Abstract Linear Algebra Problem?
Let G be a group and a ∈ G be an element. Prove that N (a)x = N (a)y if an only if
x^-1 ax = y^-1 ay. Deduce that the number of elements in the equivalence class of "a" given by the
relation above is |G| / |N(a)|
Let G be a group and a ∈ G be an element. Prove that N (a)x = N (a)y if an only if
x^-1 ax = y^-1 ay. Deduce that the number of elements in the equivalence class of "a" given by the
relation above is |G| / |N(a)|
Comments
If N(a)x = N(a)y then N(a) = N(a)yx^-1 which means that yx^-1 is in N(a) so a = (yx^-1)^-1a(yx^-1) = xy^-1ayx^-1 ==> a = xy^-1ayx^-1 ==> x^-1ax = y^-1ay. Working backwards you get the reverse.
The conjugacy class of a, Cl(a) = {gag^-1 for all g in G}. All g in N(a) give only a, and from the above if N(a)x = N(a)y then they lead to the same class x^-1ax. So the total number of elements in Cl((a) is |G|/|N(a)|.