# What does the notation $\mathbb{Z} e_n$ mean?

What does the notation in the direct sum $$\underset{n \in \delta}{\oplus} \mathbb{Z} e_n$$ means? Why is there an $$e_n$$, isn't it an element of $$\lbrace -1,1 \rbrace$$ and does not changes anything?

The $$\delta$$ is the cardinality and the context is the direct sum of rational groups.

paper

Pure subgroups of completely decomposable groups - an algorithmic approach. By Daniel Herden and Lutz Strüngmann, page 5

• This is a notation for what? Also, what is $\delta$? – Jo Be Aug 14 at 6:40
• What is the context? – Wuestenfux Aug 14 at 6:41

Whenever you have some set $$M$$ you can consider the free abelian group $$\mathbb{Z}^{\oplus M} = \bigoplus_{m \in M} \mathbb{Z}m$$. That means you take one copy of $$\mathbb{Z}$$ for every element $$m \in M$$ and treat that element like $$1$$ (or $$-1$$ if you wish). So the free abelian group with basis $$\lbrace x,y,z \rbrace$$ is given by formal expressions of the form $$ax + by + cz$$, where $$a,b,c \in \mathbb{Z}$$. For this construction it does not matter at all what type of object the elements in $$M$$ are.
I did not check what your $$e_n$$ are, but as you can see they do not have to be $$1$$ or $$-1$$. It still works if they are noodles, cars or whatever you want. It will still be useful to figure out what they are as you are using them as a basis here.
Maybe the author really just picks $$e_n \in \lbrace \pm 1 \rbrace$$ and uses that notation to have $$(e_1,0,\dots,0)$$ etc as basis vectors later on without choosing between $$\pm 1$$.
I assume the $$e_n$$ are identities of a family of algebraic objects.
Omitting the $$e_n$$ gives you an isomorphic object, but including the $$e_n$$ makes clear that you work with that object as subobject of some other object.