# 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

 $\displaystyle \forall g \in G: \ \$ $\displaystyle g \circ 1 \circ g^{-1}$ $=$ $\displaystyle g \circ g^{-1}$ Definition of Identity Element $\displaystyle$ $=$ $\displaystyle 1$ Definition of Inverse Element

 $\displaystyle \forall g \in G: \ \$ $\displaystyle \paren {g \circ \paren {-1} \circ g^{-1} }^2$ $=$ $\displaystyle g \circ \paren {-1} \circ g^{-1} \circ g \circ \paren {-1} \circ g^{-1}$ $\displaystyle$ $=$ $\displaystyle g \circ \paren {-1} \circ \paren {-1} \circ g^{-1}$ Definition of Inverse Element $\displaystyle$ $=$ $\displaystyle g \circ g^{-1}$ Order of $-1$ is $2$ $\displaystyle$ $=$ $\displaystyle 1$ 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:

 $\displaystyle g \circ \paren {-1} \circ g^{-1}$ $=$ $\displaystyle 1$ $\displaystyle g^{-1} \circ g \circ \paren {-1} \circ g^{-1} \circ g$ $=$ $\displaystyle g^{-1} \circ 1 \circ g$ applying same action on both sides $\displaystyle -1$ $=$ $\displaystyle 1$ Definition of Inverse Element, Definition of Identity Element

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.

$\blacksquare$