14

Let $G$ be a finite group and $S$ a nonempty subset of $G$. I want to prove (or disprove) that $S^{|G|}$ (that is products of length $|G|$ of elements of $S$) is a subgroup.

My work so far :

Since we are in a finite group, it suffices to show that $S^{|G|}$ is stable under the group product.

When $e\in S$, we have $S^{|G|} = \bigcup_{k=0}^{|G|}S^{k}$ and this is the subgroup generated by $S$.

However I'm having issues solving this when $e\notin S$.

I've tried to consider the monoid $C\subset \Bbb N$ of length of products of elements equal to the neutral element. If $S^{|G|}$ is not a group, then there exists $k\notin C$ smaller than $|G|$ such that $k+|G|\in C$. However I don't know where to go from here.

Any help is appreciated !

Edit : Note that I want products of exactly $|G|$ elements of $S$, not products of at most $|G|$ elements of $S$

Akiva Weinberger
  • 23,062
Astyx
  • 3,883
  • Sorry, I misunderstood the question. – verret Feb 25 '19 at 20:26
  • No problem, thank you for your time ! – Astyx Feb 25 '19 at 20:27
  • Hint: What happens if $S^k = S^{k+1}$? – j.p. Feb 25 '19 at 21:26
  • @j.p. Note that $S^k$ need not be a subset of $S^{k+1}$. In fact, there are many cases where $S^{|G|}={e}$, even if $S\ne{e}$. (This happens, for example, when $S$ consists of a single element not equal to $e$.) – Akiva Weinberger Feb 25 '19 at 21:53
  • A related question (and perhaps necessary to answer this one) is: what is the actual subgroup $H:=S^{|G|}$? Clearly $H\leq \langle S\rangle$. We may have $H<\langle S\rangle$ though. For example, if $N$ is a normal subgroup of $\langle S\rangle$ such that $\langle S\rangle/N$ is cyclic, and $S$ is contained in a single coset of $N$, then $H\leq N$. I couldn't find any other obstructions. (I think one can show that there is a smallest normal subgroup $M$ such that $\langle S\rangle/M$ is cyclic and $S$ is contained in a single coset of $M$, and I would tempted to conjecture that $H=M$.) – verret Feb 25 '19 at 22:40
  • 1
    I remember that this question has been asked here several times by different people, see for example here for a solution. – Mikko Korhonen Feb 25 '19 at 23:16
  • Also, it never became clear where the problem originates from and I would be curious about this. Is there a reference or is it homework from some group theory course? – Mikko Korhonen Feb 25 '19 at 23:22
  • Thank you for your answer, I've been pondering on this problem for hours (I did not manage to find the question you linked) ! The question originates from a friend who found the statement on reddit. – Astyx Feb 25 '19 at 23:25
  • @AkivaWeinberger: $S^k$ is not the set of $k$-th powers, so your example is wrong. I on the other hand wrongly assumed $1\in S$. But that can be fixed by taking any $s\in S$ and look at the map $S^k\to S^{k+1}$ defined by $x\mapsto x\cdot s$, which has to be injective. – j.p. Feb 26 '19 at 06:56
  • I nite that the duplicate post had an uninformative title, which makes it difficult to find. The current post, although a duplicate, has a much better title! – Derek Holt Feb 26 '19 at 08:01
  • @DerekHolt: Agreed, I edited the title of the older post to make it more useful. – Mikko Korhonen Feb 26 '19 at 09:03

0 Answers0