= Solution
Let a finite <group> $G$ act on a set $X$. For $x\in X$, define its <stabilizer subgroup> $G_x=\{g:g\cdot x=x\}$ and its <group orbit> $Gx=\{g\cdot x:g\in G\}$. The map
$$
gG_x\longmapsto g\cdot x
$$
is well-defined, because multiplying $g$ on the right by an element fixing $x$ does not change its image. It is onto by the definition of the orbit. If $g\cdot x=h\cdot x$, then $h^{-1}g\in G_x$, so $gG_x=hG_x$; hence it is one-to-one. Every <coset> has $|G_x|$ elements and the <cosets> partition $G$. This proves the <orbit-stabilizer theorem>:
$$
\boxed{|G|=|Gx|\,|G_x|.}
$$
For a <group> element $x$, its <cyclic subgroup> $H=\langle x\rangle$ has $\operatorname{ord}(x)$ elements. Apply the theorem to the action of $G$ on its left <cosets> $G/H$, whose stabilizer at $H$ is $H$. It follows that
$$
\boxed{\operatorname{ord}(x)=|H|\text{ divides }|G|.}
$$
This is the needed case of <Lagrange's theorem>, obtained directly from the action.
To prove <Cauchy's theorem for finite groups>, let $p$ be a prime divisor of $|G|$ and consider
$$
\mathcal X=\{(g_1,\ldots,g_p)\in G^p:g_1\cdots g_p=e\}.
$$
The first $p-1$ entries are arbitrary and determine the last, so $|\mathcal X|=|G|^{p-1}$ is divisible by $p$. Cyclically rotate the entries. This preserves the product constraint even if $G$ is nonabelian, since
$$
g_2\cdots g_pg_1=g_1^{-1}(g_1\cdots g_p)g_1=e.
$$
Thus a <cyclic group> of order $p$ acts on $\mathcal X$. By the <orbit-stabilizer theorem>, its orbits have size one or $p$. Its fixed tuples are exactly $(g,\ldots,g)$ with $g^p=e$. If $m$ is their number, counting the remaining size-$p$ orbits gives $m\equiv|\mathcal X|\equiv0\pmod p$. The identity provides one fixed tuple, so $m\geq p$ and a nonidentity $g$ with $g^p=e$ exists. The order of this $g$ divides $p$: division of $p$ by its order and minimality of that order prove this directly. Since $p$ is prime and $g\ne e$, the order is exactly $p$. This is the <cyclic-tuple proof of Cauchy theorem>.
Now suppose every nonidentity element has order two. Then every element equals its inverse, and for any $a,b\in G$,
$$
ab=(ab)^{-1}=b^{-1}a^{-1}=ba.
$$
Thus <group of exponent two is abelian> applies. Starting from the identity <subgroup>, choose an element outside the <subgroup> already generated. Because the <group> is abelian and that element squares to the identity, adjoining it doubles the <subgroup>: it is the disjoint union of the old <subgroup> and its translate by the new element. Repeating until the <finite group> is exhausted shows $|G|=2^r$. Conversely the <direct product of groups> $C_2^r$ has order $2^r$ and every nonidentity element has order two. Therefore
$$
\boxed{n=2^r\quad(r=0,1,2,\ldots).}
$$
The case $r=0$ is the trivial <group> and satisfies the condition vacuously. These finite examples are <elementary abelian groups> of exponent two.
An infinite example is the <group> of all binary sequences, with coordinatewise addition modulo two:
$$
\boxed{G=\prod_{j=1}^\infty\mathbb Z/2\mathbb Z.}
$$
Every sequence added to itself is zero, so every nonzero sequence has order exactly two. The sequences having a single $1$ in coordinate $j$ are all distinct, proving that the <group> is infinite.
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