# All bases for a finitely-generated abelian group have the same cardinality.

Let $B$ be a subgroup of a free abelian group $A$ with basis $(x_i)_{i=1...n}$. It has already been shown that $B$ has a basis of cardinality $\leq n$.

... We also observe that our proof shows that there exists at least one basis of $B$ whose cardinality is $\leq n$. We shall therefore be finished when we prove the last statement, that any two bases of $B$ have the same cardinality. Let $S$ be one basis, with a finite number of elements $m$. Let $T$ be another basis, and suppose that $T$ has at least $r$ elements. It will suffice to prove that $r \leq m$ (one can then use symmetry).

Let $p$ be a prime number. Then $B/pB$ is a direct sum of cyclic groups of order $p$, with $m$ terms in the sum. Hence its order is $p^m$. Using the basis $T$ instead of $S$, we conclude that $B/pB$ contains an $r$-fold product of cyclic groups of order $p$, whence $p^r \leq p^m$ and $r \leq m$, as was to be shown. (Note that we did not assume a priori that T was finite.)

I've bolded the parts I'm having trouble with. How do I show the first part and what's an $r$-fold product?

Alternative proofs to the problem welcome.

I know that $pB = \{ \sum_{i} p k_i x_i\ | \sum_{i} k_i x_i \in B\}$ and that it forms a normal subgroup, $A$ being abelian.

That $S = \{s_1,\,\dotsc,\, s_m\}$ is a basis of $B$ means that every $b \in B$ can be written in a unique way as $b = \sum\limits_{i=1}^m k_i\cdot s_i$ with all $k_i \in \mathbb{z}$. Thus $B$ is the direct sum of $m$ copies of $\mathbb{Z}$,

$$B = \bigoplus_{i=1}^m \mathbb{Z}\cdot s_i.$$

Then we have

$$pB = \bigoplus_{i=1}^m p\mathbb{Z}\cdot s_i,$$

and that yields

$$B/pB \cong \bigoplus_{i=1}^m (\mathbb{Z}/p\mathbb{Z})\cdot s_i,$$

that $B/pB$ is the direct sum of $m$ cyclic groups of order $p$.

Now $T$ is by assumption also a basis of $B$, so we also have

$$B = \bigoplus_{\tau \in T} \mathbb{Z}\cdot \tau,$$

and

$$B/pB \cong \bigoplus_{\tau \in T} (\mathbb{Z}/p\mathbb{Z})\cdot\tau.$$

If $T$ contains at least $r$ elements, say we have $t_1,\,\dotsc,\,t_r \in T$, then

$$\bigoplus_{j=1}^r (\mathbb{Z}/p\mathbb{Z})\cdot t_j$$

is a subgroup of $B/pB$, and thus $B/pB$ contains the direct sum of $r$ cyclic groups of order $p$. Since for finitely many summands/factors the direct sum and direct product of (abelian) groups are isomorphic, it contains a product of $r$ cyclic groups of order $p$, an $r$-fold product of $\mathbb{Z}/p\mathbb{Z}$.

• How can the last sum be a subgroup of $B/pB$ if we're talking about isomorphisms? – Shine On You Crazy Diamond Sep 28 '13 at 20:10
• Okay I understand that the direct product and sum coincide in this case. – Shine On You Crazy Diamond Sep 28 '13 at 20:13
• Okay, the only part I don't get now is how $r \leq m$ now. But since that wasn't part of my question I've checked your answer. – Shine On You Crazy Diamond Sep 28 '13 at 20:20
• I should have stayed with $=$ also for the quotients ;) But saying "contains an isomorphic image of $\bigoplus\limits_{j=1}^r (\mathbb{Z}/p\mathbb{Z})\cdot t_j$" would achieve the same, showing that $r \leqslant m$. – Daniel Fischer Sep 28 '13 at 20:20
• What lemma are you using to show that $r\leq m$ ? – Shine On You Crazy Diamond Sep 28 '13 at 20:21

1) "$r$-fold" means a direct sum of $r$ exemplars of a group.

2) Every element of $B/pB$ has the order $p$. It is known that such Abelian group is a direct sum of cyclic groups of order $p$.