Is the Function λg:G→G Both Onto and One-to-One?

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Homework Statement



Let G be a group and define λg:G→G to be λg(x)=g.x , x\inG.

Show that λg is onto and one-to-one.

The Attempt at a Solution


Suppose g.x=g.x' g-1.g.x=g-1.g.x' which means x=x'.

How should I show that the function is onto?
 
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If y is any member of G, then \lambda(g^{-1}y)= y
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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