when i read homomorphism concept they stated that if G is the all non zero real number and f:G -> G such that f(x)=x^2 is a homomorphism.my Q is function cannot be a one-one then how would be it is a homomorphism

A homorphism is a function \(f\) between groups such that \(f(xy) = f(x)f(y)\), and \(f:(0,\infty)\to(0,\infty)\) given by \(f(x) = x^2\) certainly is one. A homomorphism does not have to be one-to-one, or even onto.

In group theory, a bijective homomorphism is called an isomorphism. You are getting confused, perhaps, with a homeomorphism, a bijective continuous map with continuous inverse between two topological spaces.

Yes we can. In this case, being a homomorphism just means that \((xy)^2 = x^2y^2\), which is true for any ordinary numbers (\(f\) would not be a homomorphism on most matrix groups, though, since matrix multiplication is not commutative). Thus \(f\) will be a homomorphism between any pair of groups which involve ordinary numbers and ordinary multiplication. \(G = (0,\infty)\) and \(H = \mathbb{R}\backslash \{0\}\) are two such groups, and \(f\) is a homomorphism when regarded as a map from \(G\) to \(G\), or as a map from \(H\) to \(H\). It is even a homomorphism regarded as a map from \(G\) to \(H\).

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homorphismis a function \(f\) between groups such that \(f(xy) = f(x)f(y)\), and \(f:(0,\infty)\to(0,\infty)\) given by \(f(x) = x^2\) certainly is one. A homomorphism does not have to be one-to-one, or even onto.In group theory, a bijective homomorphism is called an

isomorphism. You are getting confused, perhaps, with ahomeomorphism, a bijective continuous map with continuous inverse between two topological spaces.Log in to reply

So we can't use -ve numbers.

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Yes we can. In this case, being a homomorphism just means that \((xy)^2 = x^2y^2\), which is true for any ordinary numbers (\(f\) would not be a homomorphism on most matrix groups, though, since matrix multiplication is not commutative). Thus \(f\) will be a homomorphism between any pair of groups which involve ordinary numbers and ordinary multiplication. \(G = (0,\infty)\) and \(H = \mathbb{R}\backslash \{0\}\) are two such groups, and \(f\) is a homomorphism when regarded as a map from \(G\) to \(G\), or as a map from \(H\) to \(H\). It is even a homomorphism regarded as a map from \(G\) to \(H\).

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Same function but we cannot prove to be an isomorphism. Am i right

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It is an isomorphism from \(G\) to \(G\), but not from \(H\) to \(H\), nor from \(G\) to \(H\) (although it is injective in the last case).

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