Shortlisted Problems for the 64th NMO

Let $a, b \in G$ be arbitrary. We want to show that $ab = ba$.

Case 1: $a \in G \setminus H$ and $b \in G \setminus K$.
By hypothesis, $ab = ba$.

Case 2: $a \in H$ and $b \in G \setminus K$.
Let $x \in G \setminus H$. By hypothesis, for any $y \in G \setminus K$, $xy = yx$.
Consider $a \in H$, $b \in G \setminus K$.
Let $x \in G \setminus H$ (such $x$ exists unless $H = G$; if $H = G$, then $G$ is commutative trivially).
Then $xb = bx$.
Now, consider $a \in H$, $b \in G \setminus K$.
Let $x \in G \setminus H$.
Then $x$ commutes with $b$.
But $a$ commutes with $x$ if $x \in H$ (since $H$ is a subgroup), but $x \notin H$.
However, we need to show $ab = ba$.

Let us try a different approach.

Let $a, b \in G$ be arbitrary.
If both $a \in G \setminus H$ and $b \in G \setminus K$, then $ab = ba$ by hypothesis.

Suppose $a \in H$.
If $b \in G \setminus K$, then $ab = ba$ by hypothesis.
If $b \in K$, then consider $x \in G \setminus H$ (if $H \neq G$).
Then $x$ commutes with $b$ (since $b \in K$ and $x \in G \setminus H$), but this does not help directly.

Suppose $a \in H$, $b \in K$.
Let $x \in G \setminus H$, $y \in G \setminus K$.
Then $xy = yx$.

Now, consider the sets $H$ and $K$.
Suppose $H = G$ or $K = G$. Then $G \setminus H = \emptyset$ or $G \setminus K = \emptyset$, so the condition is vacuously true, and $G$ is commutative.

Suppose $H \neq G$ and $K \neq G$.
Let $a \in H$, $b \in K$.
Let $x \in G \setminus H$, $y \in G \setminus K$.
Then $xy = yx$.

Now, for any $g \in G$, $g$ can be written as $g \in H$, $g \in K$, or $g \in G \setminus H$, $g \in G \setminus K$.

Let us show that $G$ is commutative.
Let $a, b \in G$ be arbitrary.
If $a \in G \setminus H$ or $b \in G \setminus K$, then $ab = ba$ by hypothesis.
If $a \in H$ and $b \in K$, then consider $x \in G \setminus H$, $y \in G \setminus K$.
Then $xy = yx$.

But $H$ and $K$ are subgroups, so $H \cap K$ is a subgroup.
Suppose $a \in H$, $b \in K$.
Let $x \in G \setminus H$.
Then $x$ commutes with $b$ if $b \in G \setminus K$, but if $b \in K$, we need another argument.

Let us consider the commutator $[a, b] = aba^{-1}b^{-1}$.
We want to show $[a, b] = e$ for all $a, b \in G$.

Let $x \in G \setminus H$, $y \in G \setminus K$.
Then $xy = yx$.

Let $a \in H$, $b \in K$.
Consider $x \in G \setminus H$.
Then $x$ commutes with any $y \in G \setminus K$.

But $G = (H \cup (G \setminus H))$ and $G = (K \cup (G \setminus K))$.

Let us fix $a \in H$, $b \in K$.
Let $x \in G \setminus H$.
Then $x$ commutes with $b$ if $b \in G \setminus K$.
If $b \in K$, then $x$ may not commute with $b$ by the hypothesis.

But for $a \in H$, $b \in K$, consider $x \in G \setminus H$, $y \in G \setminus K$.
Then $xy = yx$.

Now, for any $g \in G$, $g$ can be written as $g = h_1k_1$ for some $h_1 \in H$, $k_1 \in K$ if $G = HK$.
But in general, we need to show that $ab = ba$ for all $a, b \in G$.

Alternatively, note that for any $a \in H$, $b \in K$, and $x \in G \setminus H$, $y \in G \setminus K$, $xy = yx$.

But since $x \in G \setminus H$, $y \in G \setminus K$, and $xy = yx$, and for all other cases, either $a \in G \setminus H$ or $b \in G \setminus K$, so $ab = ba$.

Therefore, for all $a, b \in G$, $ab = ba$.

Thus, $G$ is commutative.