Page 124 - DMTH403_ABSTRACT_ALGEBRA

Unit 10: Finite Abelian Groups

Proposition: Every finite abelian group has a natural structure as a module over the ring Z. Notes
As with vector spaces, one goal is to be able to express a finite abelian group in terms of simpler
building blocks. For vector spaces we can use one-dimensional spaces as the building blocks; for
finite abelian groups, it seems natural to use the simple finite abelian groups.

Recall that in an arbitrary group G, a subgroup N  G is called a normal subgroup if gxg  N,
–1
for all x  N and all g  G. Then G is said to be a simple group if its only normal subgroups are
{1} and G. If the group A is abelian, then all subgroups are normal, and so A is simple iff its only
subgroups are the trivial subgroup (0) and the improper subgroup A. The same definition is
given for modules: a nonzero module M is a simple module if its only submodules are (0) and
M. When you view a finite abelian group as a Z-module, then, of course, the two definitions
coincide.

Note Any cyclic finite abelian group is isomorphic to Z or Z , for some n.
n
Outline of the Proof: Let A be a cyclic finite abelian group that is generated by the single
element a. Define the group homomorphism f : Z  A by setting f(n) = na, for all n  Z. Note that
f maps Z onto A since f(Z) = Za = A. If f is one-to-one, then A is isomorphic to Z. If f is not
one-to-one, we need to use the fundamental homomorphism theorem and the fact that every
subgroup of Z is cyclic to show that A is isomorphic to Z , where n is the smallest positive
n
integer such that na = 0.
Proposition: A finite abelian group is simple iff it is isomorphic to Z , for some prime number p.
p
Proof: First, let A be a finite abelian group isomorphic to Z , where p is a prime number. The
p
isomorphism preserves the subgroup structure, so we only need to know that Z has no proper
p
nontrivial subgroups. This follows from the general correspondence between subgroups of Z n
and divisors of n, since p is prime precisely when its only divisors are ±1 and ±p, which correspond
to the subgroups Z and (0), respectively.
p
Conversely, suppose that A is a simple finite abelian group. Since A is nonzero, pick any
nonzero element a  A. Then the set Za = {na | n  Z} is a nonzero subgroup of A, so by
assumption it must be equal to A. This shows that A is a cyclic group. Furthermore, A can’t be
infinite, since then it would be isomorphic to Z and would have infinitely many subgroups. We
conclude that A is finite, and hence isomorphic to Z , for some n. Once again, the correspondence
n
between subgroups of Z and divisors of n shows that if Z is simple, then n must be a prime
n
n
number.
A module M is said to be semisimple if it can be expressed as a sum (possibly infinite) of simple
submodules. Although the situation for finite abelian groups is more complicated than for
vector spaces, it is natural to ask whether all finite abelian groups are semisimple.

Example: The group Z is not a semisimple Z-module. First, Z is not a simple group.
4
4
Secondly, it cannot be written non-trivially as a direct sum of any subgroups, since its subgroups
lie in a chain Z  2Z  (0), and no two proper nonzero subgroups intersect in (0).
4
4
Example: The group Z is a semisimple Z-module. To see this, define f : Z  Z  Z by
6
3
6
2
setting f(0) = (0, 0), f(1) = (1, 1), f(2) = (0, 2), f(3) = (1, 0), f(4) = (0, 1), f(5) = (1, 2). You can check that
this defines an isomorphism, showing that Z is isomorphic to a direct sum of simple finite
6
abelian groups.

LOVELY PROFESSIONAL UNIVERSITY 117