Abstract Algebra: Solving Stumping Questions | αη = β and G is Abelian

m-chan
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I have 2 algebra questions which are stumping me, I just can't seem to use my notes to figure them out!

1. Let α, β ∈ S17 where α = (17 2)(1 2 15 17 ), β = (2 3 16)(6 16 17 ).
Determine η, as a product of disjoint cycles, where αη = β.

2. Let G be a group in which a^2 = 1 for all a ∈ G. Prove that G is Abelian.
Hint: Consider (ab)^2.

HELP PLEASE :(
 
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For 2, consider what (ab)2 equals.
 
Right, I've figured out 2, thanks Mark44 and I've done some of 1, but I'm stuck at the end of the question.

I have η= (2 17)(17 15 2 1)(2 3 16)(6 16 17), but I'm not sure if that's right though. I also don't know where to go from there.
 
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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