Exercise 3.1.32 (Every subgroup of $Q_8$ is normal)

Proof. The lattice of subgroups of $Q_8$ is given by

• As for every group, $⟨1⟩=\{1\}⊴Q_8$ and $Q_8⊴Q_8$, where $Q_8∕⟨1⟩\simeq Q_8$ and $Q_8∕Q_8\simeq \{1\}$.

• $N=⟨i⟩⊴Q_8$ ?

  Then $N=\{1,i,-1,-i\}$.

  Since $Q_8=⟨i,j⟩$ and $N=⟨i⟩$, there are only two verifications to do by Exercise 29

  $$\begin{array}{llll}\hfill \mathit{ii}{i}^{-1}& =i\in N,& \hfill & \\ \hfill \mathit{ji}{j}^{-1}& =(-k)(-j)=\mathit{kj}=-i\in N.& \hfill & \end{array}$$

  Therefore $N=⟨i⟩⊴Q_8$. Let $\pi$ be the natural projection $Q_8\to Q_8∕⟨i⟩$. Then $\pi (i)=\bar{i}=\bar{1}$, therefore

  $$Q_8∕⟨i⟩=⟨\pi (i),\pi (j)⟩=⟨\bar{1},\bar{j}⟩=⟨\bar{j}⟩,$$

  where

  $${\bar{j}}^2=\overline{-1}={\bar{i}}^2=\bar{1}.$$

  Therefore

  $$Q_8∕⟨i⟩\simeq Z_2.$$

  (Alternatively, $|Q_8∕⟨i⟩|=8∕4=2$, and every subgroup of order 2 is isomorphic to $Z_2$.)

• Similarly

  $$⟨j⟩⊴Q_8,Q_8∕⟨j⟩\simeq Z_2,$$
  $$⟨k⟩⊴Q_8,Q_8∕⟨k⟩\simeq Z_2.$$

  (This is also a consequence of the existence of automorphism $\phi ,\psi$ of $Q_8$ such that $\phi (i)=j,\phi (j)=i$ and $\psi (i)=k,\psi (j)=i$ (cf. Exercise 6.3.9 from future).

• $N=⟨-1⟩⊴G$.

  Then $N=\{-1,1\}$ Since $-1$ commutes with every element of $Q_8$, $N\subseteq Z(Q_8)$, hence $N⊴Q_8$ (In fact $Z(Q_9)=⟨-1⟩$).

  Moreover, $Q_8∕⟨-1⟩=\{\bar{1},\bar{i},\bar{j},\bar{k}\}$, so ${\bar{x}}^2=\bar{1}$ for every $\bar{x}\in Q_8∕⟨-1⟩$. Therefore $Q_8∕⟨-1⟩$ is not cyclic, hence

  $$Q_8∕⟨-1⟩\simeq Z_2\times Z_2\simeq V_4.$$