19
$\begingroup$

Let $G$ be any group, and $H \leq G$ a subgroup.

Suppose that for each $x \in G$, there exists a $y \in G$ such that $xH \subseteq Hy$. In other words, every left coset of $H$ is contained inside some right coset of $H$.

Question: what can we say about $H$? In particular, does this imply that $H$ is normal?

I know that if $G$ is finite, then $H$ must be normal, because then $xH \subseteq Hy$ implies $xH = Hy$, since $|xH| = |Hy|$.

$\endgroup$
24
$\begingroup$

If $xH \subseteq Hy$, then in particular $x \in Hy$. But then $Hx \cap Hy \neq \varnothing$, so $Hx=Hy$, so $xH \subseteq Hx$, so $xHx^{-1}\subseteq H$. Since this is true for every $x$, $H$ is normal.

$\endgroup$
  • $\begingroup$ If $xH \subseteq Hy$ for some $x, y \in G$ (not all), then can we still have $xH = Hy = Hx$? $\endgroup$ – Lao-tzu Dec 3 '15 at 2:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.