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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

Note by Sai Venkata Raju Nanduri
3 years, 11 months ago

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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. Mark Hennings · 3 years, 11 months ago

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@Mark Hennings So we can't use -ve numbers. Sai Venkata Raju Nanduri · 3 years, 11 months ago

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@Sai Venkata Raju Nanduri 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 \(H\). Mark Hennings · 3 years, 11 months ago

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Same function but we cannot prove to be an isomorphism. Am i right Sai Venkata Raju Nanduri · 3 years, 11 months ago

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@Sai Venkata Raju Nanduri 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). Mark Hennings · 3 years, 10 months ago

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