# The product of all the elements of a finite abelian group

I'm trying to prove the following statements.

Let $G$ be a finite abelian group $G = \{a_{1}, a_{2}, ..., a_{n}\}$.

• If there is no element $x \neq e$ in $G$ such that $x = x^{-1}$, then $a_{1}a_{2} \cdot \cdot \cdot a_{n} = e$.

Since the only element in $G$ that is an inverse of itself is the identity element $e$, for every other element $k$, it must have an inverse $a_{k}^{-1} = a_{j}$ where $k \neq j$. Thus $a_{1}a_{1}^{-1}a_{2}a_{2}^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = e$.

• If there is exactly one $x \neq e$ in $G$ such that $x = x^{-1}$, then $a_{1}a_{2} \cdot \cdot \cdot a_{n} = x$.

This is stating that $x$ is not the identity element but is its own inverse. Then every other element $p$ must also have an inverse $a_{p}^{-1} = a_{w}$ where $p \neq w$. Similarly to the first question, a rearrangement can be done: $a_{1}a_{1}^{-1}a_{2}a_{2}^{-1} \cdot \cdot \cdot xx^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = xx^{-1} = e$. And this is where I am stuck since I proved another statement.

Any comments would be appreciated for both problems.

-
We're getting a lot of variations on Wilson's theorem for groups. Time to abstract? – Bill Dubuque Jul 22 '11 at 22:04
I vote not to close, since I think this question is different for two reasons: (1) It is a generalization of the elementary number theory problem (2) The OP is asking for help with his specific approach. For the next person, instead of voting to close, please cancel my vote not to close. – Eric Naslund Jul 22 '11 at 22:08
@Bill: I completely agree that this question should be made into an FAQ question to deal with all of these types at once. – Eric Naslund Jul 22 '11 at 22:09
@Bill: I appreciate the links you have posted and your comments. But the problem is that I do not understand most of the words you use as I have just started learning about abstract algebra. – Student Jul 22 '11 at 22:54

## 3 Answers

Your approach to the first problem is correct, however let me write it slightly differently. Since for each $i$ there exists a unique $j$ such that $a_j=a_i^{-1}$ we see every element and its inverse appears in the product $$a_1a_2\cdots a_n,$$ and hence this product must be the identity. However, it is misleading to write this as $$a_1a_{1}^{-1}\cdots a_n a_n^{-1}$$ since $a_1^{-1}$ will be $a_j$ for some $j$, so we are counting the element $a_j$ twice. This product has $2n$ elements instead of $n$, and this is where your error in the second problem stems from. For the second problem, since $x$ is its own inverse, that is since $x=x^{-1}$, we won't have both $x$ and $x^{-1}$ appearing in the product. Every other element will have its inverse appear exactly once, so we are able to conclude the product $$a_1a_2\cdots a_n$$ equals $x$.

-
This and more hs already been said in many other answers - see my link above. – Bill Dubuque Jul 22 '11 at 22:06
@Jon: Yes that is correct, $xx=e$. It means that $x$ has order $2$. No element appears more then once in a group, but for elements $y$ with order greater then $2$, both $y$ and $y^{-1}$ will appear in the group. – Eric Naslund Jul 22 '11 at 22:11
@Eric: Thank you for clearly explaining everything to me. I am only starting to learn how to write proofs on my own, so forgive me if my questions seem redundant. – Student Jul 22 '11 at 22:20

Your reasoning for the first problem is right but the writeup is incorrect; in fact, $a_1a_1^{-1}a_2a_2^{-1}\cdots a_na_n^{-1}=e$ is true for any abelian group and any set of elements $a_1, a_2,\ldots,a_n$ from it - at least, with what I would take as the 'standard' definition of that product (namely $\Pi_{i=1}^n a_ia_i^{-1}$). I think you want to be more explicit about the pairing of elements in the product. When you get more explicit about this pairing, it should give you your answer for the second problem as well.

-

For the first answer, you are almost there : if $a_1 a_1^{-1} \cdots a_n a_n^{-1} = e$, since the elements $a_1, \cdots , a_n$ are all distinct, their inverses are also distinct. Since the product written above involves every element of the group, we have $a_1 a_1^{-1} \cdots a_n a_n^{-1} = (a_1 a_2 \cdots a_n) (a_1^{-1} a_2^{-1} \cdots a_n^{-1}) = (a_1 \cdots a_n)^2 = e$, and since no element is its own inverse (by hypothesis) besides $e$, you have to conclude that $a_1 \cdots a_n = e$.

For the second one, when you re-arrange the terms, $x^{-1}$ should not appear in there, since $x = x^{-1}$ and $x$ does not appear twice in the product, so all that's left is $x$.

-