# Show that in any cyclic additive group of an odd order the sum of all elements is equal to the identity element 0.

I'm trying to solve this problem, but I have a problem with understanding what is an order and generator of a cyclic group. I know that an order is just the number of elements in a group. A generator is an element which can ''produce'' all the elements in a group. Our prof has given us an example: The generator of a cyclic group $G = <\mathbb{Z}, +>$ is $1$. I can't understand how we can get negative integers from $1$.
I also doubt that I've understood the definition of the order of an element of a group correctly. Is it just the number of times we have to apply an operation to an element to get the identity element?
I'm also not sure if a group can have two or more generators.
Thank you for help in advance!

• You seem to be asking quite a lot of questions here, and while they are related, the question becomes very unfocused because of it. – Tobias Kildetoft Sep 14 '15 at 10:24
• You could start by taking some small example (say the group of integers mod $5$) and playing around with the elements there until you have some idea how the group works. Find the orders of elements, inverses and so on. You might find ideas for the proof there. – Leppala Sep 14 '15 at 10:35

• the order of a group $G$ is just the cardinality (number of elements) of that group;
• the order of an element $a\in G$ is the order of the group $\langle a\rangle$ generated by $a$,
• the group $\langle a\rangle$ generated by $a$ (or similarly for any subset of $G$) is the smallest subgroup of $G$ containing $a$.
Consequently, the element $1\in\mathbb Z$ generates $\mathbb Z$ because any subgroup of $\mathbb Z$ containing $1$ must also contain its inverse $-1$.
Now let $G$ be a finite abelian group. Then if we add all elements of $G$, we can most of the time cancel each summands with its inverse. This only fails for elements $a\in G$ that are their own inverse, i.e., with $a+a=0$. Hence for any finite abelian group $$\sum_{g\in G}g=\sum_{g\in G\atop 2g=0}g$$ If $G$ is of odd order $n=2k+1$, then $2g=0$ together with $ng=0$ implies $g=ng-2kg=0$, so that ultimately $$\sum_{g\in G}g=0$$ for any finite abelian group $G$ of odd order. (Cyclic is not needed)