In the book Algebra Ed2 by Michael Artin, in chapter 7.8 where Artin classifies the groups of order $12$, he considers a Sylow $3$-subgroup $K$ and a Sylow $2$-subgroup $H$ of order $4$.
Then $K\approx C_3$ and $H\approx C_4$ or $H\approx C_2\times C_2$. Therefore $H\cap K=1$, and then $HK=G$ as the product map is bijective. He proved that at least one of $H$ and $K$ is a normal subgroup, and consider them case by case.
In Case 3, when $K$ is normal, but $H$ is not, he writes:
Then $H$ operates by conjugation on $K = \left\{1,y,y^2\right\}$. Since $H$ is not normal, it contains an element $x$ that doesn't commute with $y$, and then $xyx^{-1} = y^2$.
Why must it be true that an element of $x$ in $H$ does not commute with $y$?
Thanks!
$HK=G$ and $H$ is abelian. If all elements of $H$ commute with those of $K$, then $G$ is abelian. This contradicts that $H$ is not normal.