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

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Comments

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 - 7 years, 8 months ago

So we can't use -ve numbers.

sai venkata raju nanduri - 7 years, 8 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 $H$ .

Mark Hennings - 7 years, 8 months ago

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

sai venkata raju nanduri - 7 years, 8 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).

Mark Hennings - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$