# Normal of finite index implies quotient-powering-invariant

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., normal subgroup of finite index) must also satisfy the second subgroup property (i.e., quotient-powering-invariant subgroup)

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## Contents

## Statement

Suppose is a group and is a normal subgroup of finite index in , i.e., is a normal subgroup of and the index of in is finite. Then, is a quotient-powering-invariant subgroup of , i.e., if is powered over a prime number , then so is the quotient group .

## Related facts

- Finite normal implies quotient-powering-invariant
- Powering-invariant and central implies quotient-powering-invariant
- Normal of finite index implies completely divisibility-closed

## Facts used

## Proof

### Proof using given facts (less explicit)

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**Given**: A group , a normal subgroup of such that the index of in is finite (in other words, the quotient group is a finite group). is powered over a prime .

**To prove**: is also powered over .

**Proof**:

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

1 | is -divisible. | is -powered. | By definition, being powered over a prime means every element has a unique root, which implies that every element has a root (the definition of -divisible). | ||

2 | is -divisible. | Fact (1) | is normal in . | Step (1) | Step-fact combination direct. |

3 | is -powered. | Fact (2) | is normal of finite index, so is a finite group. | Step (2) | Step-fact-given direct. |

### Hands-on proof

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This is the same as the preceding proof, but shows all the details explicitly without appeal to separate facts.

**Given**: A group , a normal subgroup of such that the index of in is finite (in other words, the quotient group is a finite group). A prime number such that for any , there is a unique such that .

**To prove**: For every , there is a unique such that .

**Proof**: Let be the quotient map.

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

1 | The map in descends to the corresponding map in the quotient group . | is normal in . | Direct from definition of quotient group structure. | ||

2 | The map in is surjective from to itself. In other words, for any , there exists such that . | is -powered. | For any , pick in that coset (so ). There exists such that , since is -powered. Let be the coset of . Then, we have that . | ||

3 | The map in is bijective from to itself. In other words, for any , there exists such that . | is normal of finite index, so is a finite group. | Step (2) | Follows directly from Step (2) and the observation that any surjective map from a finite set to itself must be bijective. |