Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 2

Lemma

$\paren {\pm a}^2 = \paren {\pm b}^2 = \paren {\pm c}^2 = -1$

Proof

Without loss of generality, $a$ is checked.

The proofs for other $5$ elements are similar.

\(\ds \paren {a^2}^2\)

\(=\)

\(\ds a^4\)

Powers of Group Elements: Product of Indices

\(\ds \)

\(=\)

\(\ds 1\)

Definition of Order of Group Element

So $a^2 = 1$ or $a^2 = -1$.

As the order of $a = 4$:

$a^2 \ne 1$

Hence:

$a^2 = -1$

as required.

$\blacksquare$