Coset containment implies subgroup containment

Contents

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|

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 $H,K \le G$. Then $H \le K$.

Proof

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

To prove: $H \le K$

Proof: Since $x=xe$ is in $xH$, which is contained in $yK$, there exists $k_1 \in K$ such that $x=yk_1$. Thus $yk_1H \subseteq yK$, which implies that there exist $h \in H$ and $k_2 \in K$ such that $yk_1h=yk_2$. Hence $h=k_2k_1^{-1}$, which is in $K$. So, $H\le K$.