group Prob?

Question

Let H be a proper subgroup of a finite group G.Prove G is not the set-theoretic union of all the conjugates of H

Answer 1

Let N(H) = { g in G | g^-1 H g = H }

Suppose a and b are in the same right coset of N(H), i,e, N(H) a = N(H) b. Then for some n in N(H) we have b = n a.

Then b^-1 H b = (n a)^-1 H (n a) = (a^-1 n^-1) H (n a) = a^-1 (n^-1 H n) a = a^-1 H a. So all elements of a right coset of N(H) yield the same conjugate of H.

Now H is a subgroup of N is a subgroup of G, so the number of cosets of N(H) cannot exceed the number of cosets of H. If H has n elements and there are m cosets of H (so that G has n m elements) then even if H has m conjugates -- the most it can have -- then their union can contain at most

m (n - 1) + 1 = m n - m + 1 < n m = |G|

because H is proper so m > 1, and all of the at most m conjugates of H have the same number of elements n as H, but the identity element of the group G is in all of the conjugates. So the union can have size at most m (n - 10 = 1 = m n - m + 1 = |G| - (m - 1) < |G| because m > 1.

So a finite group G is not the set-theoretic union of all the conjugates of any one of its proper subgroups.

Answer 2

~Oh G, why the H would I want to. I like G. I'm not gonna try to segregate her out.