# Union of all conjugates of subgroup of finite index is proper

## Contents |

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself

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*This article describes a result of the form that argues that a subset constructed in a certain fashion is proper, viz it is not the whole group*

## Statement

### 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 be a proper subgroup of finite index in a group . Then, the union of all conjugates of is a proper subset of . In other words:

### 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 subgroup of finite index whole group (viz improper subgroup)

## Related facts

### Breakdown for infinite groups

If is infinite and is a subgroup of infinite index, may be conjugate-dense in : every element of may be conjugate to an element of . 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

- Fundamental theorem of group actions (also see 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).
- Lagrange's theorem: This states that for , we have .
- Poincare's theorem: Any subgroup of finite index in a group contains a normal subgroup of finite index.
- Fourth isomorphism theorem: The correspondence between subgroups containing a normal subgroup and subgroups in the quotient group.

## Proof

### Finite case

**Given**: A finite group of order , and a subgroup of index .

**To prove**: The union of conjugates of in is a proper subset of .

**Proof**: We proceed as follows::

- The index is bounded by the index : (the normalizer of in ) is a subgroup of containing . Thus, the index (say ) of in is at most .
- The number of conjugate subgroups to in equals : Consider the action of on the set of conjugate subgroups to by conjugation. The stabilizer of is precisely , so by fact (1), there is a bijection between the cosets of in and the conjugate subgroups to in . Thus, the number of conjugate subgroups to in equals .
- 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 . We bound this as follows:

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

Thus, the union of all conjugates of in is a *proper* subset of .

### Case of finite index

**Given**: A group , a subgroup of finite index in .

**To prove**: The union of all conjugates of in is a proper subset of .

**Proof**:

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