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

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!

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Your statements are true. Try to improve the writing. – Sigur Feb 6 '13 at 23:59
This is essentially Wilson's theorem for finite abelian groups: en.wikipedia.org/wiki/…. – lhf Feb 7 '13 at 0:18
Thanks everyone. Makes complete sense now. Much appreciated! – Luke8ball Feb 7 '13 at 0:41

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.

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

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

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