Pronormal not implies join with any distinct conjugate is the whole group

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., pronormal subgroup) need not satisfy the second subgroup property (i.e., subgroup whose join with any distinct conjugate is the whole group)
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Statement

We can have a pronormal subgroup H of a group G, and a conjugate subgroup K \ne H of H in G such that \langle H, K \rangle is not the whole group G.

Related facts

Converse

Join with any distinct conjugate is the whole group implies pronormal

Facts used

  1. Sylow implies pronormal

Proof

Example of the symmetric group of degree four

Further information: symmetric group:S4

Suppose G is the symmetric group on \{ 1,2,3,4 \}. Let H be the 3-Sylow subgroup generated by \{ (1,2,3) \}. Then, by fact (1), we get that H is pronormal in G. On the other hand, the subgroup generated by H and its conjugate subgroup \langle (1,3,4) \rangle is the alternating group on \{ 1,2,3,4 \}, which is a proper subgroup of G.