Answer to Question #128940 in Abstract Algebra for Roland Gyasi Kyei

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\\}\\\\\na\\text{ --- rotation by angle of }\\pi\/2\\text{ counterclockwise}\\\\\nx\\text{ --- reflection about the diagonal joining vertices "2" and "4"}\\\\\nax\\text{ --- reflection about the line joining midpoints of }\\\\\n\\text{opposite sides "14" and "23"}\\\\\na^2x\\text{ --- reflection about the diagonal joining vertices "1" and "3"}\\\\\na^3x\\text{ --- reflection about the line joining midpoints of }\\\\\n\\text{opposite sides "12" and "34"}\\\\\nN=\\{e, a^2, a^4\\equiv e, a^6\\equiv a^2\\}=\\{e, a^2\\}\\text{ --- }\\\\\n\\text{subgroup of G}\\\\\n(a)\\text{ Left cosets:}\\\\\n1) e\\in G\\\\\neN\\equiv N\\\\\n2) a\\in G\\\\\naN=\\{a, a^3\\}\\\\\n3) a^2\\in G\\\\\na^2N=\\{a^2, a^4\\}\\\\\n4) a^3\\in G\\\\\na^3N=\\{a^3, a^5\\equiv a\\}\\\\\n5) x\\in G\\\\\nxN=\\{x, xa^2\\equiv a^2x\\}\\\\\n6) ax\\in G\\\\\n(ax)N=\\{ax, (ax)a^2\\equiv (xa^3)a^2\\equiv x(a^3a^2)\\equiv xa\\}\\\\\n7) a^2x\\in G\\\\\n(a^2x)N=\\{a^2x, (a^2x)a^2\\equiv (xa^2)a^2\\equiv x(a^2a^2)\\equiv x\\}\\\\\n8) xa\\in G\\\\\n(xa)N=\\{xa, (xa)a^2\\equiv xa^3\\equiv ax\\}\\\\\n\\text{Right cosets:}\\\\\n1) e\\in G\\\\\nNe\\equiv N\\\\\n2) a\\in G\\\\\nNa=\\{a, a^3\\}\\\\\n3) a^2\\in G\\\\\nNa^2=\\{a^2, a^4\\}\\\\\n4) a^3\\in G\\\\\nNa^3=\\{a^3, a^5\\equiv a\\}\\\\\n5) x\\in G\\\\\nNx=\\{x, a^2x\\}\\\\\n6) ax\\in G\\\\\nN(ax)=\\{ax, a^2(ax)\\equiv (a^2a)x\\equiv xa\\}\\\\\n7) a^2x\\in G\\\\\nN(a^2x)=\\{a^2x, a^2(a^2x)\\equiv x\\}\\\\\n8) xa\\in G\\\\\nN(xa)=\\{xa, a^2(xa)\\equiv a^2(a^3x)\\equiv ax\\}\\\\\n\\text{We can see that the corresponding left and right}\\\\\n\\text{cosets are equal. So using the criterion of a normal}\\\\\n\\text{subgroup we can conclude that }N\\text{ is a normal}\\\\\n\\text{subgroup of G}.\\\\\n(b) G\/N \\text{ has order } \\frac{|G|}{|N|}=\\frac{8}{2}=4 \\text{ (using Lagrange's theorem)}\\\\\n G\/N \\text{ is a group of all the left cosets }\\\\\n G\/N \\text{ is not cyclic because there is no element } a \\text{ such that } gN=a^i"