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 p2 is isomorphic to either Z/p2 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 p2.
- If it’s the 2nd case, G is abelian.
- If there’s an element of G of order p2, 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).
- 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. ♦
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. ♦
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 p3.
Proof. Again use semidirect product. Let H = Z/p2 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 A5 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 np 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 n2 | 3 and is 1 mod 2. If n2 = 1 then the Sylow 2-subgroup is normal, so we’ll assume n2 = 3. Similarly, n3 | 4 and is 1 mod 3, so let’s assume n3 = 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, n2 | 3 and is 1 mod 2, so let’s assume n2 = 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 → S3.
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 n5 | 24 and is 1 mod 5. If we assume G is simple, then we must have n2 = 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 → S6. 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 → S6 so we might as well assume G is a subgroup of S6.
Now, A6 is a normal subgroup of S6 and thus G ∩ A6 is a normal subgroup of G. This can only happen if G ∩ A6 = G, i.e. G is in fact a subgroup of A6. This subgroup has index 360/120 = 3. Now G acts on the set of left-cosets A6/G by left multiplication and we get a homomorphism G → S3. 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. ♦
Prove that for groups of order up to 167, the only non-abelian simple group is A5 (of order 60). [ The hardest case of 120 has already been done. ]