7
$\begingroup$

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.

  • 2
    @anon Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-22

1 Answers 1

2

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.