Visit

Groupprops, The Group Properties Wiki (pre-alpha)

Pronormality is normalizing join-closed

From Groupprops

Jump to:navigation, search
This article gives the statement, and possibly proof, of a subgroup property (i.e., pronormal subgroup) satisfying a subgroup metaproperty (i.e., normalizing 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 pronormal subgroup | Get facts that use property satisfaction of pronormal subgroup | Get facts that use property satisfaction of pronormal subgroup| Get more facts about normalizing join-closed subgroup property

Contents

Statement

Suppose H, K \le G are pronormal subgroups such that K \le N_G(H): in other words, K normalizes H. Then the join of subgroups \langle H, K \rangle (also the same as HK) is also a pronormal subgroup.

Related facts

Similar facts: statements about normalizing join-closedness

Some facts with very similar proofs:

Other facts about normalizing joins, but with a different kind of proof:

Related facts about pronormality

Join-closedness of some related properties

Proof

CONVENTION WARNING: This article/section uses the right-action convention. The left and right action conventions are equally powerful and statements/reasoning here can be converted to the alternative convention (the main reason being that every group is naturally isomorphic to its opposite group via the inverse map). For more on the action conventions and switching between them, refer to switching between the left and right action conventions.

Given: A group G, pronormal subgroups H,K \le G such that K \le N_G(H).

To prove: HK is pronormal in G. Specifically, for any g \in G, our goal is to find a z \in \langle HK, (HK)^g \rangle such that (HK)z = (HK)g.

Proof:

  1. (Given data used: H is pronormal in G): First, consider g on H. By pronormality of H, there exists x \in \langle H, H^g \rangle such that Hx = Hg. Thus, H^{gx^{-1}} = H.
  2. (Given data used: K normalizes H): Recall that K \le N_G(H), so K^{gx^{-1}} \le N_G(H^{gx^{-1}}) = N_G(H). Thus, \langle K, K^{gx^{-1}} \rangle \le N_G(H).
  3. (Given data used: K is pronormal in G): By pronormality of K, there exists y \in \langle K, K^{gx^{-1}} \rangle such that K^y = K^{gx^{-1}}. In particular, y \in N_G(H), so Hy = H.
  4. Now consider z = yx. We claim that (HK)z = (HK)x(No problem data used):. We have Hz = Hyx = (Hy)x = Hx (since Hy = H). We already know that Hx = Hg (step (1)), so Hz = Hg. Similarly, K^z = (K^y)^x = (K^{gx^{-1}})^x = K^{gx^{-1}x} = K^g. Thus, z acts like x on both H and K, and so (HK)z = (HK)x.
  5. We now check that z \in \langle (HK), (HK)^g \rangle (No problem data used): For this, observe that x \in \langle H,H^g \rangle \le \langle HK, (HK)^g \rangle. Also, y \in \langle K, K^{gx^{-1}} \rangle. The subgroup K is contained in HK. The subgroup Kg is contained in (HK)g, and the element x is in \langle HK, (HK)^g \rangle as argued already, so the subgroup K^{gx^{-1}} is also in \langle HK, (HK)^g \rangle. Thus, \langle K, K^{gx^{-1}} \rangle \le \langle HK, (HK)^g \rangle, and hence y \in \langle HK, (HK)^g \rangle. Thus, z = yx is also in the subgroup \langle HK, (HK)^g \rangle.

References

Retrieved from "http://groupprops.subwiki.org/wiki/Pronormality_is_normalizing_join-closed"
Facts about Pronormality is normalizing join-closedRDF feed
Fact aboutPronormal subgroup  +, and Normalizing join-closed subgroup property  +
Page classFact  +
Referenced inPaper:Rose-pronormal (?, ?, ?)  +
Navigation
lookup
Credits
Toolbox
request/feedback
subject wikis