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

Contents 1 Statement 1.1 Verbal statement 1.2 Property-theoretic statement 1.3 Statement with symbols 2 Related facts 2.1 Generalizations 2.2 Related facts about normality and subnormality 3 Facts used 4 Proof 4.1 Hands-on proof

Statement

Verbal statement

A subnormal subgroup of a group is also subnormal in every intermediate subgroup. In fact, its subnormal depth in any intermediate subgroup is bounded from above by the subnormal depth in the whole group.

Property-theoretic statement

The subgroup property of being a subnormal subgroup satisfies the subgroup metaproperty called the intermediate subgroup condition -- any subnormal subgroup of the whole group is also subnormal in every intermediate subgroup.

Statement with symbols

Suppose H is a subnormal subgroup of a group G. Then, for any intermediate subgroup K (i.e., ), H is subnormal in K. Moreover, if H is k-subnormal in G, H is also k-subnormal in K. (Here, when we say k-subnormal, we mean the subnormal depth is at most k).

Related facts

Generalizations

• Subnormality satisfies inverse image condition
• Subnormality satisfies transfer condition

Related facts about normality and subnormality

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

Facts used

1. Normality satisfies transfer condition: If are subgroups such that H is normal in G, then is normal in K.

Proof

Hands-on proof

Given: A group G, a k-subnormal subgroup H, a subgroup such that .

To prove: H is k-subnormal in K.

Proof: Consider a subnormal series for H of length k:

.

where H_i is normal in H_{i + 1} for each i. We claim that the series:

is a subnormal series for H in K. For this, observe that:

.

We know that H_i is normal in H_{i + 1}, so by fact (1), is normal in , yielding that is normal in , as desired.