## 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 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 $x \in G-\{e\}$. So x has order p. Pick $y \in G-\left$, which again has order p. Since x and y commute, we can prove that $x^i y^j$, 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 $y\in G-Z(G)$. Again, we can prove that $x^i y^j$, 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): $(x^i y^j) (x^a y^b) = (x^i x^a)(y^j y^b) = x^{i+a}y^{j+b} = (x^a y^b)(x^i y^j)$ Example 2. Suppose G has order pq, where pq 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 pq, 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 HN with orders qp respectively and trivial intersection. We must have HN ≡ H × N. Comparing order, HN must be the whole G. ♦ Example 3. If pq 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 HZ/q and KZ/p. Now the automorphism group Aut(H) has order q-1 which is divisible by p, so there is an element $\phi\in \text{Aut}(H)$ of order p. This means we can let K act on H by mapping 1 mod p to $\phi$. Now construct the semidirect product $G = H \rtimes K$. It’s not abelian since: $(0,1)*(m,0) = (\phi(m), 1)$ and $(m,0) * (0,1) = (m,1)$ 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 HZ/p2 and KZ/p. Now Aut(H) has order p(p-1) so it has an element $\phi\in\text{Aut}(H)$ of order p. Let K act on H by mapping 1 mod p to $\phi$ and construct the semidirect product $G = H\rtimes K$ 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 ∩ Ais a normal subgroup of G. This can only happen if G ∩ AG, 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. ♦

Exercise.

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. ]

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