 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 $|X\setminus A|\ge |A|$. 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 $(a,0)(a^{\prime },0)=(m(a,a^{\prime }),0)$. It therefore gives a strictly larger member of $\mathcal{G}$, contradicting maximality. Thus $|A|\nleqq |X\setminus A|$. Cardinal comparability, allowed as a consequence of Zorn's lemma, leaves the strict opposite inequality: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.