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.
How to prove that a mapping is homomorphic
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abstract-algebra
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1 Answers
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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.