Page 114 - DMTH403_ABSTRACT_ALGEBRA

Unit 9: Direct Products

(Recall that D = < (x, y | x = e = Y and yx = x y} > .) Notes
P
2
-l
2p
Proof: As in the proof of Theorem 9, there exists a subgroup P = < x > of order p with
P  G and a subgroup Q = < y > of order 2, since p > 2. Since (2, p) = 1,
P  Q = {e}.  o(PQ) = o(G).

 G = PQ.
Now, two cases arise, namely, when Q  G and when Q  G.

If Q  G, then G = P × Q. And then G = <xy>.
If Q is not normal in G, then G must be non-abelian.

(Remember that every subgroup of an abelian group is normal.)
-1
 xy  yx.  y xy  x.
Now, since P = < x >  G, y xy  P.  y xy = x’, for some r = 2 ,... . , p - 1.
-1
-1
Therefore, y xy = y (y xy) = y x y = (y xy) = (x ) = x r 2
-1
-2
2
-1
–1 r
-1
r r
r
r
2
 x = x , since o(y) = 2.
 x r 2 1   6 .
But o(x) = p. Therefore, by Theorem 4 of Unit 4, p | r – 1, i.e., p | (r – 1) (r + 1)
2
 p | ( r – 1 ) or p | ( r + 1 ). But 2  r  p – 1.  p = r + l,
i.e., r = p – 1. So we see that
y xy = x =’ x = x -1
r
p-1
-1
So, G = PQ = < {x, y | x = e, y = e, y xy = x > , which is exactly the same algebraic structure as
-1
p
-1
2
that of D .
2p
 G = D = {e, x, x , ... , x , y, xy, x y, . . . . , x y]
p-1
2
p-1
2
2p
Example: What are the possible algebraic structures of a group of order 6?
Solution: Let G be a group of order 6. Then, by theorem 10, G  Z or G  Ds. You must have
6
already noted that S  D . So, if G is not cyclic, then G  S .
3
3
6
Now, from Theorem 6 of Unit 4, we know that if o(G) is a prime, then G is cyclic. Thus, groups
of orders 2, 3, 5 and 7 are cyclic. This fact allows us to classify all groups whose orders are 1, 2, 3,
5, 6, 7 or 10. What about the structure of groups of order 4 = 22 and 9 = 3 ? Such groups are covered
2
by the following result.
Theorem 11: If G is a group of order p , p a prime, then G is abelian.
2
We will not prove this result, since its proof is beyond the scope of this course. But, using this
theorem, we, can easily classify groups of order p .
2
Theorem 12: Let G be a group such that o(G) = p , where p is a prime. Then either G is cyclic or
2
G = Z × Z , a direct product of two cyclic groups of order p.
p
p
Proof: Suppose G has an element a of order p . Then G = < a > .
2
On the other hand, suppose G has no element of order. Then, for any x E G, o(x) = I or o(x) = p.
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