Question

Let G=D_8, and let N={e,a^2,a^4,a^6}.
(a) Find all left cosets and all right cosets of N, and verify that N is a normal subgroup of G.
(b) Show that G/N has order 4, but is not cyclic.

Answer 1

$G=D_8=\{e, a, a^2, a^3, x, ax\equiv xa^3, a^2x, a^3x\equiv xa\}\\ a\text{ --- rotation by angle of }\pi/2\text{ counterclockwise}\\ x\text{ --- reflection about the diagonal joining vertices "2" and "4"}\\ ax\text{ --- reflection about the line joining midpoints of }\\ \text{opposite sides "14" and "23"}\\ a^2x\text{ --- reflection about the diagonal joining vertices "1" and "3"}\\ a^3x\text{ --- reflection about the line joining midpoints of }\\ \text{opposite sides "12" and "34"}\\ N=\{e, a^2, a^4\equiv e, a^6\equiv a^2\}=\{e, a^2\}\text{ --- }\\ \text{subgroup of G}\\ (a)\text{ Left cosets:}\\ 1) e\in G\\ eN\equiv N\\ 2) a\in G\\ aN=\{a, a^3\}\\ 3) a^2\in G\\ a^2N=\{a^2, a^4\}\\ 4) a^3\in G\\ a^3N=\{a^3, a^5\equiv a\}\\ 5) x\in G\\ xN=\{x, xa^2\equiv a^2x\}\\ 6) ax\in G\\ (ax)N=\{ax, (ax)a^2\equiv (xa^3)a^2\equiv x(a^3a^2)\equiv xa\}\\ 7) a^2x\in G\\ (a^2x)N=\{a^2x, (a^2x)a^2\equiv (xa^2)a^2\equiv x(a^2a^2)\equiv x\}\\ 8) xa\in G\\ (xa)N=\{xa, (xa)a^2\equiv xa^3\equiv ax\}\\ \text{Right cosets:}\\ 1) e\in G\\ Ne\equiv N\\ 2) a\in G\\ Na=\{a, a^3\}\\ 3) a^2\in G\\ Na^2=\{a^2, a^4\}\\ 4) a^3\in G\\ Na^3=\{a^3, a^5\equiv a\}\\ 5) x\in G\\ Nx=\{x, a^2x\}\\ 6) ax\in G\\ N(ax)=\{ax, a^2(ax)\equiv (a^2a)x\equiv xa\}\\ 7) a^2x\in G\\ N(a^2x)=\{a^2x, a^2(a^2x)\equiv x\}\\ 8) xa\in G\\ N(xa)=\{xa, a^2(xa)\equiv a^2(a^3x)\equiv ax\}\\ \text{We can see that the corresponding left and right}\\ \text{cosets are equal. So using the criterion of a normal}\\ \text{subgroup we can conclude that }N\text{ is a normal}\\ \text{subgroup of G}.\\ (b) G/N \text{ has order } \frac{|G|}{|N|}=\frac{8}{2}=4 \text{ (using Lagrange's theorem)}\\ G/N \text{ is a group of all the left cosets }\\ G/N \text{ is not cyclic because there is no element } a \text{ such that } gN=a^i$