Coset Product on Non-Normal Subgroup is not Well-Defined

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Theorem

Let $\struct {G, \circ}$ be a group.

Let $H$ be a subgroup of $G$ which is not normal.

Let $a, b \in G$.


Then it is not necessarily the case that the coset product:

$\paren {a \circ H} \circ \paren {b \circ H} = \paren {a \circ b} \circ H$

is well-defined.


Proof

Proof by Counterexample:

Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:

$\begin{array}{c|cccccc}\circ & e & (123) & (132) & (23) & (13) & (12) \\ \hline e & e & (123) & (132) & (23) & (13) & (12) \\ (123) & (123) & (132) & e & (13) & (12) & (23) \\ (132) & (132) & e & (123) & (12) & (23) & (13) \\ (23) & (23) & (12) & (13) & e & (132) & (123) \\ (13) & (13) & (23) & (12) & (123) & e & (132) \\ (12) & (12) & (13) & (23) & (132) & (123) & e \\ \end{array}$


Consider the subgroups of $S_3$:

The subsets of $S_3$ which form subgroups of $S_3$ are:

\(\ds \) \(\) \(\ds S_3\)
\(\ds \) \(\) \(\ds \set e\)
\(\ds \) \(\) \(\ds \set {e, \tuple {123}, \tuple {132} }\)
\(\ds \) \(\) \(\ds \set {e, \tuple {12} }\)
\(\ds \) \(\) \(\ds \set {e, \tuple {13} }\)
\(\ds \) \(\) \(\ds \set {e, \tuple {23} }\)


Let $H = \set {e, \tuple {12} }$.

From Normal Subgroups in Symmetric Group on 3 Letters, $H$ is not normal.


Let $A = \set {\tuple {123}, \tuple {23} }$.

We have:

\(\ds \tuple {123} H\) \(=\) \(\ds \set {\tuple {123}, \tuple {23} }\)
\(\ds \tuple {23} H\) \(=\) \(\ds \set {\tuple {123}, \tuple {23} }\)

and so $A$ is the left coset of $H$ by both $\tuple {123}$ and $\tuple {23}$.


Let $B = \set {\tuple {132}, \tuple {13} }$.

\(\ds \tuple {132} H\) \(=\) \(\ds \set {\tuple {132}, \tuple {13} }\)
\(\ds \tuple {13} H\) \(=\) \(\ds \set {\tuple {132}, \tuple {13} }\)

and so $B$ is the left coset of $H$ by both $\tuple {132}$ and $\tuple {13}$.


Now consider:

\(\ds A \circ B\) \(=\) \(\ds \tuple {123} H \tuple {132} H\)
\(\ds \) \(=\) \(\ds \tuple {123} \tuple {132} H\)
\(\ds \) \(=\) \(\ds e H\)
\(\ds \) \(=\) \(\ds H\)


But:

\(\ds A \circ B\) \(=\) \(\ds \tuple {23} H \tuple {13} H\)
\(\ds \) \(=\) \(\ds \tuple {23} \tuple {13} H\)
\(\ds \) \(=\) \(\ds \tuple {132} H\)
\(\ds \) \(=\) \(\ds B\)
\(\ds \) \(\ne\) \(\ds H\)

So two different evaluations of the coset product give two different result.

Hence by definition the coset product is not well-defined on $H$.

The result follows.

$\blacksquare$


Sources

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