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

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.

• 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 $n$. – palio Feb 15 '14 at 19:26
• 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. – Hagen von Eitzen Feb 15 '14 at 19:31
• 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. – David Peterson Feb 15 '14 at 19:32
• @DavidPeterson aka. $|G|$ is odd – Hagen von Eitzen Feb 15 '14 at 19:34
• 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. – Hagen von Eitzen Feb 15 '14 at 20:48

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:
So the answer is $|G'|$.
• Can I have $31.50 to see the rest of the answer. Thanks. – David Peterson Feb 16 '14 at 0:50 • @DavidPeterson No you can't. But you can visit books.google.com/… – user33321 Feb 16 '14 at 0:53 • To summarize (and assuming the links are not necessarily perennial): answering a question of L. Fuchs, J. Denes and P. Hermann proved (Annals of Discrete Math. 15 (1982) 105-109) that the set$S$is equal to a coset of$G'$for every finite group$G$. – YCor Feb 16 '14 at 15:25 The answer to your question is even more subtle. The set of all the possible products is always a coset of the commutator subgroup. Theorem Let$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\$.