Union of all conjugates of subgroup of finite index is proper

Verbal statement
The union of conjugates of a proper subgroup of finite index in a group, is a proper subset. In particular, in a finite group, the union of all conjugates of any proper subgroup is a proper subset.

Symbolic statement
Let $$H$$ be a proper subgroup of finite index in a group $$G$$. Then, the union of all conjugates of $$H$$ is a proper subset of $$G$$. In other words:

$$\bigcup_{g \in G} gHg^{-1} \ne G$$

Property-theoretic statement
Call a subgroup conjugate-dense if the union of its conjugates is the whole group. Then, the result states that:

conjugate-dense $$\land$$ subgroup of finite index $$\implies$$ whole group (viz improper subgroup)

Breakdown for infinite groups
If $$G$$ is infinite and $$H$$ is a subgroup of infinite index, $$H$$ may be conjugate-dense in $$G$$: every element of $$G$$ may be conjugate to an element of $$H$$. Examples include upper triangular matrices in the general linear group over an algebraically closed field. For more information, see:

Category:Instances of conjugate-dense subgroups

Related notion of contranormality
Note that the result only says that the union of the conjugates must be a proper subset. It does not state that the subgroup generated by the union of the conjugates must be a proper subgroup. The subgroup generated by the union of the conjugates of a proper subgroup, which is also called its normal closure, may well be the whole group. A subgroup wholse normal closure is the whole group is termed a contranormal subgroup.

Any conjugate-dense subgroup is contranormal, but while no proper subgroup of finite index can be conjugate-dense, there can exist contranormal proper subgroups of finite index (for instance, any maximal non-normal subgroup).

Similar facts

 * Product of conjugates is proper
 * Union of two subgroups is not a subgroup unless they are comparable
 * Directed union of subgroups is subgroup

Other facts about conjugates and their union

 * Finite non-Abelian and every proper subgroup is Abelian implies not simple: The proof technique uses ideas of unions of conjugates of subgroups and counting arguments.
 * The concept of a Frobenius group, where there is a Frobenius kernel and a Frobenius complement. The union of conjugates of the Frobenius complement intersects the Frobenius kernel in the identity element, and they together cover the whole group. Further, all the conjugates of the Frobenius complement intersect trivially.
 * Results such as Artin's induction theorem work for a setup where there is a collection of subgroups such that the union of all conjugates of all subgroups in that collection equals the whole group.

Facts used

 * 1) uses::Fundamental theorem of group actions (also see uses::Group acts on set of subgroups by conjugation for an explanation of the application of the fundamental theorem of group actions to this concrete situation).
 * 2) uses::Lagrange's theorem: This states that for $$H \le G$$, we have $$|G| = |H|[G:H]$$.
 * 3) uses::Poincare's theorem: Any subgroup of finite index in a group contains a normal subgroup of finite index.
 * 4) uses::Fourth isomorphism theorem: The correspondence between subgroups containing a normal subgroup and subgroups in the quotient group.

Finite case
Given: A finite group $$G$$ of order $$n$$, and a subgroup $$H$$ of index $$r > 1$$.

To prove: The union of conjugates of $$H$$ in $$G$$ is a proper subset of $$G$$.

Proof: We proceed as follows::

$$1 + s(|H| - 1) \le 1 + r(|H| - 1) = 1 + r|H| - r = 1 + [G:H]|H| - r = |G| - (r - 1) < |G|$$
 * 1) The index $$s = [G:N_G(H)]$$ is bounded by the index $$r = [G:H]$$: $$N_G(H)$$ (the normalizer of $$H$$ in $$G$$) is a subgroup of $$G$$ containing $$H$$. Thus, the index (say $$s$$) of $$N_G(H)$$ in $$G$$ is at most $$r$$.
 * 2) The number of conjugate subgroups to $$H$$ in $$G$$ equals $$s$$: Consider the action of $$G$$ on the set of conjugate subgroups to $$H$$ by conjugation. The stabilizer of $$H$$ is precisely $$N_G(H)$$, so by fact (1), there is a bijection between the cosets of $$N_G(H)$$ in $$G$$ and the conjugate subgroups to $$H$$ in $$G$$. Thus, the number of conjugate subgroups to $$H$$ in $$G$$ equals $$s = [G:N_G(H)]$$.
 * 3) Since all the conjugates have equal size and already intersect in the identity element, the total number of elements in the union is at most $$1 + s(|H| - 1)$$. We bound this as follows:

(An intermediate step uses Lagrange's theorem, and the final step uses the fact that $$r > 1$$).

Thus, the union of all conjugates of $$H$$ in $$G$$ is a proper subset of $$G$$.

Case of finite index
Given: A group $$G$$, a subgroup $$H$$ of finite index in $$G$$.

To prove: The union of all conjugates of $$H$$ in $$G$$ is a proper subset of $$G$$.

Proof:


 * 1) By fact (3), there is a normal subgroup $$N$$ of finite index in $$G$$ contained inside $$H$$. Let $$\varphi:G \to G/N$$ be the quotient map. Thus, $$G/N$$ is a finite group and $$H/N$$ is a proper subgroup of $$G/N$$.
 * 2) By fact (4) (the fourth isomorphism theorem) or a direct checking, we see that the conjugates of $$H$$ in $$G$$ are precisely the inverse images under $$\varphi$$ of the conjugates of $$H/N$$ in $$G/N$$.
 * 3) Applying the result in the finite case to the group $$G/N$$ and the proper subgroup $$H/N$$, we obtain that the union of all conjugates of $$H/N$$ in $$G/N$$ is a proper subset of $$G/N$$. Combining this with step (2) yields that the union of all conjugates of $$H$$ in $$G$$ is a proper subgroup of $$G$$.