y))=(g\circ f)(x)\times(g\circ f)(y).$$ That's all you have to do, right? – anon Aug 17 '12 at 0:52

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$(g\circ f)(x\cdot y)=g(f(x\cdot y))=g(f(x)\ast f(y))=g(f(x))xg(f(y))=(g\circ f)(x)x(g\circ f)(y)$ – Sigur Aug 17 '12 at 0:53

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It would be better to use `\times` ($\times$) instead of overuse the letter $x$... – anon Aug 17 '12 at 0:53
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