# Finite Abelian Group

Let $G$ be a finite abelian group, $G = \{e, a_{1}, a_{2}, ..., a_{n} \}$. Prove that $(a_{1}a_{2}\cdot \cdot \cdot a_{n})^{2}$ = $e$.

I've been stuck on this problem for quite some time. Could someone give me a hint?

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every element has an inverse, and you can reorder terms –  RHP Jul 22 '11 at 4:50
Can I assume $(a_{1}a_{2} \cdot \cdot \cdot a_{n})^{2} = e$ is true? If I can, then $(a_{1}a_{2} \cdot \cdot \cdot a_{n})(a_{1}a_{2} \cdot \cdot \cdot a_{n}) = e$. If $ab = e$, then $b = a^{-1}$. So $(a_{1}a_{2} \cdot \cdot \cdot a_{n}) = (a_{1}a_{2} \cdot \cdot \cdot a_{n})^{-1}$. Then $(a_{1}a_{2} \cdot \cdot \cdot a_{n})(a_{1}a_{2} \cdot \cdot \cdot a_{n})^{-1} = e$. Thus $e = e$??? –  Student Jul 22 '11 at 4:59
@Jon: You cannot assume that what you are trying to prove is true; that leads to a circular argument. –  Arturo Magidin Jul 22 '11 at 5:00
@ Arturo: Thanks for the clear up :) –  Student Jul 22 '11 at 5:12

Here is a hint: for any given $a_i\in G$, there are two possibilities: either

• $a_i$ is its own inverse, or
• $a_i$ is not its own inverse, but rather $a_j=a_i^{-1}$ for some $j\neq i$.
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So for the first case where $a_{i}$ = $a_{i}^{-1}$. $(a_{1}a_{2} \cdot \cdot \cdot a_{n})^{2} = (a_{1}a_{2} \cdot \cdot \cdot a_{n})(a_{1}a_{2} \cdot \cdot \cdot a_{n}) = (a_{1}a_{2} \cdot \cdot \cdot a_{n})(a_{1}^{-1}a_{2}^{-1} \cdot \cdot \cdot a_{n}^{-1}) = a_{1}a_{1}^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = e$ –  Student Jul 22 '11 at 5:09
No, that reasoning isn't correct; for example, it might be the case that $a_3=a_3^{-1}$, but $a_4=a_5^{-1}$ and $a_5=a_4^{-1}$. In other words, the two cases I described above concern any single given $a_i\in G$, but there is no reason that every $a_i\in G$ should fall into the same case. –  Zev Chonoles Jul 22 '11 at 5:14
To explain it in yet another manner, it is false that either $$\bullet\quad\text{For every }a_i\in G,\,\,a_i\text{ is its own inverse.}$$ or $$\bullet\quad\text{For every }a_i\in G,\,\,a_i\text{ is not its own inverse.}$$ There may be some mixture. Combine the two arguments you have thought of for each case, and you will have an argument that will work in general. –  Zev Chonoles Jul 22 '11 at 5:22
$(a_{1}a_{2} \cdot \cdot \cdot a_{n})(a_{1}a_{2} \cdot \cdot \cdot a_{n})$. Then each $a_{i}$ in the second set of parentheses can be replaced appropriately by $a_{i}^{-1}$ if $a_{i}$ is its own inverse or by $a_{i}^{-1} = a_{j}$ for some $j \neq i$. Then since $G$ is Abelian, $a_{1}a_{1}^{-1} \cdot \cdot \cdot a_{n}a_{n}^{-1} = e$. Is this better? –  Student Jul 22 '11 at 5:28
This is precisely the right idea! However, there is one small detail you should also prove to make a fully rigorous argument: you should show that an element and its inverse are paired uniquely. Supposing it were possible that $a_k=a_i^{-1}$ and $a_k=a_j^{-1}$ for $i\neq j$, then after rewriting the terms in the second set of parentheses, we'd have two copies of $a_k$, and be missing some other element of $G$. However, this cannot happen. Do you see why? –  Zev Chonoles Jul 22 '11 at 5:31

The map $\phi:x\in G\mapsto x^{-1}\in G$ is an automorphism of $G$ so, in particular, it induces a bijection $G\setminus\{e\}\to G\setminus\{e\}$. It maps $b=a_1\cdots a_n$ to itself, so that $b=b^{-1}$ and, therefore, $b^2=e$.

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Automorphism? What's an automorphism? Well, you know, and I know, but do you reckon OP knows? –  Gerry Myerson Jul 22 '11 at 5:57
@Gerry: Not yet, but hopefully I'll read about automorphisms soon enough :) –  Student Jul 22 '11 at 6:04
@Gerry: he'll know soon enough :) –  Mariano Suárez-Alvarez Jul 22 '11 at 6:05
+1: this seems to be a genuinely different argument. –  Pete L. Clark Jul 22 '11 at 6:59
This is very clever and succinct, +1! –  Zev Chonoles Jul 22 '11 at 7:22

Note that a question which is in some sense more natural is: "What is the product of all the elements of a finite abelian group?" The given question gives some information about this product, namely that it is an element of order at most $2$.

The answer to the more general question is given by the following result.

Wilson's Theorem in a Finite Abelian Group: Let $G$ be a finite abelian group. Then the product of all elements of $G$ is equal to the identity unless $G$ has exactly one element $t$ of order $2$, in which case the product is $t$. For a proof see e.g. $\S 6$ of these notes.

In fact one can view the answer to the OP's question as 2/3 of the way towards a proof of WTFAG. Namely, we take the product of all of the elements in the group, and note that the elements which have order greater than $2$ occur in pairs $x,x^{-1}$. Therefore the product of all the elements in a finite abelian group $G$ is also equal to the product of all the elements of $G[2]$, i.e., the product of all elements of order at most $2$. In an abelian group, the subset $G[2]$ is a subgroup, so the product has order $2$ and hence squares to the identity element $e$, completing the answer to the OP's question.

But now, on to WTFAG: First of all, if there are no elements of order $2$ then $G[2] = \{e\}$, verifying WTFAG in that case. Second of all, if there is exactly one element $t$ of order $2$, then $G[2] = \{e,t\}$ and the product of all elements in $G[2]$ is equal to $t$.

To complete the proof of WTFAG, one needs to consider the case in which $G[2]$ has more than two elements. In my notes, I do this by first establishing the CLAIM that $G[2]$ (or really any finite group in which every element is self-inverse) is isomorphic to a finite direct product of copies of groups of order $2$, and then using some simple counting arguments.

The CLAIM follows either from the structure theory for finite abelian groups -- which is proved in the same set of notes, but nevertheless the emphasis in the notes is on the fact that in many applications, especially in number theory, the big structure theorem can be avoided -- or from the fact that a group in which each element has order $2$ is necessarily a vector space over the field $\mathbb{F}_2$ of two elements (and then we invoke the structure theorem for finite-dimensional vector spaces). I would actually prefer to have a third, more elementary proof of this fact. Perhaps someone can suggest one?

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HINT:

An element and its inverse are unique, and each element is either its own inverse or not.

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In addition to Pete Clark's answer, there is also a very neat answer to the question What is the set of all different products of all the elements of a finite group $G$? So $G$ not necessarily abelian. Well, if a 2-Sylow subgroup of $G$ is trivial or non-cyclic, then this set equals the commutator subgroup $G'$. If a 2-Sylow subgroup of $G$ is cyclic, then this set is the coset $xG'$ of the commutator subgroup, with $x$ the unique involution of a 2-Sylow subgroup.

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$@$Nicky: +1! This is a very nice generalization of the abelian case (which, in fact, I had been wondering about). Do you have any references to where these facts are discussed? –  Pete L. Clark Jul 22 '11 at 7:59
@Pete: have to look it up, author was the Hungarian mathematician J. Dénes. See also J. Dénes and P. Hermann, `On the product of all elements in a finite group', Ann. Discrete Math. 15 (1982) 105-109. The theorem also connects to the theory of Latin Squares and so-called complete maps. –  Nicky Hekster Jul 22 '11 at 9:40
Thanks for the reference! –  Mariano Suárez-Alvarez Jul 22 '11 at 23:45