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

Solution 1:

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.