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

\(\ds \forall g \in G: \, \)

\(\ds g \circ 1 \circ g^{-1}\)

\(=\)

\(\ds g \circ g^{-1}\)

Definition of Identity Element

\(\ds \)

\(=\)

\(\ds 1\)

Definition of Inverse Element

\(\ds \forall g \in G: \, \)

\(\ds \paren {g \circ \paren {-1} \circ g^{-1} }^2\)

\(=\)

\(\ds g \circ \paren {-1} \circ g^{-1} \circ g \circ \paren {-1} \circ g^{-1}\)

\(\ds \)

\(=\)

\(\ds g \circ \paren {-1} \circ \paren {-1} \circ g^{-1}\)

Definition of Inverse Element

\(\ds \)

\(=\)

\(\ds g \circ g^{-1}\)

Order of $-1$ is $2$

\(\ds \)

\(=\)

\(\ds 1\)

Definition of Inverse Element

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:

\(\ds g \circ \paren {-1} \circ g^{-1}\)

\(=\)

\(\ds 1\)

\(\ds g^{-1} \circ g \circ \paren {-1} \circ g^{-1} \circ g\)

\(=\)

\(\ds g^{-1} \circ 1 \circ g\)

applying same action on both sides

\(\ds -1\)

\(=\)

\(\ds 1\)

Definition of Inverse Element, Definition of Identity Element

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.

$\blacksquare$