If $S$ is a nonempty subset of group $G$, then $S^{|G|}$ is a subgroup of $G$.

Question

Let $G$ be a group with $|G| = n$ and let $ \emptyset \ne S \subseteq G$.

I want to show that $S^n$ is a subgroup of $G$ where by $S^n$ I mean the set $\lbrace s_1\cdots s_n \; | \; s_i \in S\rbrace$.

@RicPed - if it were, then $S^n = S$ and the problem would be trivial. — Prahlad Vaidyanathan, Aug 25 '13 at 10:45
Yeah, you are right, just thought about later that would be obvious :) — Riccardo, Aug 25 '13 at 10:55
@DerekHolt: If $G= \mathbb{Z}_3$ and $S= {2}$, then $S^{3}= {0}$ but $\langle S \rangle= \mathbb{Z}_3$. — Seirios, Aug 25 '13 at 11:19
We can focus only on the case $2|S| \leq |G|$, because if we have $|G| < |S|+|S|$ then it can be proved that $G=SS$ and inductively we can write every element $g \in G$ as a product of an arbitrary ($\geq 2$) number of elements of $S$. (for more detail see exercise 2.29 Rotman "An Introduction to the theory of Groups" 4 ed. — Riccardo, Aug 25 '13 at 13:47
Here is a bit of a reduction (still no idea how to get any further though): We can assume that $S$ generates $G$ and then the statement is that in a group of order $n$, generated by a set $S$, any word of length $2n$ in $S$ is equal to a word in $S$ of length $n$ (where equals means when evaluated in the group). — Tobias Kildetoft, Aug 26 '13 at 07:53
@Tobias Kildetoft: I would be interested to know how you reduce to $\langle S \rangle = G$. Even when $|S|>1$, the subgroup in question is not always equal to $\langle S \rangle$. For example for $S = { g, g^4 } \subset C_6 = \langle g \rangle$, we get $S^6 = {1,g^3}$. — Derek Holt, Aug 26 '13 at 08:24
@DerekHolt So assume we can show what I wrote above. Now $\left<S\right>$ is a subgroup of $G$ of order $d$ which divides $n$, and $S^n\subseteq \left<S\right>$. So the above would show that any word of length $2d$ in $S$ is equal to a word of length $d$. Now, we want to show that any word of length $2n$ in $S$ is equal to a word of length $n$, but since $n = kd$ for some $k$, this follows directly. — Tobias Kildetoft, Aug 26 '13 at 08:28
OK, thanks! I have a suspicion that the intention was to consider products of length at most $n$ rather than exactly $n$. — Derek Holt, Aug 26 '13 at 08:35
@DerekHolt I was thinking about that too, but since I have so far been unable to find a counterexample to the statement as given, I am not sure. — Tobias Kildetoft, Aug 26 '13 at 08:52
@TobiasKildetoft I heard about it from a friend of mine. Isn't it better if I move this question to Mathoverfolow !? — Robert M, Aug 26 '13 at 09:42
History: OP deleted this question, asked it on MO; then undeleted it here; meanwhile, it got migrated here from MO, so there are now two copies here, this one and http://math.stackexchange.com/questions/476531/special-subgroup-of-a-group-of-order-n — Gerry Myerson, Aug 26 '13 at 13:21

score 23 · Accepted Answer · answered Aug 26 '13 at 15:12

23

Consider the powers of the subset $S$:

$$S, S^2, S^3, S^4, \ldots$$

Because $G$ is finite, there is eventually some repetition. Let $S^r = S^{r+s}$ where $s > 0$ and $r+s$ is as small as possible. Then

$$S, S^2, S^3, \ldots, S^{r-1}, S^r, S^{r+1}, \ldots, S^{r+s-1}$$

are distinct subsets of $G$. Furthermore, $\{S^r, S^{r+1}, \ldots, S^{r+s-1}\}$ is a (cyclic) subgroup of order $s$ in the semigroup of nonempty subsets of $G$. Thus it is equal to $H/K$ for some $K \trianglelefteq H \leq G$ (this is not too difficult to prove, see 3.57 in "A Course in Group Theory" by Rose). Hence $s$ divides the order of $G$.

Next note that for each $t \geq r$, we have that $S^t = S^{t+s}$. Choose $r \leq t < r+s$ so that $s$ divides $t$. Then $S^t = S^{t+t} = (S^t)^2$, so $S^t$ is a subgroup (in fact, it is the only power of $S$ that is a subgroup). Now if would suffice to prove that $|G| \geq r$. If this is the case, then $|G| - t = ks$ for some $k \geq 0$ and so $S^{|G|} = S^{t+ks} = S^t$.

Let $x \in S$. Now $xS^i \subseteq S^{i+1}$, so $|S^i| = |xS^i| \leq |S^{i+1}|$. Thus if $|S^i| = |S^{i+1}|$, then $xS^i = S^{i+1}$. Hence

$$S^{i+2} = S^{i+1}S = xS^iS = xS^{i+1} = x^2S^{i}$$

and similarly $S^{i+k} = x^kS^i$ for all $k \geq 1$. In particular when $k = |G|$, we see that $S^{i+|G|} = S^i$. Thus $i \geq r$, since $S^r$ is the first power of $S$ that repeats. So when $i < r$, we have $|S^i| < |S^{i+1}|$ which gives

$$|S| < |S^2| < |S^3| < \ldots < |S^r|$$

and so we can find at least $r$ distinct elements in $|S^r|$, which proves that $|G| \geq r$.

Related to this answer: cyclic semigroups.

answered Aug 26 '13 at 15:12

Mikko Korhonen

25,224

In the beginning of 2nd paragraph, is it obvious that we can choose $t$ between $r$ and $r+s$ such that $s$ divides $t$? – Prism Aug 26 '13 at 17:25
@Prism: There are $s$ integers $t$ such that $r \leq t < r+s$. Any set of $s$ consecutive integers is a complete set of representatives modulo $s$. This is fairly intuitive and not very difficult to prove. – Mikko Korhonen Aug 26 '13 at 17:43
Ah of course! Thanks for clarifying. I have got one more quick question. You write "If this is the case, then $|G|-t=ks$ for some $k\ge 0$." I understand $|G|-t$ is multiple of $s$ (because both $|G|$ and $t$ are multiple of $s$) but how do we know $|G|\ge t$? Is it because $t$ is the smallest multiple of $s$ that is greater than $r$? – Prism Aug 26 '13 at 17:49
That's basically the reason, yes. Here $|G|$ is a multiple of $s$ and $\geq r$, and $t$ is the unique multiple of $s$ such that $r \leq t < r+s$. If you have a set of $s$ consecutive integers, then distinct elements of that set are not congruent modulo $s$. – Mikko Korhonen Aug 26 '13 at 17:52
Makes sense :) I like this answer a lot. Very constructive! I am going to give bounty to this once 48 hours has passed. – Prism Aug 26 '13 at 18:02
Thanks! What we talk about here is also useful for proving that ${S^r, S^{r+1}, \ldots, S^{r+s-1} }$ is a group, basic fact about cyclic semigroups. – Mikko Korhonen Aug 26 '13 at 19:12

If $S$ is a nonempty subset of group $G$, then $S^{|G|}$ is a subgroup of $G$.

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