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3 added 112 characters in body

Consider the sets $X, X^2, \dots$. We claim that $|X^i|\leq |X^{i+1}|$, and, moreover, if $|X^i|=|X^{i+1}|$ then $|X^{i+1}|=|X^{i+2}|$. Actually, under the mapping $X^i\times X\to X^{i+1}$, $(b,x)\mapsto bx$, the preimage of any element of $X^{i+1}$ has the cardinality at most $|X|$ since all $x$-coordinates in this preimage should be distinct. Thus, $|X^i|\cdot |X|\leq |X^{i+1}|\cdot |X|$; hence $|X^i|\leq |X^{i+1}|$, and $|X^{i+1}|=|X^i|$ iff this cardinality is always $|X|$, that is -- iff $bxy^{-1}\in X^i$ for all $b\in X^i$ and $x,y\in X$. This obviously implies $cxy^{-1}\in X^{i+1}$ for all $c\in X^{i+1}$ and $x,y\in X$, and, conversely, this means that $|X^{i+1}|=|X^{i+2}|$.

Now, if $|X^n|=n$ then $X^n=G$, and the claim is trivial. Otherwise, $|X^i|=|X^{i+1}|$ for some $i\leq n-1$, and hence $|X^i|=|X^{i+1}|=\dots=|X^n|=\dots=|X^{2n}|$. Since $X^n\subseteq X^{2n}$, the latter implies that $X^n$ is a subgroup.

NB. Some background is left aside this proof. Let $H=\langle X^n\rangle$, $K=\langle X\rangle$. Since $X^{-1}\subseteq X^{n-1}$, we have $H\triangleleft K$; moreover, $XX^{-1}\subset H$, so $X$ lies in one coset modulo $H$. Hence $K/H$ is cyclic, and $X^i$ also lies in one coset modulo $H$.

Now the arguments above show that $|X^i|=|X^{i+1}|$ iff $X^i$ is a coset modulo $H$. Hence, if $k$ is the least multiple of $|K/H|$ which is not less than $|H|$, then even $X^k=H$.

2 added 492 characters in body

Consider the sets $X, X^2, \dots$. We claim that if $|X^i|=|X^{i+1}|$ then $|X^{i+1}|=|X^{i+2}|$. Actually, under the mapping $X^i\times X\to X^{i+1}$, $(b,x)\mapsto bx$, the preimage of any element of $X^{i+1}$ has the cardinality at most $|X|$ since all $x$-coordinates in this preimage should be distinct. Thus, $|X^{i+1}|=|X^i|$ iff this cardinality is always $|X|$, that is -- iff $bxy^{-1}\in X^i$ for all $b\in X^i$ and $x,y\in X$. This obviously implies $cxy^{-1}\in X^{i+1}$ for all $c\in X^{i+1}$ and $x,y\in X$, and, conversely, this means that $|X^{i+1}|=|X^{i+2}|$.

Now, if $|X^n|=n$ then $X^n=G$, and the claim is trivial. Otherwise, $|X^i|=|X^{i+1}|$ for some $i\leq n-1$, and hence $|X^i|=|X^{i+1}|=\dots=|X^n|=\dots=|X^{2n}|$. Since $X^n\subseteq X^{2n}$, the latter implies that $X^n$ is a subgroup.

NB. Some background is left aside this proof. Let $H=\langle X^n\rangle$, $K=\langle X\rangle$. Since $X^{-1}\subseteq X^{n-1}$, we have $H\triangleleft K$; moreover, $XX^{-1}\subset H$, so $X$ lies in one coset modulo $H$. Hence $K/H$ is cyclic, and $X^i$ also lies in one coset modulo $H$.

Now the arguments above show that $|X^i|=|X^{i+1}|$ iff $X^i$ is a coset modulo $H$. Hence, if $k$ is the least multiple of $|K/H|$ which is not less than $|H|$, then even $X^k=H$.

1

Consider the sets $X, X^2, \dots$. We claim that if $|X^i|=|X^{i+1}|$ then $|X^{i+1}|=|X^{i+2}|$. Actually, under the mapping $X^i\times X\to X^{i+1}$, $(b,x)\mapsto bx$, the preimage of any element of $X^{i+1}$ has the cardinality at most $|X|$ since all $x$-coordinates in this preimage should be distinct. Thus, $|X^{i+1}|=|X^i|$ iff this cardinality is always $|X|$, that is -- iff $bxy^{-1}\in X^i$ for all $b\in X^i$ and $x,y\in X$. This obviously implies $cxy^{-1}\in X^{i+1}$ for all $c\in X^{i+1}$ and $x,y\in X$, and, conversely, this means that $|X^{i+1}|=|X^{i+2}|$.

Now, if $|X^n|=n$ then $X^n=G$, and the claim is trivial. Otherwise, $|X^i|=|X^{i+1}|$ for some $i\leq n-1$, and hence $|X^i|=|X^{i+1}|=\dots=|X^n|=\dots=|X^{2n}|$. Since $X^n\subseteq X^{2n}$, the latter implies that $X^n$ is a subgroup.