Trying to understand the definition of a free Abelian group.

Let $B$ be a subset of an (additive) abelian group $F$. Then $F$ is free abelian with basis $B$ if the cyclic subgroup $\langle b \rangle$ is infinite cyclic for each $b \in B$ and $F=\sum_{b \in B} \langle b \rangle$ (direct sum). A free abelian group is thus a direct sum of copies of $\Bbb Z$. A typical element $x \in F$ has a unique expression $$x = \sum m_b b$$ where $m_b \in \Bbb Z$ and almost all $m_b$ (all but a finite number) are zero.

I understand that by $F = \sum_{b \in B} \langle b \rangle$ (direct sum), they mean an external direct product of the infinite cyclic subgroups $\langle b \rangle$.

But what does $x=\sum m_bb$ mean?

Since $F = \sum_{b \in B} \langle b \rangle = \{ ...,-b_1, 0, b_1, ...\} \oplus \dots \oplus \{\dots, -b_n, 0, b_n, \dots\}$ and $x \in F$, then $x=(m_1b_1, \dots, m_nb_n)$. But each $b$ is an element of $F$ and therefore by the same logic shouldn't each $b$ be of the form $(j_1b_1, \dots, j_nb_n)$? And therefore $x=(m_1(j_1b_1, \dots, j_nb_n), etc \dots)$?

What exactly am I misunderstanding here? This seems like a circular definition.

• Direct sums are only direct products in the special case of finitely many terms. Internal direct sums are simply a special form for a direct sum. – Excluded and Offended May 30 '17 at 15:30

I understand that by $F = \sum_{b \in B} \langle b \rangle$ (direct sum), they mean an external direct product of the infinite cyclic subgroups $\langle b \rangle$.

No. They mean an internal direct sum of these cyclic subgroups.

But what does $x=\sum m_bb$ mean?

It means that every $x$ may be written as an integral linear combination of the elements of $B$. Moreover, because the sum is direct, the coefficients are uniquely determined by $x$.

This is just the generalisation of the concept of a vector space with a fixed basis if we allow the coefficients (scalars) to lie in the ring $\mathbb{Z}$ instead of in a field. The price to pay is Abelian groups are generally not free, in opposition with vector spaces (which are always free).

Example. Consider $F = \mathbb{Z}^2$, and $B = \{b_0,b_1 \} = \{ (1,0),(-1,1)\}$. Then for example $$x = (2,-1) = b_0 - b_1 = 1b_0 + (-1)b_1,$$ so $m_{b_0} = 1$, $m_{b_1} = -1$ in this case. Note that the addition operation $+$ is the one in $F$ (hence the internal direct sum).

• So for groups, direct sum means internal direct product? – Oliver G May 30 '17 at 15:29
• @OliverG For Abelian groups, or more generally, modules, the direct sum and direct product differ only when the number of summands is infinite. In the direct sum, almost all coefficients are zero (as otherwise we cannot literally add together an infinite number of elements). – Alex Provost May 30 '17 at 15:32
• In the book Contemporary Abstract Algebra by Gallian, he defines a direct sum of two groups to be the external direct product of two abelian groups with additive operations. And this is the same definition used in A first course in abstract algebra by Fraleigh. So their definitions are wrong? – Oliver G May 30 '17 at 16:02
• @OliverG As I said in my previous comment, the distinction between a direct sum and a direct product only becomes essential with an infinite number of summands. Also note that these authors are defining a so-called "external" direct sum; the direct sum in your question is "internal" because we are adding subgroups inside a larger group. We are not defining a new addition operation on a direct product of two groups; rather, we are using the addition already present in the large group $F$. – Alex Provost May 30 '17 at 16:17
• @OliverG Multiplicative notation seems inappropriate here. To say that $F = \bigoplus_{b \in B} \langle b \rangle$ (standard notation for the direct sum) means that a) $F$ is a sum of the cyclic subgroups, i.e., any element in $F$ may be written as $\sum_b m_b b$ for some $b \in B$ and $m_b \in \mathbb{Z}$, and b) that the sum is direct, i.e., that each cyclic subgroup $\langle b \rangle$ intersects the rest of the sum $\sum_{b' \in B \setminus \{ b \}} \langle b' \rangle$ trivially. This last condition is equivalent to the fact that the expression $\sum_b m_b b$ is unique for any element. – Alex Provost May 30 '17 at 23:43

There is problema already when you write that $F = \sum_{b \in B} \langle b \rangle = \{\ldots,-b_1, 0, b_1,\ldots\} \oplus \dots \oplus \{\dots, -b_n, 0, b_n, \dots\}$, because you assume that $B$ is countable. But the greatest problem comes later. What follows from this is that each element of $F$ is a sum of an element of $\{\ldots,-b_1, 0, b_1,\ldots\}$ with an element of $\{\ldots,-b_2, 0, b_2,\ldots\}$ with an element of $\{\ldots,-b_3, 0, b_3,\ldots\}$ and so on, where all but finitely many of the summands is $0$.

The coordinate expression of $b$,

$$b = \sum_{c \in B} m_c c$$

is the one where

$$m_c = \begin{cases} 1 & c = b \\ 0 & c \neq b \end{cases}$$

The formula for $F$ is actually an internal direct sum.

However, in a related construction of the external direct sum, (when it doesn't lead to confusion) one usually uses the same letter $b$ both for the element of the given set and its image in the group (i.e. is the tuple with a $1$ in the $b$-th coordinate and $0$ elsewhere).