Contranormality is UL-join-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., contranormal subgroup) satisfying a subgroup metaproperty (i.e., UL-join-closed subgroup property)
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Statement with symbols

Suppose G is a group and H_i \le K_i \le G is a family of subgroups indexed by i \in I, and indexing set. Further, suppose each H_i is a contranormal subgroup of K_i. Then, the join of all the H_is is contranormal in the join of all the K_is.

Related facts

Weaker facts


Definitions used

Contranormal subgroup

Further information: Contranormal subgroup

Given a subgroup H \le K we say that H is contranormal in K if any normal subgroup of K containing H must equal the whole of K. In other words, the normal closure of H in K is K.

Facts used

  1. Normality satisfies transfer condition: If L is normal in G and K \le G, then L \cap K is normal in K.


Given: Group G, indexing set I, H_i \le K_i \le G for all i \in I, and each H_i is contranormal in K_i.

To prove: The join of the H_i (which we denote as H) is contranormal in the join of the K_is (which we denote as K).

Proof: By the definition of contranormality, we need to show that if L is a normal subgroup of the join of K, and L contains each H, then L = K.

So suppose L is normal in K with H \le L. For each i \in I, L \cap K_i is normal in K_i by fact (1). Also, H_i \le H \le L, and H_i \le K_i so H_i \le L \cap K_i. Thus, L \cap K_i is a normal subgroup of K_i containing H_i. By contranormality of H_i in K_i, we get L \cap K_i = K_i, so K_i \le L. Since this holds for each i \in I, we get K \le L, forcing K = L.