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

No vote yet

0 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestA

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. – Mark Hennings · 3 years, 11 months agoLog in to reply

– Sai Venkata Raju Nanduri · 3 years, 11 months ago

So we can't use -ve numbers.Log in to reply

– Mark Hennings · 3 years, 11 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 \(H\).Log in to reply

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

Log in to reply

– Mark Hennings · 3 years, 10 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).Log in to reply