Definition of a group with redundancy?

  • Thread starter Thread starter Syrus
  • Start date Start date
  • Tags Tags
    Definition Group
Syrus
Messages
213
Reaction score
0

Homework Statement



I would define a group as follows:

A group consists of a non-empty set G together with a binary operation (say, *) on G such that the following axioms hold:

G1...G2...G3...



Homework Equations





The Attempt at a Solution



I know this is a trivial question, but doesn't a binary operation on a set (here G) necesserily imply that the set is closed under that binary operation? The only reason i ask is that many of the definitions of groups I have come across include both that * is a binary operation under which G is closed. Isn't this redundant?
 
Last edited:
Physics news on Phys.org
It all depends on how you define a binary operation. I would define it as a function

*:G\times G\rightarrow G

If you define a binary operation like that, then the property you mention is indeed redundant.

I guess, that many authors include that property because most readers aren't yet ready to view a binary operation as a function. So to make it easy on them, they include the axiom that G is closed.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top