# Finite index implies powering-invariant

From Groupprops

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., subgroup of finite index) must also satisfy the second subgroup property (i.e., powering-invariant subgroup)

View all subgroup property implications | View all subgroup property non-implications

Get more facts about subgroup of finite index|Get more facts about powering-invariant subgroup

## Statement

Suppose is a group and is a subgroup of finite index in . Then, is a powering-invariant subgroup of . In other words, if is a prime number such that is powered over , then is also powered over .

## Facts used

- Poincare's theorem states that any subgroup of finite index contains a normal subgroup of finite index
- Normal of finite index implies quotient-powering-invariant
- Finite implies powering-invariant
- Powering-invariant over quotient-powering-invariant implies powering-invariant

## Proof

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

**To prove**: is powering-invariant in .

**Proof**:

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | Let be the normal core of in . Then, is a normal subgroup of finite index in . | Fact (1) | has finite index in . | Fact-given direct. | |

2 | is a quotient-powering-invariant of . | Fact (2) | Step (1) | Step-fact direct. | |

3 | is a powering-invariant subgroup of . | Fact (3) | Step (1) | By Step (1), is finite, hence , as a subgroup of it, is finite. Thus, by Fact (3), it is powering-invariant in . | |

4 | is powering-invariant in . | Fact (4) | Steps (2) and (3) | Step-combination direct. |