Page 123 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Since 27 is a power of an odd prime, it follows that Z is cyclic. This can also be shown directly

27
by guessing that 2 is a generator.
Since Z has order 3 - 3 = 18, an element can only have order 1, 2, 3, 6, 9 or 18. We have
2

3
27
2 = 4,
2
2 = 8,
3
2  8  10, and
2
6
2  2 · 2  8 · 10  -1,
3
6
9
so it follows that 2 must be a generator.
We conclude that Z  216  Z × Z × Z .
2
2
18
To give the first decomposition, states that any finite abelian group is isomorphic to a direct
product of cyclic groups of prime power order. In this decomposition we need to split Z up into
18
cyclic subgrops of prime power order, so we finally get the decomposition
Z  216  Z × Z × Z × Z .
2
9
2
2
On the other hand, the second decomposition, where any finite finite abelian group is written as
a direct product of cyclic groups in which the orders any component is a divisor of the previous
one. To do this we need to group together the largest prime powers that we can. In the first
decomposition, we can combine Z and Z to get Z as the first component. We end up with
9
2
18
Z  216  Z × Z × Z 2
2
18
as the second way of breaking Z  up into a direct product of cyclic subgroups.
216
Example: Let G and H be finite abelian groups, and assume that they have the following
property. For each positive integer m, G and H have the same number of elements of order m.
Prove that G and H are isomorphic.
Solution: We give a proof by induction on the order of |G|. The statement is clearly true for
groups of order 2 and 3, so suppose that G and H are given, and the statement holds for all
groups of lower order. Let p be a prime divisor of |G|, and let G and H be the Sylow
p
p
p-subgroups of G and H, respectively. Since the Sylow subgroups contain all elements of order
a power of p, the induction hypothesis applies to G and H . If we can show that G H for all p,
p
p
p
p
then it will follow that G H, since G and H are direct products of their Sylow subgroups.
Let x be an element of G with maximal order q = p . Then < x > is a direct factor of G , so there
m
p
p
is a subgroup G’ with G = < x > × G’. By the same argument we can write H = < y > × H’, where
p
p
y has the same order as x.
Now consider < x > × G’ and < y > × H’. To construct each of these subgroups we have removed
p
p
elements of the form (x , g’), where x has order q and g’ is any element of G’. Because x has
k
k
maximal order in a p-group, in each case the order of g’ is a divisor of q, and so (x , g’) has order
k
q since the order of an element in a direct product is the least common multiple of the orders of
the components. Thus to construct each of these subgroups we have removed (p – p ) · |G’|
m
m-1
elements, each having order q. It follows from the hypothesis that we are left with the same
number of elements of each order, and so the induction hypothesis implies that < x > × G’ and
p
< y > × H’ are isomorphic. But then G’  H’, and so G  H , completing the proof.
p
p
p
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