2
$\begingroup$

If $G$ be a group can I define the element $$\prod_{g \in G} g.$$ I think this is defined when $G$ is a finite group but not defined when $G$ is infinite. So please help me for this.

$\endgroup$
4
  • 6
    $\begingroup$ The definition only works if $G$ is finite and abelian, If $G$ is not abelian, then the product depends on the order of the elements. $\endgroup$
    – Derek Holt
    Commented Aug 6, 2017 at 16:12
  • $\begingroup$ If $G$ is finite but not abelian then can we define this product $\endgroup$ Commented Aug 6, 2017 at 16:16
  • $\begingroup$ You can define the product if you specify an order in which to multiply them. $\endgroup$
    – anon
    Commented Aug 6, 2017 at 16:21
  • $\begingroup$ See also en.wikipedia.org/wiki/… $\endgroup$
    – user261263
    Commented Aug 6, 2017 at 17:02

2 Answers 2

2
$\begingroup$

In addition to the answer of @Arnaud D. and remarks, 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.

$\endgroup$
2
  • $\begingroup$ Nice, I was wondering wether the set of such products was always a coset of $G'$. Thanks for the reference ! $\endgroup$
    – Arnaud D.
    Commented Aug 7, 2017 at 10:46
  • $\begingroup$ Thanks and yes you see with the example below of $S_3$ that this set equals the coset $(12)A_3$. $\endgroup$ Commented Aug 7, 2017 at 11:15
1
$\begingroup$

You are correct that taking the product of all elements in the group is not well-defined if the group is infinite, but actually even for a finite group it is not always possible to define it. The problem is that if your group is not abelian, then you would need to specify in what order you are taking the elements : so if you write your group as $\{g_1,\cdots ,g_n\}$ then you can define $\prod_{k=1}^n g_k$, but this would not be the same as $\prod_{k=1}^n g_{\sigma(k)}$ for all permutations $\sigma\in S_k$.

Example : let us take $G=S_3$, then we have for example $$(1,2)(1,3)(2,3)(1,2,3)(1,3,2)=(1,3),$$ $$(1,3,2)(1,2,3)(2,3)(1,2)(1,3)=(1,2),$$ $$(1,2,3)(2,3)(1,2)(1,3)(1,3,2)=(2,3)$$

On the other hand, if $G$ is abelian, then there is no problem, as every ordering would give the same result.

$\endgroup$
1
  • $\begingroup$ +1 for the specific example (though the commas aren't generally part of the notation for a permutation group element) $\endgroup$ Commented Aug 6, 2017 at 20:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .