1Groups
IB Groups, Rings and Modules
1.6 Finite abelian groups
We now move on to a small section, which is small because we will come back to
it later, and actually prove what we claim.
It turns out finite abelian groups are very easy to classify. We can just write
down a list of all finite abelian groups. We write down the classification theorem,
and then prove it in the last part of the course, where we hit this with a huge
sledgehammer.
Theorem (Classification of finite abelian groups). Let
G
be a finite abelian
group. Then there exist some d
1
, ··· , d
r
such that
G
∼
=
C
d
1
× C
d
2
× ··· × C
d
r
.
Moreover, we can pick
d
i
such that
d
i+1
| d
i
for each
i
, and this expression is
unique.
It turns out the best way to prove this is not to think of it as a group, but
as a Z-module, which is something we will come to later.
Example. The abelian groups of order 8 are C
8
, C
4
× C
2
, C
2
× C
2
× C
2
.
Sometimes this is not the most useful form of decomposition. To get a nicer
decomposition, we use the following lemma:
Lemma. If n and m are coprime, then C
mn
∼
=
C
m
× C
n
.
This is a grown-up version of the Chinese remainder theorem. This is what
the Chinese remainder theorem really says.
Proof.
It suffices to find an element of order
nm
in
C
m
×C
n
. Then since
C
n
×C
m
has order nm, it must be cyclic, and hence isomorphic to C
nm
.
Let
g ∈ C
m
have order
m
;
h ∈ C
n
have order
n
, and consider (
g, h
)
∈ C
m
×C
n
.
Suppose the order of (
g, h
) is
k
. Then (
g, h
)
k
= (
e, e
). Hence (
g
k
, h
k
) = (
e, e
).
So the order of
g
and
h
divide
k
, i.e.
m | k
and
n | k
. As
m
and
n
are coprime,
this means that mn | k.
As
k
=
ord
((
g, h
)) and (
g, h
)
∈ C
m
× C
n
is a group of order
mn
, we must
have k | nm. So k = nm.
Corollary. For any finite abelian group G, we have
G
∼
=
C
d
1
× C
d
2
× ··· × C
d
r
,
where each d
i
is some prime power.
Proof.
From the classification theorem, iteratively apply the previous lemma to
break each component up into products of prime powers.
As promised, this is short.