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If G is a finite abelian group of odd order prove that the product of all elements in G is equal to the identity element of G.Well , I can prove this with an example but is there any general proof for this?

Shashank Dwivedi
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    You just search "abelian product odd" and the top hit has 27 upvotes and 7 answers for exactly your question. Seriously, you have to stop writing posts that are just problem statements, and you have to stop writing posts when the search feature already shows you there are duplicates. (This is apparently the fifth question in a row you've posed like that.)You are positioning yourself to get truckloads of downvotes and near-instant closures if you continue misusing the site this way. – rschwieb Mar 08 '17 at 14:17

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Simply note that if $a \in G$, with $a \ne 1$, then $a^{-1} \ne a$.

In fact, if $a = a^{-1}$, then $a^{2} = 1$. If $a \ne 1$, then $a$ has period $2$, which cannot occur in a group of odd order.

Alternatively, let $P = \prod_{a \in G} a$. Since $a \mapsto a^{-1}$ is a bijection, we have $P = \prod_{a \in G} a^{-1}$. Thus $P^{2} = 1$, as all terms cancel in pairs. By the same argument as above, $P = 1$.

Andreas Caranti
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Assume that $G$ is not the trivial group, where this is trivially true. First, we note that since the order of $G$ is not even, and since the order of each element divides the order of the group, there is no element $a$ such that $a^2=e$.

Then, the product of all elements of $G$ may be rewritten as

$$a_1a_2a_3\ldots a_n = (a_1a_1^{-1})(a_2a_2^{-1})\ldots (a_na_n^{-1}) = e\cdot\ldots\cdot e=e$$

where we simply "pair up" all of the inverses of $G$. Note that this can be done strictly because $G$ is abelian.

You should think about whether this is true when $G$ is not a finite group. Can this argument be used in the infinite case? Why or why not? To give an example, consider the group $G=\mathbb{Z}$.

Santana Afton
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