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Unformatted text preview: 188 3. PRODUCTS OF GROUPS combination of ﬁnitely many elements of B . Since S is ﬁnite, it is contained in the subgroup geneated by a ﬁnite subset B0 of B . But then
G D ZS Â ZB0 . So B0 generates G . It follows from the previous
lemma that B0 D B .
I Proposition 3.5.5. Any two bases of a ﬁnitely generated free abelian group
have the same cardinality. Proof. Let G be a ﬁnitely generated free abelian group. By the previous
lemma, any basis of G is ﬁnite. If G has a basis with n elements, then
G Š Zn , by Proposition 3.5.2. Since Zn and Zm are nonisomorphic if
m ¤ n, G cannot have bases of different cardinalities.
I Deﬁnition 3.5.6. The rank of a ﬁnitely generated free abelian group is the
cardinality of any basis. Proposition 3.5.7. Every subgroup of Zn can be generated by no more
than n elements. Proof. The proof goes by induction on n. We know that every subgroup of
Z is cyclic (Proposition 2.2.21), so this takes care of the base case n D 1.
Suppose that n > 1 and that the assertion holds for subgroups of Zk for
k < n. Let F be the subgroup of Zn generated by fe1 ; : : : ; en 1 g; thus, F
is a free abelian group of rank n 1.
Let N be a subgoup of Zn . By the induction hypothesis, N 0 D N \ F
has a generating set with no more than n 1 elements. Let ˛n denote the
nt h co-ordinate function on Zn . Then ˛n is a group homomorphism from
Zn to Z, and ˛n .N / is a subgroup of Z. If ˛n .N / D f0g, then N D N 0 ,
so N is generated by no more than n 1 elements. Otherwise, there is a
d > 0 such that ˛n .N / D d Z. Choose y 2 N such that ˛n .y/ D d . For
every x 2 N , ˛n .x/ D kd for some k 2 Z. Therefore, ˛n .x ky/ D 0,
so x ky 2 N 0 . Thus we have x D ky C .x ky/ 2 Zy C N 0 . Since N 0
is generated by no more than n 1 elements, N is generated by no more
than n elements.
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.
- Fall '08