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Let $G$ be a finite group.

How many different elements can we obtain by multiplying all element in a group?

Of course, if $G$ is abelian the answer is one but when G is non-abelian, changing the order of the multiplication may produce new elements.

My second question is actually related to my attempt to solve the first one.

Let $S$ be set of all elements produced by multiplying all elements in $G$. Then, it is easy to show that $Aut(G)$ acts on $S$ naturally. I wonder whether this can be transitive.

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  • $\begingroup$ If $|G|=n$ then there is $n!$ ways to multiply the elements of $G$ now among these $n!$ elements there must be equal elements as we can't exceed the number of elements of $G$ whic is exactly $n$. $\endgroup$
    – palio
    Feb 15, 2014 at 19:26
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    $\begingroup$ With $G=S_3$ the answer is $S_3\setminus A_3$, i.e. the set of 2-cycles, $S=\{(1\,2),(1\,3),(2\,3)\}$. One inclusion is clear as the product is always an odd permutation. The other follows by symmetry. $\endgroup$ Feb 15, 2014 at 19:31
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    $\begingroup$ Small subcase: If $x^2\neq e$ for all $x\neq e$ every element has a distinct inverse, so pick any $g\in G$ such that we can find $h,k$ with $g=hkh^{-1}k^{-1}$. Multiply this by the rest of the elements in the order $aa^{-1}$ and we can get all $g$ of this form. $\endgroup$
    – David P
    Feb 15, 2014 at 19:32
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    $\begingroup$ @DavidPeterson aka. $|G|$ is odd $\endgroup$ Feb 15, 2014 at 19:34
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    $\begingroup$ With a bit of work one can see that $G=S_4$ leads to $S=A_4$. In this case the operation is not transitive. $\endgroup$ Feb 15, 2014 at 20:48

2 Answers 2

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All products are equal modulo the commutator subgroup, so $S$ is contained in a coset of $G'$. It turns out that $S$ is equal to this coset:

http://www.sciencedirect.com/science/article/pii/S0304020808732572

So the answer is $|G'|$.

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    $\begingroup$ Can I have $31.50 to see the rest of the answer. Thanks. $\endgroup$
    – David P
    Feb 16, 2014 at 0:50
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    $\begingroup$ @DavidPeterson No you can't. But you can visit books.google.com/… $\endgroup$
    – user33321
    Feb 16, 2014 at 0:53
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    $\begingroup$ To summarize (and assuming the links are not necessarily perennial): answering a question of L. Fuchs, J. Denes and P. Hermann proved (Annals of Discrete Math. 15 (1982) 105-109) that the set $S$ is equal to a coset of $G'$ for every finite group $G$. $\endgroup$
    – YCor
    Feb 16, 2014 at 15:25
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The answer to your question is even more subtle. The set of all the possible products is always a coset of the commutator subgroup.

Theorem Let $G$ be a finite group of order $n$, say $G=\{g_1, \dots, g_n\}$ and let $P(G)=\{g_{\sigma(1)}\cdot g_{\sigma(2)} \dots g_{\sigma({n-1})} \cdot g_{\sigma(n)}: \sigma \in S_n\}$.

(a) If $|G|$ is odd, then $P(G)=G'$

(b) If $|G|$ is even, let $S \in Syl_2(G)$. Then $P(G)=G'$ in case $S$ is non-cyclic. If $S$ is cyclic, then $P(G)=xG'$, where $x$ is the unique element of order $2$ of $S$.

This higly non-trivial and beautiful result relates to combinatorics - the construction of Latin Squares. It heavily relies on the proof of the so-called Hall-Paige conjecture (Hall, Marshall; Paige, L. J. Complete mappings of finite groups. Pacific Journal of Mathematics 5 (1955), no. 4, 541--549.). This could be proved thanks to the classification of the finite simple groups.

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