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

From Groupprops

This article gives the statement, and possibly proof, of a group property (i.e., group having subgroups of all orders dividing the group order)notsatisfying a group metaproperty (i.e., subgroup-closed group property).

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Get more facts about group having subgroups of all orders dividing the group order|Get more facts about subgroup-closed group property|

## Statement

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

- has subgroups of all orders dividing the order of
- does not have subgroups of all orders dividing the order of

## Facts used

- Finite solvable not implies subgroups of all orders dividing the group order
- 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 to be symmetric group:S4, and to be the subgroup that is alternating group:A4. has subgroups of every order dividing the group order, whereas , that has order 12, does not have a subgroup of order 6.