# Subnormality satisfies transfer condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) satisfying a subgroup metaproperty (i.e., transfer condition)

View all subgroup metaproperty satisfactions | View all subgroup metaproperty dissatisfactions |Get help on looking up metaproperty (dis)satisfactions for subgroup properties

Get more facts about subnormal subgroup |Get facts that use property satisfaction of subnormal subgroup | Get facts that use property satisfaction of subnormal subgroup|Get more facts about transfer condition

## Contents

## Statement

### Verbal statement

The intersection of a subnormal subgroup with any subgroup is subnormal in that subgroup. Moreover, the subnormal depth of the intersection is bounded from above by the subnormal depth of the original subgroup.

### Statement with symbols

Suppose is a subnormal subgroup of a group , and is a subgroup. Then, is a subnormal subgroup of . Moreover, if is -subnormal in , is -subnormal in (i.e., its subnormal depth is at most .

## Related facts

### Related facts about subnormality

- Subnormality satisfies intermediate subgroup condition
- Subnormality satisfies inverse image condition
- Subnormality satisfies image condition

### Related facts about normality

- Normality is strongly UL-intersection-closed
- Normality satisfies transfer condition
- Normality satisfies intermediate subgroup condition
- Normality satisfies inverse image condition

## Facts used

- Normality satisfies transfer condition: If is a normal subgroup of a group , and , then is a normal subgroup of .
- Transfer condition is subordination-closed: If is a subgroup property satisfying the transfer condition, the subordination of .

## Proof

### Hands-on proof

**Given**: A group , a -subnormal subgroup of , and a subgroup of .

**To prove**: is a -subnormal subgroup of .

**Proof**: Since is -subnormal subgroup of , we have a subnormal series:

.

We claim that the following is a subnormal series for in :

.

For this, we need to show that each is normal in . For this, note that:

.

Since is normal in , fact (1) tells us that is normal in , completing the proof.

### Property-theoretic proof

This follows directly from facts (1) and (2).