Page 113 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes
Example: Show that a group G of order 30 either has a normal subgroup of order 5 or a
normal subgroup of order 3, i.e. G is not simple. A group G is called simple if its only normal
subgroups.
Solution: Since 30 = 2 × 3 × 5, G has a Sylow 2-subgroup, a Sylow 3-subgroup and a Sylow
5-subgroup. The number of Sylow 5-subgroups is of the form 1 + 5k and divides 6. Therefore, it
can be 1 or 6. If it is 1, then the Sylow 5-subgroup is normal in G.
On the other hand, suppose the number of Sylow 5-subgroups is 6. Each of these subgroups are
distinct cyclic groups of order 5, the only common element being e. Thus, together they contain
24 + 1 = 25 elements of the group. So, we are left with 5 elements of the group which are of order
2 or 3. Now, the number of Sylow 3-subgroups can be 1 or 10. We can’t have 10 Sylow
3-subgroups, because we only have at most 5 elements of the group which are of order 3. So, if
the group has 6 Sylow 5-groups then it has only 1 Sylow 3-subgroup.
Now let us use the powerful Sylow theorems to classify groups of order 1 to 10. In the process we
will show you the algebraic structure of several types of finite groups.

9.3 Groups of Order 1 to 10

Here, we will apply the results of the above discussion to study some finite groups. In particular,
we will list all the groups of order 1 to 10, up to isomorphism.
We start with proving a very useful result.

Theorem 9: Let G be a group such that o(G) = pq, where p, q are primes such that p > q and
q | p – 1. Then G is cyclic.

Proof: Let P be a Sylow p-subgroup and Q be a Sylow q-subgroup of G. Then o(P) = p and
o(Q) = q. Now, any group of prime order is cyclic, so P = < x > and Q = < y > for some x, y  G.
By the third Sylow theorem, the number n of subgroups of order p can be 1 , 1+ p, 1 + 2p, . . . , and
p
it must divide q. But p > q. Therefore, the only possibility for n is 1. Thus, there exists only one
p
Sylow p-subgroup, i.e., P. Further, by Sylow’s second theorem P  G.
Again, the number of distinct Sylow q-subgroups of G is n , = 1 + kq for some k, and n, | p. Since
q
p is a prime, its only factors are 1 and p.  n, = 1 or n = p. Now if 1 + kq = p, then q | p – 1. But
q
we started by assuming that 9 | p - 1. So we reach a contradiction. Thus, n = 1 is the only
q
possibility. Thus, the Sylow q-subgroup Q is normal in G.
Now we want to show that G = P × Q. For this, let us consider P  Q. The order of any element of
P  Q must divide p as well as q, and hence it must divide (p, q) = 1.
P  Q = {e}.  o(PQ) = o(P) o(Q) = pq = o(G).  G = PQ.
So we find that G = P × Q  Z × Z  Z ,
pq
p
p
Therefore, G is cyclic of order pq.
Using Theorem 9, we can immediately say that any group of order 15 is cyclic. Similarly, if
o(G) = 35, then G’is cyclic.
Now if q | p – 1, then does o(G) = pq imply that G is cyclic? Well, consider S . You know that o(S )
3
3
= 6 = 2.3, but S is not cyclic. In fact, we have the following result.
3
Theorem 10: Let G be a group such that o(G) = 2p, where p is an odd prime, Then either G is cyclic
or G is isomorphic to the dihedral group D of order 2p.
2p

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