Coset containment implies subgroup containment

DIRECT: The fact or result stated in this article has a trivial/direct/straightforward proof provided we use the correct definitions of the terms involved
View other results with direct proofs
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this

Statement

Verbal statement

If a Left coset (?) of one subgroup is contained in a left coset of another subgroup, then the subgroup is also contained in the other.

Statement with symbols

Suppose $xH\subseteq yK$ are cosets of subgroups $H,K\leq G$. Then $H\leq K$.

Related facts

• Subgroup containment implies coset containment: If one subgroup of a group is contained in another, then every left coset of the subgroup is contained in a left coset of the other subgroup.
• Nonempty intersection of cosets is coset of intersection: If the intersection of a collection of left cosets of subgroups is nonempty, it is a coset of the intersection of the corresponding subgroups.

Proof

Given: $xH\subseteq yK$ for some $x,y\in G$.

To prove: $H\leq K$

Proof: Since $x=xe$ is in $xH$, which is contained in $yK$, there exists $k_{1}\in K$ such that $x=yk_{1}$. Thus $yk_{1}H\subseteq yK$, which implies that there exist $h\in H$ and $k_{2}\in K$ such that $yk_{1}h=yk_{2}$. Hence $h=k_{2}k_{1}^{-1}$, which is in $K$. So, $H\leq K$.