21
$\begingroup$

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 $|xH| = |Hy|$.

$\endgroup$
26
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
0
$\begingroup$

Theorem 1 Let $H,K \leq G$ and assume that a left coset of $H$ is contained in a left coset of $K$. Then $H \subseteq K$.

Proof Assume first that $x \in G$ with $xH \subseteq K$. Then $x=x \cdot 1 \in K$. Pick an $h \in H$, then $xh=k$ for some $k \in K$, so $h=x^{-1}k \in K$, and it follows that $H \subseteq K$.
Now suppose $xH \subseteq yK$ for some $x,y \in G$. Then $y^{-1}xH \subseteq K$ and from the previous it now follows that $H \subseteq K$.

Corollary 1 Let $H,K \leq G$ and assume that a right coset of $H$ is contained in a right coset of $K$. Then $H \subseteq K$.

Proof One can mimic the proof above, or, in general, if $Hx \subset Ky$ for some $x,y \in G$, this is equivalent to $x^{-1}H \subseteq Ky^{-1}$.

So what happens if we mix left and right.

Theorem 2 Let $H,K \leq G$, $x,y \in G$ and assume $xH \subseteq Ky$. Then $Kx=Ky$.

Proof Since $xH \subseteq Ky$, $H^{x^{-1}} \subseteq Kyx^{-1}$, so Corollary 1 yields $H^{x^{-1}} \subseteq K$, that is, $H \subseteq K^x$ and equivalently, $xH \subseteq Kx$. Hence, $\emptyset \neq xH \subseteq Ky \cap Kx$, implying $Kx=Ky$.

Corollary 2 Let $H \leq G$ and $x,y \in G$ with $xH \subseteq Hy$. Then $Hx=Hy$. In particular if $H$ is finite then $xH=Hx=Hy$.

Proof Take $H=K$ in Theorem 2, yielding $Hx=Hy$. If $H$ is finite then $xH \subseteq Hy$ implies $xH=Hy$ and the result follows.

$\endgroup$

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.