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

Lemma

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

Proof
So $g \circ \left({-1}\right) \circ g^{-1} = 1$ or $g \circ \left({-1}\right) \circ g^{-1} = -1$.

Suppose $g \circ \left({-1}\right) \circ g^{-1} = 1$, then we have:

which is a contradiction.

So it is the another case: $g \circ \left({-1}\right) \circ g^{-1} = -1$.

Therefore, $\forall g \in G, g \circ \left\{ {\pm 1}\right\} \circ g^{-1} = \left\{ {\pm 1}\right\}$.

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