Product of Elements in a Finite Cyclic Group with Odd or Even Order

Question

Assuming that G is a finite cyclic group, let "a" be the product of all the elements in the group.

i. If G has odd order, then a=e. Is this because there are an even number of non-trivial elements must have their inverses within the non-trivial factors within the product?

ii. If G has even order then a is not equal to e. Here there an odd number of non-trivial elements. There must also be an odd number of elements who are their own inverses, and therefore a cannot equal e?

Thanks for any help! I'm just reading a textbook and trying practice problems!

This is essentially Wilson's theorem for finite abelian groups: http://en.wikipedia.org/wiki/Wilson's_theorem#Gauss.27s_generalization. — lhf, Feb 07 '13 at 00:18
Thanks everyone. Makes complete sense now. Much appreciated! — Luke8ball, Feb 07 '13 at 00:41

score 4 · Answer 1 · answered Feb 07 '13 at 00:00

Since we know it is a finite cyclic group,

Hint: $ 1 + 2 + \ldots + n = \frac {n (n+1)}{2}$

This is a multiple of $n$ if $n$ is odd, and not a multiple of $n$ if $n$ is even.

Your statements are true, and don't just apply to a finite cyclic group. I would prefer your solution over mine, but I think the above is what the textbook had in mind.

score 3 · Answer 2 · answered Feb 07 '13 at 00:03

Let $\,G=\langle x\rangle =\{1\,,\,x\,,\,x^2\,,\,\ldots\,,\,x^{n-1}\}\,$:

$$x\cdot x^2\cdot\ldots\cdot x^{n-1}=x^{\frac{n(n-1)}{2}}=\begin{cases}x^{(n-1)k}\neq 1&\,,\,\,\text{ if}\,\;\;\;n=2k\,\,\,\text{ is even}\\{}\\x^{kn}=1&\,,\,\text{ if}\,\;\;n-1=2k\,\,\,\text{is even}\end{cases}$$

Product of Elements in a Finite Cyclic Group with Odd or Even Order

2 Answers2