= Solution
Apply <Zorn's lemma> using the chain upper bounds just proved, obtaining a maximal partial <group operation> $m$ on a nonempty carrier $A\subseteq X$. Suppose $|X\setminus A|\geq|A|$. Then choose a <subset> $B\subseteq X\setminus A$ and a <bijection> $b:A\to B$. The <set> $A\cup B$ can be identified with $A\times C_2$ by sending $(a,0)$ to $a$ and $(a,1)$ to $b(a)$.
Use <group extension by a duplicate carrier> to transport the direct-product <group operation> along this <bijection>. Its restriction to $A\times A$ is exactly the original $m$, because $(a,0)(a',0)=(m(a,a'),0)$. It therefore gives a strictly larger member of $\mathcal G$, contradicting maximality. Thus $|A|\nleq|X\setminus A|$. Cardinal comparability, allowed as a consequence of <Zorn's lemma>, leaves the strict opposite inequality:
$$
\boxed{|X\setminus A|<|A|.}
$$
The argument uses choice through <Zorn's lemma> and cardinal comparability; the extension itself is an explicit transported direct-product operation once the <bijection> with $B$ is chosen.
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