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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.

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    possible duplicate of [Prove that the product of coprimes must be congruent to $-1$ or $1 \pmod{m}$](http://math.stackexchange.com/questions/23937/prove-that-the-product-of-coprimes-must-be-congruent-to-1-or-1-pmodm)2011-07-22
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    We're getting a lot of variations on Wilson's theorem for groups. Time to abstract?2011-07-22
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    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.2011-07-22
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    @Bill: I completely agree that this question should be made into an FAQ question to deal with all of these types at once.2011-07-22
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    @Eric I disagree. With that logic we'll soon have hundreds of minor variations on such. Better to refer the OP to the other answers, and if they don't suffice then the OP can ask more specific questions. Nothing asked above is new.2011-07-22
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    Jon: Recall that I referred you to other answers in my answer to your prior question. Before posting a new (duplicate) question as above, you should refer to these prior questions. In fact they answer your questions and then some.2011-07-22
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    @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.2011-07-22
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    @Jon I'm having difficulty finding even a single word you might not understand - let alone most words - since the linked answers involve only very basic group theory. Could you please elaborate why you couldn't deduce your answer from these prior answers.2011-07-22
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    Jon: Abstract algebra is about concrete objects, just many of them at once. You need experience working out what happens in a number of concrete cases. I suggest as a starting example the "numbers" $0$, $1$, $2$, $3$, $4$, $5$ with the "sum" of $a$ and $b$ being the remainder when the ordinary $a+b$ is divided by $6$.2011-07-23
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    @Bill: Statements like "The map $\rm n \mapsto n^{-1}$ is a permutation of order 2 so it decomposes into cycles of length 1 or 2. So non-fixed-points are paired with their inverse. This leads to a proof of Wilson's theorem for groups" honestly confuse me. And yes, I did look at Wilson's theorem on wikipedia, but I just don't see how that relates to my question at the moment. Wilson's theorem may be very basic for you, but it is too deep for me to understand at this point in time.2011-07-23
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    @Jon But it is explained in more elementary terms in the linked post. Ditto in the link I gave above. If you follow the links (and their links) to all the prior questions you will find many elementary examples that should quickly reveal the essence of the matter. Almost everything mentioned in the answers to your recent questions has already been said in prior answers - often multiple times. If you want to really understand this topic I highly recommend that you read the answers to these prior related questions.2011-07-23

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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 $x$.

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    This and more hs already been said in many other answers - see my link above.2011-07-22
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    @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.2011-07-22
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    @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.2011-07-22
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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.

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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 $x$.