Group Actions and Normal Subgroups

  • Thread starter Thread starter Kalinka35
  • Start date Start date
  • Tags Tags
    Group Normal
Kalinka35
Messages
48
Reaction score
0

Homework Statement


Let H be a subgroup of G and let X be the set of left cosets of H in G.
Show that the formula g(xH) = gxH is an action of G on X.
Prove that H is a normal subgroup of G if and only if every orbit of the induced action of H on X contains just one point.


Homework Equations





The Attempt at a Solution


I've shown that the formula defines an action.
For the second part, since H is normal its left and right cosets are equivalent. Thus we consider h(xH) for h in H. So h(xH) = hxH = Hhx. But since h is in H this is the same as Hx. So for all h, the orbit contains only the point Hx.
However, the converse of the second part is what is giving me trouble. We know that each orbit contains only one element, but I'm not sure what else we can can from that.
 
Physics news on Phys.org
H is a subgroup, so it includes e. For all h, h(xH)=e(xH)=xH. Left-multiplying by xi (x inverse) on both sides gives xihxH=H, so xihx belongs to H (for all x in G and h in H).
 
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...
Back
Top