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

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Lemma

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


Proof

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


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

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:

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

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.

$\blacksquare$