Parity Group is a Group

Theorem

The parity group is in fact a group.

Proof

We can completely describe the parity group by showing its Cayley table:

$\begin{array}{r|rr} \left({\left\{{1, -1}\right\}, \times}\right) & 1 & -1\\ \hline 1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{array} \qquad \begin{array}{r|rr} \left({\Z_2, +_2}\right) & \left[\!\left[{0}\right]\!\right]_2 & \left[\!\left[{1}\right]\!\right]_2\\ \hline \left[\!\left[{0}\right]\!\right]_2 & \left[\!\left[{0}\right]\!\right]_2 & \left[\!\left[{1}\right]\!\right]_2 \\ \left[\!\left[{1}\right]\!\right]_2 & \left[\!\left[{1}\right]\!\right]_2 & \left[\!\left[{0}\right]\!\right]_2 \\ \end{array}$

It is easily checked that the four group axioms hold for both instances.

As there is only one group of order 2, up to isomorphism, from Group of Prime Order Cyclic, we see that both examples of the parity group are isomorphic to $C_2$.

Sources

• Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.3$: Example $11$