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

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 $H$

## Proof

### Direct proof using the facts

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

### Concrete example

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.