Exercise 3.2.3 (Lattice of subgroups of $Q_8$)

Proof. The given lattice of $Q_8$ is

By Lagrange’s Theorem, every proper non trivial subgroup has order $2$ or $4$.

A subgroup of order $2$ is cyclic, and there is only one element of order $2$, which is $-1$. Therefore the only subgroup of order $2$ is $⟨-1⟩$.

Consider now a subgroup $H$ of order $4$. Then $H\simeq Z_4$ or $H\simeq Z_2\times Z_2$. In this last case, every element of $H$ distinct of $1$ has order $2$. This is impossible, because there is only one element of order $2$. Therefore $H$ is cyclic. Since the elements of order $4$ are $\pm i,\pm j,\pm k$, and $⟨i⟩=⟨i^{-1}⟩=⟨-i⟩$ (and similar results for $j$ and $k$), where conclude that there are exactly three subgroups of order $4$: $⟨i⟩$, $⟨j⟩$, $⟨k⟩$. So there are no other subgroup than those written in the preceding lattice.

Since $i^2=j^2=k^2=-1$, we obtain the inclusions $⟨-1⟩\subseteq ⟨x⟩$, where $x\in \{i,j,k\}$, and no intermediate subgroup between these subgroups. Since the list of subgroups is complete, there is no other direct inclusion than the seven inclusions of the lattice. □