Set of Transpositions is not Subgroup of Symmetric Group

Theorem

Let $S$ be a finite set with $n$ elements such that $n > 2$.

Let $G = \struct {\map \Gamma S, \circ}$ denote the symmetric group on $S$.

Let $H \subseteq G$ denote the set of all transpositions of $S$ along with the identity mapping which moves no elements of $S$.

Then $H$ does not form a subgroup of $G$.

Proof

First it is noted that $H \subseteq G$, and that the identity mapping is an element of $H$.

Hence to demonstrate that $H$ is a subgroup of $G$, one may use the Two-Step Subgroup Test.

Let $\phi \in H$ be a transposition.

Then from Transposition is Self-Inverse, $\phi^{-1} \in H$.

So $H$ is closed under inversions.

Let $a, b, c \in S$.

Let $\pi, \phi \in H$ such that:

$\pi = \tuple {a \ b}$

$\phi = \tuple {b \ c}$

Then by inspection:

$\phi \circ \pi = \tuple {a \ c \ b}$

Thus $\phi \circ \pi$ is a $3$-cycle and so not a transposition.

Hence $H$ is not closed under composition.

So, by the Two-Step Subgroup Test, $H$ is not a subgroup of $G$.

$\blacksquare$

Sources

• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Exercise $\text{E iv}$