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I have an interesting question from my abstract-algebra book. I would like to understand the full solution of it.

Let $X$ be a a set, $G$ a group and $f\colon G \to X$ bijection.

On $X$ we define $x \circ y = f\bigl( f^{-1}(x) *f^{-1}(y)\bigr) $

What structure does $(X,\circ)$ have and why?

Rasmus
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Learner
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  • See also http://math.stackexchange.com/questions/373731/showing-that-g-is-a-group-under-an-alternative-operation/373743#373743. – lhf Nov 27 '13 at 23:01

1 Answers1

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The bijection allows you to transport the group structure from $G$ to $X$. By precisely the formula you have written down, $X$ is a group such that the map $f$ becomes a group isomorphism. It is a good exercise to check these elementary facts.

Rasmus
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