Coset of Subgroup of Subgroup

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Theorem

Let $G$ be a group.

Let $H, K \le G$ be subgroups of $G$.

Let $K \subseteq H$.


Let $x \in G$.

Then either:

$x K \subseteq H$

or:

$x K \cap H = \O$

where $x K$ denotes the left coset of $K$ by $x$.


Proof

Suppose $x K \cap H \ne \O$.

Then:

\(\ds x K \cap H\) \(\ne\) \(\ds \O\)
\(\ds \leadsto \ \ \) \(\ds \exists y \in G: \, \) \(\ds y\) \(\in\) \(\ds x K \cap H\)
\(\ds \leadsto \ \ \) \(\ds y\) \(\in\) \(\ds x K\)
\(\ds \leadsto \ \ \) \(\ds x K\) \(=\) \(\ds y K\) Left Cosets are Equal iff Element in Other Left Coset


We have that:

$y \in H$

and:

$K \subseteq H$

As $H$ is a group, it is closed.

Hence:

$\forall x \in K: y x \in H$

which means, by definition of subset product:

$y K \subseteq H$

Hence the result.

$\blacksquare$


Sources

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