Page 127 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Using o(x) for the order of an element x, let s be the order of b + Za in the factor group G/Za.
Then sb  Za, and we can write sb = (qt)a for some exponent qt such that t = p for some  and

p | q. Then qa is a generator for Za, since q is relatively prime to o(a). Since s is a divisor of the
.
.
order of b, we have o(b)/s = o(sb) = o((qt)a) = o(a)/t, or simply, o(b) t = o(a) s. All of these are
powers of p, and so o(b)  o(a) implies that s|t, say t = ms. Then x = (qm)a is a solution of the
equation sb = sx. If d = b – x, then d + Za = b + Za and so sd = sb – sx = sb – sb = 0. Therefore,
Zd  Za = (0), since nd  Za implies n(b – x) = nb – nx  Za. Thus, nb  Za implies n(b + Za) = Za
in G/Za, so s|n and nd = 0.
(b) The outline of this part is to factor out Za and use induction to decompose A/Za into a direct
sum of cyclic groups. Then part (a) can be used to choose the right preimages of the generators
of A/Za to generate the complement B of Za.
We use induction on the order of A. If |A| is prime, then A is cyclic and there is nothing to prove.
Consequently, we may assume that the statement of the lemma holds for all groups of order less
than |A| = p . If A is cyclic, then we are done. If not, let Za be a maximal cyclic subgroup, and use

the induction hypothesis repeatedly to write A/Za as a direct sum B  B  . . .  B of cyclic
1
n
2
subgroups.
We next use part (a) to choose, for each i, a coset a + Za that corresponds to a generator of A such
i
i
that Za  Za = (0). We claim that A  Za  B for the smallest subgroup B = Za + Za + · · · + Za n
i
1
2
that contains a , a , . . ., a .
1
n
2
First, if x  Z  (Za +· · · + Za ), then x = m a +· · · + m a  Za for some coefficients m , . . . ,m .
n
n
a
1
1
n n
1 1
Thus x + Za = (m a + · · · + m a ) + Za = Za, and since A/Za is a direct sum, this implies that
n n
1 1
m a + Za = Za for each i. But then m a  Za, and so m a = 0 since Za  Za = (0). Thus x = 0.
i i
i i
i
i i
Next, given x  A, express the coset x + Za as (m a +· · · + m a ) + Za for coefficients m , . . ., m .
1
n n
n
1 1
Then x  xZa, and so x = ma + m a + · · · + m a for some m.
1 1
n n
Thus, we have shown that Za  B = (0) and A = Za + B, so A  Za  B.
Theorem 2 (Fundamental Theorem of Finite Abelian Groups): Any finite abelian group is
isomorphic to a direct sum of cyclic groups of prime power order. Any two such decompositions
have the same number of factors of each order.
Proof: We first decompose any finite abelian group A into a direct sum of p-groups, and then we
can use the previous lemma to write each of these groups as a direct sum of cyclic subgroups.
Uniqueness is shown by induction on |A|. It is enough to prove the uniqueness for a given
p-group. Suppose that
Z   Z   · · ·  Z  = Z   Z   · · ·  Z 
p 1 p 2 p n p 1 p 2 p m
where     . . .   and     . . .   . Consider the subgroups in which each element has
1
2
2
1
n
m
been multiplied by p. By induction,  – 1 =  – 1, . . ., which gives  =  , . . ., with the possible
1
1
1
1
exception of the  ’s and  ’s that equal 1. But the groups have the same order, and this determines
i
j
that each has the same number of factors isomorphic to Z . This completes the proof.
p
Self Assessment
1. A .................. is a set, A together with an operations “.”. That combines any two elements a
and b to form another element denoted a.b.
(a) cyclic (b) permutation
(c) abelian (d) normal
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