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suppose you have two sets $G_1$ and $G_2$ with same cardinality

in $G_2$ you have the group structure and there is a bijective map from $G_1$ to $G_2$ this is just a set map. can we define a binary operation on $G_1$ with the help of binary operation on $G_2$ so that $G_1$ also become a group?

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This is a standard "transport of structure" argument. Conceptually, $G_1$ and $G_2$ are the "same" for all (non-set-theoretic) intents and purposes because they are in bijection. – Zhen Lin May 20 '12 at 10:31
Yes. If $f:G_1\to G_2$ is the given bijection, then letting $\circ_1$ be defined by $x\circ_1 y=f^{-1}(f(x)\circ_2 f(y))$ gives you a group structure on $G_1$. – Michael Greinecker May 20 '12 at 10:32
but how I will be sure that $x._1y$ is in $G_1$?I mean closoure property holds? – Un Chien Andalou May 20 '12 at 10:38
I don't see your problem. $f(x)$ and $f(y)$ are in $G_2$, so $f(x)circ_2 f(y)$ is in $G_2$, since $G_2$ has a group structure. And since $f$ is bijective, $f^{-1}(f(x)circ_2 f(y))$ is a unique element in $G_1$. – Michael Greinecker May 20 '12 at 10:41
thank you dear sir :) – Un Chien Andalou May 20 '12 at 10:42
up vote 6 down vote accepted

HINT: There’s really only one reasonable thing to try, and it works. Let $f:G_1\to G_2$ be your bijection, and let $\cdot$ be the group operation in $G_2$. You want to use $f$ to define a group operation $\odot$ on $G_1$. Suppose that $x,y\in G_1$; what would $x\odot y$ have to be in order for $f$ to be a group isomorphism?

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