Having subgroups of all orders dividing the group order is not subgroup-closed

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This article gives the statement, and possibly proof, of a group property (i.e., group having subgroups of all orders dividing the group order) not satisfying a group metaproperty (i.e., subgroup-closed group property).
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Statement

It is possible to have a finite group G and a subgroup H such that:

  • G has subgroups of all orders dividing the order of G
  • H does not have subgroups of all orders dividing the order of H

Facts used

  1. Finite solvable not implies subgroups of all orders dividing the group order
  2. Every finite solvable group is a subgroup of a finite group having subgroups of all orders dividing the group order

Proof

Direct proof using the facts

The proof follows directly from Facts (1) and (2).

Concrete example

Further information: symmetric group:S4, subgroup structure of symmetric group:S4, alternating group:A4, subgroup structure of alternating group:A4

We can take G to be symmetric group:S4, and H to be the subgroup that is alternating group:A4. G has subgroups of every order dividing the group order, whereas H, that has order 12, does not have a subgroup of order 6.