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> [!proposition] Proposition. ([[quaternion group is nonabelian, but all its subgroups are normal]])
> Every [[subgroup]] of the [[quaternion group]] $Q$ of [[order of a group|order]] $8$ is a [[normal subgroup]], but the [[center of a group|center]] of $Q$ is $\{ \v 1, -\v 1 \} \neq Q$.
> [!proof]- Proof. ([[quaternion group is nonabelian, but all its subgroups are normal]])
> ~
The [[subgroup]] lattice for $Q$ is ![[CleanShot 2023-09-16 at 14.12.34.jpg]]
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Clearly $Q$ and $\{ \v 1 \}$ are [[normal subgroup]]s of $Q$. Also, $-\v1 \v g=\v g \v 1$ for all $\v g \in G$ so $\{ \v 1, -\v 1 \}$ is as well.
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Let $\v g$ and $\b \ell$ be arbitrary elements in $Q \cut \{ \v 1, -\v 1 \}$. In general:
>- $\v g \b \ell=- \b \ell \v g= \b \ell^{3} \v g$;
>- $\v g \b \ell^{2}=-\v g = \b \ell^{2} \v g$;
>- $\v g \b \ell^{3}=- \v g \b \ell =\b \ell \v g$.
Thus for any $C:=\langle \v c \rangle$, $\v c \in \{ \v i, \v j, \v k \}$, we have for arbitrary $\v g \in Q \cut \{ \v -1 \}$ that $\begin{align}
\v gC = & \{ \v g\v c , \ \v g \v c ^{2}, \ \v g \v c ^{3}, \v 1\} \\ & \ \ \downarrow \ \ \ \downarrow \ \ \ \ \ \downarrow \ \ \ \downarrow
\\= & \{ \v g \v c^{3}, \v g \v c ^{2} , \v g \v c, \ \ \ \v 1 \} \\
= & \{ \v c \v g, \v c ^{2} \v g, \v c ^{3} \v g, \v 1 \} \\
= & C\v g,
\end{align}$
and also that $(- \v 1 )C = \{ -\v c, - \v c^{2}, - \v c^{3}, -\v 1 \}=\{ \v c^{3}, \v 1, \v c, - \v 1 \},$
from which we conclude $\langle \v i \rangle, \langle \v j \rangle, \langle \v k \rangle$ are [[normal subgroup]]s of $Q$.
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However, $Z(G)=\{ \v 1, - \v 1 \}$, since for all $\v g$ , $\b \ell$ in $Q \cut \{ \v 1, -\v 1 \}$ we have $\v g \b \ell = -\b \ell \v g$.
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