While working on some group theory problems today a friend and I came up with the following result.

**Lemma. **Let be a normal subgroup of a finite group such that . If the order of divides , then .

*Proof.* Let be the order of . Then the order of in divides both and . But . Hence , i.e., , i.e., .

**Corollary 1.** Let be a normal subgroup of a finite group such that . If such that divides , then .

*Proof.* Apply the lemma to the elements of .

**Corollary 2. **Let be a normal subgroup of a finite group such that . Then is the unique subgroup of of order .

*Proof.* Use **Corollary 1**.

Here is an example of the lemma in action.

**Problem.** Show that has no normal subgroup of order or .

*Solution.* If is a normal subgroup of of order , then . Hence every element of order or in must lie in . In particular, . By a result in the previous post, , a contradiction.

Likewise, if is a normal subgroup of of order , then must contain every 3-cycle; in particular, . Hence . But this has order 2, and , a contradiction.

More generally, we can prove the following.

**Corollary 3.** for has no non-trivial proper normal subgroup with .

*Proof. *Suppose otherwise and let divide . Then must contain all -cycles. So if is even then taking gives . If is odd, it contains the cycles and for some . Then has order 3. So contains all 3-cycles, i.e., . Since is maximal, either or , a contradiction.

### Like this:

Like Loading...

*Related*

Filed under Algebra

Tagged as group, permutation