Thread: Lagrange's Thm and normal subgroup

1. Lagrange's Thm and normal subgroup

If H is a subgroup of a group G, we define:
G/H = {xH : x belongs to G}
H\G = {Hx : x belongs to G}
Let G = D10 = {x, y| x^5 = y^2 = 1, yx =x^4(y)}
and K = {1, x, x^2,x^3,x^4} and H ={1,y}

How do you find G/K? and find the normal subgroup of D10

Can you give me some hints how to do this question plz?

Cheers

2. Cosets

To find $G/K$ one needs to just consider the cosets. By Lagrange you know that $|G/K|=\frac{|G|}{|K|}=\frac{10}{5}=2$ so in this case we should expect 2 distinct cosets. One is always $1K$ and this will be the same coset as when you translate K by anything in K. So to find the other coset try transalating (multiplying on the left) by one of the other elements in $D_{10} - K$ and see what happens.

Also notice why this particular subgroup is always Normal, and see that there is nothing particularly special about this group, but more importantly that it has size one half as big as the mother group. In fact this is always the case that every subgroup of index 2 is normal, and in more generality if p is the smallest prime that divides the order of the group, any subgroup of index p is normal (note Lagrange does not guarantee the existance of such a subgroup). But in the Dihedral group of order 2n there always is one of index 2 because there is always a cyclic group of order n (the group of rotations of the n-gon in this case the regular pentagon)

Hope this helps.