Is the set of words generated by a non-empty subset $S$ of a finite group $G$ of order $n$ of length $n$ always be a (normal) subgroup? Let S be a non-empty subset of a group $\mathbf G$, and $\mathbf \vert G \vert$ = n. For each $k$, let $S^{k}$ be the set with elements of form $\{s_1 \dotsm s_k \mid s_i \in S\}$. Prove that $S^{n}$ is a subgroup G. Or going more further, a normal subgroup of G?
Let $H$ be the group generated by $S$. I have two observations:

*

*For any $h ∈ H$, $h \in S^{m}$ for some $m$ with $1\le m\le n$. That is, each element of $H$ is a word of length less than $n$.


*For $1\le i\le j\le n$, we have $\vert S^{i} \vert\le\vert S^{j} \vert$. That is, the cardinality of the chain $S^{i}$ is non-decreasing.
So from above, if $\vert S^{n}\vert = \vert H\vert$, we are done. But there still is a case when for some $i$, the case $\vert S^{i}\vert = \vert S^{i+1}\vert$. Anyone can help?
 A: There is an idea comes into my mind, I hope this would work:
Let us suppose the second case occur. So there exists an index i such that $\vert S^{i}\vert = \vert S^{i+1}\vert$, we consider two cases seperately:

*

*$1 ∈ S^{i}$, then as $\vert S^{i}\vert = \vert S^{i+1}\vert$, we see that $M = S^{i} = S^{i+1} = \ldots = S^{2i} = \ldots$ which implies the closure property of $S^{i}$, so $M = S^{n}$ is a subgroup of H.

2.$1 \notin S^{i}$, we shall derive a contradiction in this case. Again, by the property $\vert S^{i}\vert = \vert S^{i+1}\vert$, it is easy to check that for any $s_i, s_j ∈ S$, $s_iS = s_jS$ since every set formed by left action on $S^{i}$ by S coincide as $S^{i+1}$ is the union of this sets and $\vert S^{i}\vert = \vert S^{i+1}\vert$. So we can choose a representive element $s \in S$ with $S^{i+1} = sS^{i}$, and similarly, $S^{i+1} = S^{i}s$, so combine this two identity we get $S^{i+1} = sS^{i} = S^{i}s$. For the avoidance of doubt I shall also show for $S^{i+2}$. Now, $S^{i+2} =SS^{i+1} = SsS^{i} = SS^{i}s = S^{i+1}s = S^{i}s^{2}$. Now it is more clear to see that the induction can play a role in here to derive that $S^{i+k} = S^{i}s^{k}$. By the finite property of G, we see that there exists $m \in \mathbb N$ such that $1 \in S^{i+m}$, typically $1 \in S^{n}$. But $S^{i+k} = S^{i}s^{k}$ implies $|S^{i}| = |S^{j}| for j \geqslant i$ and $1 \notin S^{i}$ which is clearly impossible. So $1 \in S^{i}$ and we turn to case 1.
A: It is not necessarily the case that $\lvert S^i\rvert$ reaches $\lvert H\rvert$. For example, if $S=\{g\}$ for some $g\neq1$.
What you do know is that eventually you will have $\lvert S^i\rvert=\lvert S^{i+1}\rvert$. Once this happens, you know that $S^{i+1}=gS^i$ for any $g\in S$. Indeed, if you fix an element $g\in S$, then $gS^i\subseteq S^{i+1}$ but $\lvert gS^i\rvert=\lvert S^i\rvert=\lvert S^{i+1}\rvert$. Then $S^{i+2}=S^{i+1}S=gS^iS=gS^{i+1}$ so $\lvert S^{i+2}\rvert=\lvert S^{i+1}\rvert$. Thus, the cardinality $\lvert S^i\vert$ is strictly increasing until it stabilizes.
Moreover, the cardinality $\lvert S^i\rvert$ must stabilize by the time $i=n$. Then $\lvert S^{2n}\rvert=\lvert S^n\rvert$. However, $1\in S^n$, so $S^n\subseteq S^{2n}$. Then $S^n=S^{2n}$, so $S^n$ is closed under multiplication. Since $G$ is finite, this is enough to say that $S^n$ is a subgroup of $G$.
