# A group of finite order can be infinitely large?

• jessicaw
In summary, a group of finite order means a group with a finite number of elements. Every member of this group must have finite order. Repeating elements in a group will eventually lead back to the identity element and not add anything new. The proof for this is simple and can be seen shortly after learning the definition of order. An infinite group can have finite order for each element, but repeating elements will not enlarge the group.
jessicaw
a group or a cyclic group of finite order can i just repeatedly write down the repeated elements and form a very large even infinite group?

What do you mean by "write down repeating elements". If the group is of finite order, then every member of it has finite order. Eventually, "repeating elements" get you back to the identity and then you get nothing new.

HallsofIvy said:
What do you mean by "write down repeating elements". If the group is of finite order, then every member of it has finite order. Eventually, "repeating elements" get you back to the identity and then you get nothing new.

why" If the group is of finite order, then every member of it has finite order."? The group may not be a cyclic group. For example {1,2,3} the group is finite the member may have infinite order like the element {2}?

"get you back to the identity and then you get nothing new." is what i am confused; i am always wondering if i can write a group called {1,1,1,1,1,1,1,1,1,1,1,1....}. Please help. Thanks

What do you mean by the group {1, 2, 3}? Surely not the group with three members since that is a cyclic group.

In any case, the proof that every element of a finite group has finite order is elementary. I would be surprised if you didn't see it shortly after learning the definition of "order".

Let a be any member of the finite group, G. Then, for all n, $a^n\in G$. Since there are only a finite number of members of G while there are an infinite number of positive integers, by the pigeon hole principal, we must have $a^k= a^j$ for some distinct j and k. If j< k, then $a^ka^{-j}= a^{k- j}= a^ja^{-j}= e$ where e is the group identity and I am using "$a^{-j}$ to represent the inverse of $a^j$. If k< j, then $a^ka^{-k}= e= a^{j}a^{-k}= a^{j-k}$.

In either case, a has finite order.

Usually what is meant by a group of finite order is a group with finitely many elements so by definition it can not be infinite.

But if you just require that each element be of finite order then one can have an infinite group with no problem

take the infinite direct product of Z/2 with itself.

I think the OP had in mind something like, for example, with Z_3, writing {1,2,3,1,2,3,1,2,3,1,2,3,1,...}. And the answer is, no, that is actually equal to {1,2,3}, i.e. "repeating elements" does not enlarge a set.