Are Cosets of Normal Subgroups Always Normal in Quotient Groups?

  • Thread starter Thread starter ehrenfest
  • Start date Start date
  • Tags Tags
    Normal
ehrenfest
Messages
2,001
Reaction score
1
[SOLVED] normal subgroups

Homework Statement


My book states the following without any justification right before proving the Third Isomorphism Theorem: "If H and K are two normal subgroups of G and K \leq H, then H/K is a normal subgroup of G/K."
The elements of H/K are cosets of K in H. The elements of G/K are cosets of K in G. Therefore I think that statement is simply absurd. That is, the elements of H/K are not even contained in the quotient group G/K; therefore, they cannot possibly form a normal subgroup in G/K.

EDIT: wait, never mind, the cosets of K in H are also cosets of K in G; sorry
EDIT: and the reason H/K is normal in G/K is that gK*hK*g^(-1)K = (ghg^{-1})K = h' K since H is normal in G. Very EDIT: cool.

Homework Equations


The Attempt at a Solution

 
Last edited:
Physics news on Phys.org
The elements in H/K are, indeed, contained in G/K since H is contained in G: A general element in H/K is hK which is in G/K since h is in G.
 
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...

Similar threads

Back
Top