Non-Abelian Group: (ab)^2 = (b^2)(a^2)?

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in a non abelian group does (ab)^2 = (b^2)(a^2)?
this isn't homework, my semester is over.
 
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1. Just expand (ab)^2

2. Or look in the smallest non abelian group there is: S_3
 
it expands to abab , but how come (ab)^-1 = (b^-1)(a^-1)?
 
because inversion reverses the order of the elements, it is an anti-automorphism and has nothing to do with the commutativity or otherwise of the group.

we know the inverse of ab is a c such that abc=1, bc=a^{-1}, and then c=b^{-1}a^{-1}
 
ok, thanks.
 
math-chick_41 said:
it expands to abab , but how come (ab)^-1 = (b^-1)(a^-1)?

What do you get if you multiply (ab)(b-1a-1)?
 
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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