Proof of the inverse of an inverse

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Homework Statement


In any group, verify directly from the axioms that
(a) inverse of the inverse of x= x
(b) (xy)^inverse = (inverse y)(inverse x) for all x,y in G. (note the reversal here)


The Attempt at a Solution


(a) I tried to use the axiom that xe=x=ex but I don't know where to go from there.
(b) I don't know how to start it.
 
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If b is the inverse of a then ab=ba=e. If a is the inverse of b then ba=ab=e. They are the SAME THING. Think of what that means if a=x and b=x^(-1).
 
So my proof should conclude with noticing that x is the inverse of x-inverse?
 
fk378 said:
So my proof should conclude with noticing that x is the inverse of x-inverse?

Well, yes. It is, isn't it?
 
For the second question - what happens if you multiply xy with the object you need to show is its inverse?
 
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...
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