Contranormality is upper 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., upper join-closed subgroup property)
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 contranormal subgroup |Get facts that use property satisfaction of contranormal subgroup | Get facts that use property satisfaction of contranormal subgroup|Get more facts about upper join-closed subgroup property


Statement with symbols

Suppose H \le G is a subgroup, and K_i, i \in I, is an indexed family of subgroups with H \le K_i for each i \in I. Then, if H is contranormal in each K_i, H is also contranormal in the [join of subgroups|join]] of the K_is.

Definitions used

Contranormal subgroup

Further information: contranormal subgroup

H \le K is a contranormal subgroup if for any L \le K containing H such that L is normal in K, L = K.

Related facts

Stronger facts


Facts used

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


Given: H \le G, family of subgroups K_i, i \in I with H \le K_i, and H contranormal in each K_i.

To prove: H is normal in the join of all the K_is.

Proof: Suppose L is a normal subgroup of the join of the K_is, containing H. Then, for each K_i, L \cap K_i is a subgroup of K_i containing H, and by fact (1), it is normal in K_i. Since H is contranormal in K_i, L \cap K_i = K_i for each i \in I, so K_i \le L for each i \in I. Thus, L must equal the join of the K_is.