Let be a group, a subgroup of , and fix .

- Prove that and that .
- Deduce that if and is the unique subgroup of of order then is normal in .

- Note that is not empty since . Now let and be in , where . Then , and by the Subgroup Criterion we have .
We define a mapping by . is clearly surjective. Now suppose ; then , so that . Hence is injective. Thus .

- Suppose now that and that has a unique subgroup of order . For all , is a subgroup of order , hence . Thus is normal in .

Advertisements

## Comments

You’ve got a typo just before claiming you’ve satisfied the Subgroup Criterion. It should be g(ab^(-1))g^(-1) in gHg^(-1).

Fixed. Thanks!