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Let $ G=${$a_1,a_2,..a_n$}, show $(a_1.....a_n)^2=e$ where e is the identity element, G is an abelian group of order n.

I dont want the solution , i just want a tip to get me started

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    Tip: For each $a_i$ there is a $a_i^{-i}$ in there as well. You are listing multiplying every element twice. So ... – 2017-02-11

4 Answers 4

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Hint: Consider these three sets:

$H = \{g| g \in G\}$ and $J = \{g^{-1}| g \in G\}$ and $G = \{a_1, .... ,a_n\}$.

Do the sets have any elements in common? Are there any elements in one set that are not in the other?

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Hint:

Rearrange as $(a_1\ldots a_n)(a_1\ldots a_n)$. Now, for each element $a_i$ in the first grouping, there should be a unique element in the second grouping which cancels it. Rearrange to put them next to eachother.

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    o so i can say for example because of the order n, every element has an inverse inside the group, it could be either itself or another element, so i can arrange them together and cancel? – 2017-02-10
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    Not exactly. The problematic ones are precisely the elements (other than the identity) that _don't_ pair up (i.e., the elements other than the identity that are their own inverse). – 2017-02-10
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    My previous comment was for the product of all the elements, not the square of the product. The square of the product automatically takes care of the self-inverse elements as well. – 2017-02-11
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    Note that while the square of the product is the identity, the actual product (not squared) need not be the identity. For example, for a cyclic group of order $2$, the product of the elements is not the identity. – 2017-02-11
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Since $G$ is abelian, you may arrange the factors in any order without changing the product $(a_1\cdots a_n)^2$.

Now split $G$ into two parts :

  • the elements that are equal to their own inverse

  • the other elements

EDIT :

As mentioned in one of the comments below, there is no need to separate the elements of $G$ into two classes !

It is sufficient to observe that the map $i:G\to G,x\mapsto x^{-1}$ is a bijection (since $i\circ i=id_G$).

Hence, each factor in the product $a_1\cdots a_n$ is the inverse of exactly one factor in the same product so that, since multiplication is commutative, $(a_1\cdots a_n)^2=e$.

REMARK - 1

We may reach $e$ without squaring, in some finite abelian groups.

For example in $\mathbb{Z}_2^2$ :

$$(0,0)+(1,0)+(0,1)+(1,1)=(0,0)$$

But with some other groups, squaring becomes necessary ...

For example in $\mathbb{Z}_4$ :

$$0+1+2+3=2\neq0$$

REMARK - 2

Without the assumption of commutativity, the product of all elements is not well defined since order counts !

But given some suitable order $a_1,\cdots,a_n$ in some suitable finite non abelian group of order $n$, there are examples where $(a_1\cdots a_n)^2\neq e$.

I tried first with $\mathfrak{S}_3$, but it doesn't work because whatever order we consider for the six permutations of $\{1,2,3\}$, the signature of the product will be $-1$, hence that product will be some transposition : $(12)$, $(13)$ or $(23)$, whose square is the identity !

Let's try with another non-abelian finite group ... say $\mathfrak{S}_4$

If we denote by $[1324]$ the permutation $s$ defined by :

$$s(1)=1\quad s(2)=3\quad s(3)=2\quad s(4)=4$$

and similar notations for the $23$ other permutations, we get (after a tedious computation) the following counterexample :

$$p=[1234] \circ [1324] \circ [1243] \circ [1342] \circ [1423] \circ [1432]$$ $$\circ [2134] \circ [2143] \circ [2314] \circ [2341] \circ [2413] \circ [2431]$$ $$\circ [3124] \circ [3142] \circ [3214] \circ [3241] \circ [3412] \circ [3421] $$ $$\circ [4123] \circ [4132] \circ [4213] \circ [4231] \circ [4312] \circ [4321] $$

(note that the 24 permutations are enumerated in lexicographic order except that the second and third have been exchanged)

It can be checked (fingers crossed) that :

$$p\circ p=[3124]\neq[1234]$$

REMARK-3

Suppose $G$ abelian and let $A$ be the subset of all elements of $x\in G$ such that $x^{-1}\neq x$.

The product of all elements in $A$ is $e$, because the bijection $G\to G,x\mapsto x^{-1}$ induces a bijection from $A$ to itself.

Now, if we remove the assumption of commutativity, this product (which is defined only after having chosen an explicit order) has no reason to be equal to $e$.

Here is an example, again in $\mathfrak{S}_4$ (same notations as in remark 2) :

$$[1342]\circ[1423]\circ[2314]\circ[2341]\circ[2413]\circ[2431]\circ[3124]$$ $$\circ[3142]\circ[3241]\circ[3421]\circ[4123]\circ[4132]\circ[4213]\circ[4312]$$ $$=[4321]\neq[1234]$$

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    is $(a_1.......a_n)=e$ if G has no self inverse element ? – 2017-02-10
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    Yes, except that strictly speaking you can't have a group $G$ without an element that is its own inverse. Take $e$. $e^2=e$, so $e$ is its own inverse. If you have a group where no non-identity element is its own inverse, then yes, $a_1 \cdots a_n = e$. – 2017-02-11
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    yes, thanks you ! – 2017-02-11
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    Don't need to separate the self inverting from the non. Every element in {a1, a2..... an} has an inverse in {a1,a2,..... an} whether that inverse is itself or not. – 2017-02-11
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    You need the separation if you're considering the product of the elements rather than the square of the product. – 2017-02-11
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    But we aren't considering the product. We are considering the square of the product. – 2017-02-11
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    A good question similar to this would be proving the product of the non-self inverting elements is the identity. Unfortunately can not reach the same conclusion for the product of all elements. (product, not square of product). – 2017-02-11
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    @fleablood: Yes, the squared version has already been well answered (more than once in this thread), but asddf, in one of his comments, asked about the product. – 2017-02-11
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    @fleablood: Absolutely right ! No need to separate into two classes of elements. I will edit my post right now ! – 2017-02-11
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    For the non-squared version, unless I'm missing something, the following claim holds: The product of the elements of a finite abelian group is the identity if and only if the set of non-identity elements which are self-inverse is either empty or has cardinality greater than $1$. The proof, while elementary, is a little harder than the one needed to show that the square of the product is always the identity. – 2017-02-11
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    I just finished to edit my answer and tried to answer to some of the questions above – 2017-02-11
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    Thanks all of you I got it :)) – 2017-02-11
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hint: for each $a_i\in G$ there is an $a^{-1}_i, \ $ which must be one of the $a_j;\ 1\le j\le n.$

Now, write $(a_1.....a_n)^2=(a_1.....a_n)(a_1.....a_n),\ $ open parentheses, and use the fact that $G$ is abelian.