Exercise 3.3.5 ($QD_{16}$, the return)

Proof. In Exercise 2.5.11, we have proved that the center of $Q{D}_{16}$ is $Z(Q{D}_{16})=⟨{\sigma }^4⟩$, so $⟨{\sigma }^4⟩$ is normal in $Q{D}_{16}$.

The lattice of subgroups of $Q{D}_{16}$ is

By the Lattice Isomorphism Theorem, the lattice of subgroups of $Q{D}_{16}∕⟨{\sigma }^4⟩$ is isomorphic to the lattice of subgroups of $Q{D}_{16}$ above $⟨{\sigma }^4⟩$, which gives, using ${\bar{\sigma }}^4=\bar{1}$,

(But in the solution of Exercise 2.5.11, we used the knowledge of the isomorphism type of $Q{D}_{16}∕⟨{\sigma }^4⟩$ to prove that there is no other subgroup of $Q{D}_{16}$ above $⟨{\sigma }^4⟩$.)

We recognize in this diagram the lattice of $D_8$.

I repeat the arguments to prove that $Q{D}_{16}∕⟨{\sigma }^4⟩\simeq D_8$:

$Z=⟨{\sigma }^4⟩$ is a normal subgroup of $Q{D}_{16}$, so there is a surjective homomorphism $\pi :Q{D}_{16}\to Q{D}_{16}∕⟨{\sigma }^4⟩$. Moreover, the classes $\bar{\sigma }=\mathit{𝜎𝑍}$ and $\bar{\tau }=\mathit{𝜏𝑍}$ are generators of $Q{D}_{16}∕Z$ which satisfy

Since $D_8=⟨r,s\mid r^4=s^2=1,\mathit{𝑠𝑟}=r^{3}s⟩$, there is a surjective homomorphism $\lambda :D_8\to Q{D}_{16}∕Z$ such that $\lambda (r)=\bar{\sigma }$ and $\lambda (s)=\bar{\tau }$. Moreover, $|D_8|=|Q{D}_{16}∕Z|=8$, therefore $\lambda$ is an isomorphism, so

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