Page 112 - DMTH403_ABSTRACT_ALGEBRA

Unit 9: Direct Products

mathematician, proved a remarkable extension of Cauchy’s result. This result, called the first Notes
Sylow theorem, has turned out to be the basis of finite group theory. Using this result we can
say, for example, that any group of order 100 has subgroups of order 2, 4, 5 and 25.
Theorem 6 (First Sylow Theorem): Let G be a finite group such that o (G) = p m, where p is a
n
prime, n  1 and (p, m) = 1. Then G contains a subgroup of order pk  k = 1, . . . , n.
We shall not prove this result or the next two Sylow theorems either. But, after stating all these
results we shall show how useful they are.
The next theorem involves the concepts of conjugacy and Sylow p-subgroups which we now
define.
Definition: Two subgroups H and K of a group G are conjugate in G if  g  G such that
K = g Hg and then K is called a conjugate of H in G.
-1
Now we define Sylow p-subgroups.
Definition: Let G be a finite group and p be a prime such that p | o(G) but p o(G), for some n
n
n+1
 1. Then a subgroup of G of order pn is called a Sylow p-subgroup of G.
So, if o(G) = p m, (p, m) = I, then a subgroup of G of order p” is a Sylow p-subgroup. Theorem 6
n
says that this subgroup always exists. But, a group may have more than one Sylow p-subgroup.
The next result tells us how two Sylow p-subgroups of a group are related.
Theorem 7 (Second Sylow Theorem): Let G be a group such that o(G) = p m, (p, m) = 1, p a prime.
n
Then any two Sylow p-subgroups of G are conjugate in G.
And now let us see how many Sylow p-subgroups a group can have.
Theorem 8 (Third Sylow Theorem): Let G be a group of order p m, where (p, m) = 1 and p is a
n
prime. Then n , the number of distinct Sylow p-subgroups of G, is given by n = 1 + kp for some
p
p
k  0. And further, n | o(G).
p
We would like to make a remark about the actual use of Theorem 8.
Remark: Theorem 8 says that n  1(mod p). (n , pn) = 1. Also, since np | o(G), using Theorem 9
p,

p
of Unit 1 we find that n | m. This fact helps us to cut down the possibilities for n,, as you will see
p
in the following examples.
Example: Show that any group of order 15 is cyclic.
Solution: Let G be a group of order 15 = 3 × 5. Theorem 6 says that G has a Sylow 3-subgroup.
Theorem 8 says that the number of such subgroups must divide 15 and must be congruent to
l(mod 3). In fact, by Remark 3 the number of such subgroups must divide 5 and must be congruent
to l(mod 3). Thus, the only possibility is 1. Therefore, G has a unique Sylow 3-subgroup, say H.
Hence, by Theorem 7 we know that H  G. Since H is of prime order, it is cyclic.
Similarly, we know that G has a subgroup of order 5. The total number of such subgroups is 1,6
or 11 and must divide 3. Thus, the only possibility is 1. So G has a unique subgroup of order 5, say
K. Then K  G and K is cyclic.

Now, let us look at H  K. Let x  H  K. Then x  H and x  K.
 o(x) | o(H) and o(x) | o(K) i.e., o(x) | 3 and o(x) | 5.
 o(x) = 1.  x = e. That is, H  K = {e}. Also,

 G = HK.
So, G = H × K  Z X Z = Z ,
5
3
15
LOVELY PROFESSIONAL UNIVERSITY 105