# Join of normal and subnormal implies subnormal of same depth

From Groupprops

This article describes a computation relating the result of the Join operator (?) on two known subgroup properties (i.e., Normal subgroup (?) and Subnormal subgroup (?)), to another known subgroup property (i.e., Subnormal subgroup (?))

View a complete list of join computations

## Statement

Let be a natural number. Then, the join of any normal subgroup of a group with any -subnormal subgroup is -subnormal (Note: When we say a subgroup is -subnormal, we mean that its subnormal depth is at most equal to ).

In symbols, if is a -subnormal subgroup of and is a normal subgroup of , then the join of subgroups (which in this case is also the product of subgroups ) is also a -subnormal subgroups.

## Facts used

- Join lemma for normal subgroup of subgroup with normal subgroup of whole group: This states that if and , then .

## Proof

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

**To prove**: is -subnormal in .

**Proof**:

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

1 | There exists a subnormal series from to given as follows: . Here, means is normal in |
is -subnormal in | Using the definition of -subnormal. | ||

2 | For , . | is normal in . | (1) | (1) | For fact (1), set , , and . |

3 | The series is a subnormal series of length for in . | (2) | Follows directly. | ||

4 | is -subnormal in . | (3) | Follows directly. |