## Applications

We’ll use the results that we obtained in the previous two posts to obtain some very nice results about finite groups.

**Example 1**. A finite group *G* of order *p*^{2} is isomorphic to either **Z**/*p*^{2} or (**Z**/*p*) × (**Z**/*p*). In particular, it is abelian.

**Proof. **Since *G* is a *p*-group, it has non-trivial centre, so *Z*(*G*) has order *p* or *p*^{2}.

- If it’s the 2nd case,
*G*is abelian.- If there’s an element of
*G*of order*p*^{2}, then*G*is cyclic. - Else, pick . So
*x*has order*p*. Pick , which again has order*p*. Since*x*and*y*commute, we can prove that , for 0 ≤*i*,*j*≤*p*-1, are all distinct. This shows that*G*is isomorphic to (**Z**/*p*) × (**Z**/*p*).

- If there’s an element of
- Suppose it’s the first case. So
*Z*(*G*) is cyclic of order*p*, generated by (say)*x*. Now we pick . Again, we can prove that , for 0 ≤*i*,*j*≤*p*-1, are all distinct. So every element of*G*can be written in this form, but this shows that*G*is abelian since*x*commutes with everything (including*y*):

which is a contradiction. ♦

**Example 2**. Suppose *G* has order *pq*, where *p* < *q* are primes with *p* not dividing (*q*-1). Then *G* is cyclic.

**Proof**. By Cauchy’s theorem, *G* has a subgroup *N* of order *q*. This is a Sylow *q*-subgroup and all such subgroups are conjugate. The number of such subgroups divides *pq* and is 1 mod *q* – thus it divides *p* and can only be 1 or *p*. Since *p* < *q*, it is 1. Thus, *N* is the unique Sylow *q*-subgroup of *G*, which means *N* is normal in *G* (since conjugation takes *N* to a Sylow *q*-subgroup, which has to be *N*).

By Cauchy’s theorem again, *G* has a subgroup *H* of order *p*. As before, the number of such subgroups divides *q* and is 1 mod *p*. Thus it equals 1 or *q*. Since *q* is not 1 mod *p*, it must be 1, so *H* is normal in *G*.

Thus, we have normal subgroups *H*, *N* with orders *q*, *p* respectively and trivial intersection. We must have *HN* ≡ *H* × *N*. Comparing order, *HN* must be the whole *G*. ♦

**Example 3**. If *p*, *q* are primes and *p* divides *q*-1, then there is a non-abelian group *G* of order *pq*.

**Proof**. The construction is via semidirect product. Let *H* = **Z**/*q* and *K* = **Z**/*p*. Now the automorphism group Aut(*H*) has order *q*-1 which is divisible by *p*, so there is an element of order *p*. This means we can let *K* act on *H* by mapping 1 mod *p* to . Now construct the semidirect product . It’s not abelian since:

and are not equal. ♦

[ *Note: when p*=2, we obtain the **dihedral group**. It’s the group of symmetries (including reflections) of the regular *q*-gon, and can be generalised to the case when *q* is any integer > 1. ]

**Example 4**. If *p* is a prime, then there is a non-abelian group *G* of order *p*^{3}.

**Proof**. Again use semidirect product. Let *H* = **Z**/*p*^{2} and *K* = **Z**/*p*. Now Aut(*H*) has order *p*(*p*-1) so it has an element of order *p*. Let *K* act on *H* by mapping 1 mod *p* to and construct the semidirect product as before. ♦

[ *From the above two examples, we see that the semidirect product is a useful method for constructing unusual groups.* ]

Exercise: construct a non-abelian group *G* with normal subgroup *N* such that *N* and *G*/*N* are both isomorphic to **Z**.

Recall that *A*_{5} is a simple group. It turns out it’s the smallest non-abelian simple group since any group of order 1-59 is either abelian or has a non-trivial normal subgroup.

Let’s consider some examples. Subsequently, we’ll denote *n _{p}* for the number of Sylow

*p*-subgroups.

**Example 5**. Prove that a group of order 12 has a non-trivial normal subgroup.

**Proof**. We have *n*_{2} | 3 and is 1 mod 2. If *n*_{2} = 1 then the Sylow 2-subgroup is normal, so we’ll assume *n*_{2} = 3. Similarly, *n*_{3} | 4 and is 1 mod 3, so let’s assume *n*_{3} = 4. But since a Sylow 2-subgroup and Sylow 3-subgroup intersect trivially, upon counting we get > 12 elements, which is impossible. ♦

**Example 6**. Prove that a group of order 24 has a non-trivial normal subgroup.

**Proof**. Again, *n*_{2} | 3 and is 1 mod 2, so let’s assume *n*_{2} = 3 (or the Sylow 2-subgroup would be normal). Consider the conjugation action of *G* on itself: this acts on the set of Sylow 2-subgroups, i.e. we get a homomorphism *G** → S*_{3}.

If we assume *G* is simple, the kernel (being a normal subgroup) is trivial or the entire group. It cannot be trivial since the map is not injective. So the homomorphism is trivial, i.e. conjugation leaves the three Sylow 2-subgroups invariant and all three are normal subgroups, which contradicts the fact that the three Sylow 2-subgroups are conjugate. ♦

**Example 7 (Tricky!)**. Prove that a group of order 120 has a non-trivial normal subgroup.

**Proof**. Now *n*_{5} | 24 and is 1 mod 5. If we assume *G* is simple, then we must have *n*_{2} = 6. Now, the conjugation action of *G* on itself also acts on the set of the Sylow 5-subgroups transitively. Hence, we get a transitive action *G* → *S*_{6}. Once again, since *G* is simple, we reason that the kernel is either trivial or the whole group. For transitive action, the only possibility is trival kernel, so we get an injective transitive action *G* → *S*_{6} so we might as well assume *G* is a subgroup of *S*_{6}.

Now, *A*_{6} is a normal subgroup of *S*_{6} and thus *G* ∩ *A*_{6 }is a normal subgroup of *G*. This can only happen if *G* ∩ *A*_{6 }= *G*, i.e. *G* is in fact a subgroup of *A*_{6}. This subgroup has index 360/120 = 3. Now *G* acts on the set of left-cosets *A*_{6}/*G* by left multiplication and we get a homomorphism *G* → *S*_{3}. Since *G* is simple, we’re forced to conclude the kernel is the entire group *G*, which is absurd since *G* acts transitively on the left-cosets. ♦

**Exercise**.

Prove that for groups of order up to 167, the only non-abelian simple group is *A*_{5 }(of order 60). [ The hardest case of 120 has already been done. ]