1
$\begingroup$

I had this problem that said given a finite ring with unity R with 1+1 ≠ 0 such that for a permutation of its invertible elements, their product is equal to 1. It asked to show that 4 divides the order of its invertible elements subgroup.

I was able to show that 2 must divide the order of U(R), but now i’d be curious if given a group of order 2k, where k is odd, is the product of it’s elements always different from 1.

$\endgroup$
6
  • 1
    $\begingroup$ What is $2$ in your first sentence? Is the question "is it true that if group $G$ has order $2k$ where $k$ is odd, then product of all elements of $G$ in any order is different from $1$"? $\endgroup$
    – mihaild
    Commented Nov 13, 2023 at 14:28
  • 1
    $\begingroup$ Yes, it's true, and it's a consequence of something called Burnside's transfer theorem. Concretely, consider the regular representation of $G$ on itself, where $g \in G$ acts as $h \mapsto hg$. You can check that an element of order $2$ acts as an odd permutation, and therefore we have a nontrivial homomorphism $G \to C_2$, and the product of all the elements of $G$ (in any order) is again an odd permutation, hence nontrivial. $\endgroup$
    – Sean Eberhard
    Commented Nov 13, 2023 at 14:33
  • 1
    $\begingroup$ Also, what does $1+1\ne 2$ mean? As far as I know $2$ is by definition $1+1$ in any unital ring. $\endgroup$
    – Sean Eberhard
    Commented Nov 13, 2023 at 14:35
  • $\begingroup$ It was meant to be 0 instead of 2, that was my mistake, oops $\endgroup$
    – FireWolf15G8
    Commented Nov 13, 2023 at 14:38
  • 1
    $\begingroup$ Then isn't $\mathbf Z/3\mathbf Z$ a counterexample? $\endgroup$
    – Sean Eberhard
    Commented Nov 13, 2023 at 16:57

1 Answer 1

Reset to default
1
$\begingroup$

In addition to the comments above, there is also a very neat answer to the general question $\color{blue} {\text{What is the set of all different products of all the elements of a finite group $G$? }}$ So $G$ not necessarily abelian and in a product each element appears exactly one time.

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 (non-trivial) 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.

$\endgroup$

You must log in to answer this question.

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