Non-abelian subgroup of size 6 in A_6

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I need to find just one non-abelian subgroup of size 6 in A_6.

I have started by noting that the subgroup must be isomorphic to D_6 and then tried to use the permutations in D_6 that sends corners to corners. I then came across the problem that the reflection elements in D_6 consist of 2-cycles and hence are not elements of A_6.
I am now very stuck... please help :)
 
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There's really only one nonabelian group of order 6, S_3. Can you think of a sort of natural way to find a copy of S_3 in A_6? Hint: interchange elements two at a time.
 
Thanks a lot. I have found what the subgroup is now: {e, (23)(56), (13)(46), (12)(45), (123)(456), (132)(465)}. I need to justify my answer so I need to show that this is isomorphic to D_6 and therefore is a subgroup. I'll have a go.
 
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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