# Complemented normal implies quotient-powering-invariant

## Contents

## Statement

Suppose is a group and is a complemented normal subgroup of (i.e., there exists a permutable complement to in , i.e., is an internal semidirect product involving ). Then, is a quotient-powering-invariant subgroup of : for any prime number such that is powered over (i.e., every element of has a unique root), is also powered over .

## Facts used

## Proof

### Proof using given facts

The proof follows directly from Facts (1) and (2).

### Hands-on proof

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**Given**: A group , a normal subgroup of with permutable complement . is powered over a prime number .

**To prove**: is powered over .

**Proof**:

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

1 | Let be the retraction that sends an element of to the unique element of in its coset with respect to . This corresponds to the quotient map and . | is normal in and is complemented, with complement . | |||

2 | For , there exists a unique such that . | is powered over . | |||

3 | The obtained in Step (2) is actually an element of . In other words, for , there exists a unique such that . So, is powered over . | Steps (1), (2) | We have that . Thus, is an element whose power is . Since is the unique element of such that , we must have , forcing . | ||

4 | is powered over . | Steps (1), (3) | Step (1) says that . Step (3) says that is powered over . Thus, is powered over . |