Subnormality is not finite-upper join-closed

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This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) not satisfying a subgroup metaproperty (i.e., finite-upper join-closed subgroup property).
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Suppose G is a group, H \le G is a subgroup and K,L \le G are subgroups containing H. Then, it can happen that H is a subnormal subgroup of K and of L, but H is not a subnormal subgroup of the join of subgroups \langle K, L \rangle.

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Example of the symmetric group

Further information: symmetric group:S5

Let G be the symmetric group on the set \{ 1,2,3,4,5 \}. Let K and L be the dihedral groups given as follows:

K = \langle (1,3,2,4), (1,2) \rangle; \qquad L = \langle (1,3,2,5), (1,2) \rangle

Define H = K \cap L. Then, H is a two-element subgroup comprising (1,2) and the identity permutation.

Observe that: