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Question: If $G$ is a finite group such that the product of its elements (each chosen only once) is always $1$, independent of the ordering in the product, what can we say about $G$?

I was trying to conclude that $G$ must be abelian (I am not sure about this). I proceed as follows: if $|G|\leq 4$, then obviously, $G$ is abelian. Suppose $|G|>4$. Then consider products of all elements in the order $a^{-1}b^{-1}ab cc^{-1}dd^{-1}....$. This, being identity, implies that $a^{-1}b^{-1}ab=1$, i.e. $G$ is abelian.

BUT, the problem is in $a^{-1}b^{-1}ab$, the elements $a$ and $a^{-1}$ should be distinct, and similarly for $b$. Otherwise, the product $a^{-1}b^{-1}ab cc^{-1}dd^{-1}....$ is not in the prescribed form as in question.

Also, note that the cyclic group of order $4$ doesn't satisfy that $abcd=1$ where $\{a,b,c,d\}\cong \mathbb{Z}_4$.

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    $\begingroup$ Am I missing something? If a and b do not commute, so ab ≠ ba then abP ≠ baP for every element P in the group. So take P equal to the product of all the elements besides a and b, in any order. $\endgroup$ – lulu Jul 8 '15 at 13:40
  • $\begingroup$ How do you know that $a^{-1}\neq a$? $\endgroup$ – Thomas Andrews Jul 8 '15 at 17:39
  • $\begingroup$ It may or may not happen that $a\neq a^{-1}$. In $\mathbb{Z}_2=\{1,x\}$, $x$ is inverse of itself. $\endgroup$ – Groups Jul 9 '15 at 3:07
  • $\begingroup$ @Groups. This is a good remark. See lhf's answer below. $\endgroup$ – Tom-Tom Jul 9 '15 at 11:51
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(as per my earlier comment): let $a$ and $b$ be arbitrary elements of your group and let $P$ be the product of all the other elements, in any order. Then, by assumption, $abP = baP$ so $ab = ba$.

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The group must be Abelian.

Fix an ordering of the elements of $G$: $g_1, g_2, \dots, g_n$.

Then $g_1 g_2 g_3 \cdots g_n = 1 = g_2 g_1 g_3 \cdots g_n$

Now cancel $g=g_3 \cdots g_n$ to get $g_1 g_2 = g_2 g_1$.

You don't need that the product is always $1$, just that it does not depend on the order. Moreover, this is the right hypothesis because of Wilson's theorem for finite Abelian groups:

The product of all elements in a finite Abelian group is either $1$ or the element of order $2$ if there is only one such element.

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  • $\begingroup$ Very nice ! The first part is the same as in @lulu 's answer, but Wilson's theorem allows to say more than just abelian. +1 $\endgroup$ – Tom-Tom Jul 9 '15 at 11:50
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In addition to all the answers, there is also a very neat answer to the general 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.

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 connects to the theory of Latin Squares and so-called complete maps.

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    $\begingroup$ This is a good link to question. Thanks for the nice reference. $\endgroup$ – Groups Jul 10 '15 at 4:02

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