If $X$ is an infinite set, can we always make it into a group in which every element has order $\leq 2$?
Just for a countable set, I tried to define product with required property, but, because of ``infiniteness'', I didn't find any direction to proceed further.
If every element of a finite group has order $2$, the group is of the form $\bigoplus C_2$. We can extend this idea to infinite sets: $\bigoplus_X C_2$ (the direct sum of $|X|$-many $C_2$s) has cardinality $X$ if $|X|\ge\aleph_0$.
Proof. One may identify $\bigoplus_X C_2$ with the set of finite subsets of $X$ (each finite subset $S$ yields a vector with $1$s in the coordinates $s\in S$ and $0$ in other coordinates). This in turn may be split up into the parts $X_0\cup X_1\cup X_2\cup\cdots$ where $X_k$ denotes the set of $k$-subsets of $X$. It should be clear that there is an onto map $X^k\to X_k\cup\cdots\cup X_1$ given by $(x_1,\cdots,x_k)\mapsto\{x_1,\cdots,x_k\}$ and so we conclude $|X_k|\le |X^k|$. By cardinal arithmetic, we know that $|X^k|=|X|$ and we have
$$|G|=|X_0\cup X_1\cup\cdots|\le |X_0|+|X_1|+\cdots=|X|+|X|+|X|+\cdots=|X|\cdot\aleph_0=|X|$$
and so there exists a bijection $X\cong G$. Then just transport the structure.
