A subgroup such that every left coset is contained in a right coset.

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 $|xH| = |Hy|$.

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.

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

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.