Non-Abelian Order 8 Group with One Order 2 Element is Quaternion Group/Lemma 1

Lemma

 * $\set {\pm 1}$ is a normal subgroup of $G$.

Proof
So:
 * $g \circ \paren {-1} \circ g^{-1} = 1$

or:
 * $g \circ \paren {-1} \circ g^{-1} = -1$

Suppose $g \circ \paren {-1} \circ g^{-1} = 1$.

Then:

which is a contradiction.

So it is the other case:
 * $g \circ \paren {-1} \circ g^{-1} = -1$

Therefore:
 * $\forall g \in G: g \circ \set {\pm 1} \circ g^{-1} = \set {\pm 1}$

By definition, $\set {\pm 1}$ is a normal subgroup.