Let G be a finite group and H be a proper subgroup. Prove that the union of the conjugates of H is not the whole of G.

Thanks for any help

**Answer**

(Note: Finite was not specified when I wrote this answer; I’ll keep the more general answer, though)

The result is true if we assume that H is of finite index. It may be false if H is of infinite index.

For a counterexample in the infinite index case, let F be an algebraically closed field, let G be the group of all n×n invertible matrices with coefficients in F, and let H be the subgroup of upper triangular matrices. Since every matrix over an algebraically closed field is similar to an upper triangular matrix (e.g., the Jordan canonical form), it follows that the union of conjugates of H equals the whole group, even though H does not equal all of G.

For a proof in the finite index case, let [G:H]=n. Then the action of G on the cosets H by left multiplication gives a homomorphism G→Sn with kernel K⊆H. This reduces to the *finite* case.

In the finite case, let |H|=k; then |G|=kn. There are at most n distinct conjugates. Since the identity element is in all of the conjugates, the union of the conjugates of H has at most

n(k−1)+1=nk−n+1 element

and since we are assuming n>1, it follows that

|⋃g∈GgHg−1|≤nk−(n−1)<nk=|G|,

so the union cannot equal all of G.

**Attribution***Source : Link , Question Author : hmmmm , Answer Author : Arturo Magidin*