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.

  • $\begingroup$ 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

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.


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