# group

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
4 years, 9 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.

- 4 years, 9 months ago

So we can't use -ve numbers.

- 4 years, 9 months ago

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 $$H$$.

- 4 years, 9 months ago

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

- 4 years, 9 months ago

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).

- 4 years, 9 months ago