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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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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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Thanks guys for helping me out :D – math101 Aug 17 '12 at 0:56
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@anon Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Jun 22 '13 at 8:55
up vote 2 down vote accepted

The composition, in order to be "operation preserving", must fulfill $$(g \circ f)(x \cdot y) = g(f(x\cdot y)) = (g \circ f)(x) \times (g \circ f)(y).$$

Since $f$ preserves the operation $\cdot$, we have $g(f(x\cdot y) = g(f(x) \ast f(y))$, where $f(x)$ and $f(y)$ are elements of $B$ and since $g$ preserves the operation $\ast$, we get $g(f(x) \ast f(y)) = g(f(x)) \times g(f(y))$ and we're done.

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