Applications of the Sylow theorems Flashcards
Theorem 20.1.
Let G be a group and let P be a Sylow p-subgroup of G. Then P is normal in G if and only if
P is a unique Sylow p-subgroup of G.
Used in a variety of examples. Important thm.
Example 20.2. There exists no simple group of order 200. Indeed, let G be a group of order 200 = 2352. By the Third Sylow’s Theorem, the number of Sylow 5-subgroups of G divides 200 and has the form 5k + 1. The divisors of 200 are 1, 5, 25, 2, 10, 50, 4, 20, 100, 8, 40, 200. The only number from the above list that has the form 5k + 1 is 1. Therefore, G has a unique Sylow 5-subgroup, which is then normal in G by Theorem 20.1. This proves that G is not simple.
Example 20.3. There exists no simple group of order 12. For, assume to the contrary that G is a simple group of order 12 = 223. By the Third Sylow’s Theorem the number of Sylow 3-subgroups of G is either 1 or 4. Since G is simple, this number must be four because of Theorem 20.1. Each of these four subgroups has order 3, and is therefore isomorphic to Z3. By Lagrange’s Theorem, the intersection of any two of these subgroups must be trivial. Therefore, G has eight elements of order 3. Next, since G is not simple, it must contain at least two Sylow 2-subgroups H1 and H2, each of them having order 4. By Lagrange’s Theorem H1 ∩ H2 has either one or two elements. Hence we conclude that G contains at least six elements of order a power of 2, which contradicts the fact that |G| = 12.
It is possible to classify all the groups of order pq, where p and q are distinct primes: there are at most two such groups, one of them is Zpq, and the other (if it exists) is a non-abelian group obtained by extending Zp by Zq in a way similar to direct products. Here, however, we shall not give this full classification, but shall rather restrict our attention to two special cases.
Theorem 21.1.
Let p and q be two distinct prime numbers, with p > q. If p − 1 is not divisible by q then
Zpq is the only group of order pq.
Review proof of this: Alot of different understanding required, so good exercise
Theorem 21.2. Let p > 2 be a prime. The only groups of order 2p are
the cyclic group Z2p and the dihedral group Dp.
Groups of small orders
In this section we shall classify all the groups of order less than 16. In doing this we shall make use of the following general classification theorems proved so far:
- Zp is the only group of order p (Theorem 6.11);
- Zp2 and Zp ⊕ Zp are the only two groups of order p2 (Theorem 18.4);
- Zpq is the only group of order pq if p − 1 is not divisible by q (Theorem 21.1);
- Z2p and Dp are the only two groups of order 2p (Theorem 21.2);
(p and q denote primes, with p > q).
Actually, the above results cover all the orders less than 16, apart from 8 and 12. We now deal with these two cases.
Theorem 22.1.
How many non-isomorphic groups are there of order 8.
What are they? (Think abelian and non_abelian)
There are five nonisomorphic groups of order 8, namely Z8, Z4⊕ Z2, Z2 ⊕Z2 ⊕Z2, Q8 andD4.
Now we turn our attention to groups of order 12. We already know that D6 and A4 are non-abelian of this order. The following example exhibits another group of order 12.
Example 22.2. Let T be the subgroup of S12 generated by the permutations (123456)(789101112)and(17410)(21259)(31168). Byusingthe standard method for calculating a group given by its generators we can see that
T ={id, (123456)(789101112), (135)(246)(7911)(81012), (1 4)(2 5)(3 6)(7 10)(8 11)(9 12), (1 5 3)(2 6 4)(7 11 9)(8 12 10), (165432)(712111098), (17410)(21259)(31168), (18411)(27510)(31269), (19412)(28511)(37610), (11047)(29512)(38611), (11148)(21057)(39612), (11249)(21158)(31067)}.
The orders of elements of T are 1, 2, 3, 3, 4, 4, 4, 4, 4, 4, 6, 6. Similarly, the ordersof elements of D6 and A4 are 1,2,2,2,2,2,2,2,3,3,6,6 and 1,2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3 respectively. Therefore, we see that T is a group of order 12 which is not isomorphic to D6 and A4.
Theorem 22.3.
The only groups of order 12 are
Z12, Z2 ⊕ Z2 ⊕ Z3, D6, A4 and T.
