Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is written on Page 93 of Derek J.S. Robinson's A Course in the Theory of Groups:

Let $G$ be an arbitrary abelian group, $T$ its torsion-subgroup. For an arbitray prime $p$, denote $G_p$ as the $p$-primary component of $G$. Then $T$ is the sum of all the $G_p$.

Isn't $T$ the direct sum of $G_p$? For any element $a \in T$, $a$ is of finite order. If the order of $a$ equals $p_1^{n_1} \cdots p_k^{n_k}$ where $p_i$'s are primes and $n_i$'s are positive integers. Then there must exist $a_i \in G_{p_i}$, $i=1, \cdots k$, such that $a = \sum_i a_i$. Because if $q \notin \{p_1, \cdots p_k \}$, and a nonzero $b \in G_q$ appears in the summation form of $a$, then the order of $a$ must be divisible by $q$. But in fact it isn't.

In my understanding, a direct sum is like the coproduct in category language, where every element written in the summation form has only finite nonzero components. But a sum can take the sum of infinitely many elements. In fact, picking infinitely elements of $G$, one per $G_p$, and adding them together may get an element of infinite order, lying outside $T$.

Where am I wrong?

Then I considered $\mathbb{Q} / \mathbb{Z}$. What is the sum of inverse of primes $\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \cdots$? Is it still in $\mathbb{Q}$? The definition of a group $(G, *, 1)$ only says for finitely many $a_i \in G$, $a_1 * \cdots * a_n$ is in $G$; it doesn't guarantee the closedness of $G$ under infinitely many times of applying the operation $*$.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

No, a "sum of subgroups $H_i$" is the subgroup generated by the subgroups, which consists only of all finite sums of elements of $\cup H_i$; there is no such thing as "infinite sums" in general groups (you need some notion of convergence in order to talk about infinite sums).

In particular, the abelian group $\mathbb{Q}$ does not contain infinite sums. With repeated application of a binary operation you can get sums with arbitrarily large but finite number of summands, not infinite sums/series.

There are two types of "direct sum". The so-called "External Direct Sum" (which is a subgroup of the direct product, etc), and the so-called "Internal Direct Sum".

An abelian group $G$ is the external direct sum of groups $\{G_i\}_{i\in I}$ if $G$ is exactly the set of $I$-tuples, $\prod_{i\in I}G_i$, with almost all entries trivial.

If, on the other hand, $\{H_i\}_{i\in I}$ are a family of subgroups of $G$ such that $\sum H_i = G$ and $H_i\cap\left(\sum_{j\neq i}H_j\right)=\{0\}$, then $G$ is isomorphic to $\oplus H_i$, but is not actually equal to this direct sum. Then we say that $G$ is the internal direct sum of the $H_i$.

Technically, internal and external direct sums are different types of constructions. Morally and in practice, they are the same, because if $G$ is the internal direct sum of its subgroups $H_i$, then it is isomorphic to the external direct sum of the $H_i$; and the external direct sum of the $H_i$ is an internal direct sum of the subgroups $\mathcal{H}_i$, where $\mathcal{H}_i$ is the subgroup of all elements whose $j$th entries are equal to $0$ for all $j\neq i$. So in practice, the distinction is dropped.

Robinson is perfectly correct in saying that $T$ is the sum of the subgroups (i.e., the subgroup generated by them). You are, also, correct in noting that this sum is in fact "direct" (that is, that $T$ is the internal direct sum of the $p$-torsion parts of $G$). If a group is the internal direct sum of some subgroups, then it is also the sum of these subgroups (the converse does not hold, if there are intersections between the subgroups).

share|improve this answer
I misunderstood the term sum. Thank you very much for your perfect explanation! –  ShinyaSakai Nov 20 '11 at 15:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.