Exercise 2.5.11 (Quasi Dihedral Group $QD_{16}$)

Proof. We can complete the lattice without any other knowledge about $Q{D}_{16}$ than the data given in the statement, i.e.,

and every proper subgroup is contained in one of the subgroups $⟨{\sigma }^2,\tau ⟩,⟨\sigma ⟩,⟨{\sigma }^2,\mathit{\tau \sigma }⟩$, which we assume in the first part.

For this purpose, we use the three sublattices of these subgroups of order $8$. Later, in the second part, we will study the group $Q{D}_{16}$, and prove (1). (From a logical point of view, part (2°) precedes part (1°).)

Note: Since

we obtain

It is preferable to consider the subgroup $⟨{\sigma }^2,\mathit{\tau \sigma }⟩$ which is in the lattice, rather than $⟨{\sigma }^2,\mathit{\sigma \tau }⟩$ which is in the statement.

(1°)

We assume (1) in this first part.

(a)

Lattice of the subgroups of $⟨{\sigma }^2,\tau ⟩$.

Note first that the relations defining $Q{D}_{16}$ imply $\mathit{\tau \sigma }={\sigma }^3\tau$, thus $\tau {\sigma }^2={\sigma }^3\mathit{\tau \sigma }={\sigma }^6\tau$.

We name $r,s$ two generators of $D_8=⟨r,s\mid r^4=s^2=1,\mathit{sr}=r^{3}s⟩$. Since ${({\sigma }^2)}^4={\tau }^2=1$ and $\tau {\sigma }^2={({\sigma }^2)}^3\tau$, by van Dyck’s Theorem (see the note in Ex. 2.4.7) there is a surjective homomorphism $\phi :D_8=⟨r,s⟩\to ⟨{\sigma }^2,\tau ⟩$ such that $\phi (r)={\sigma }^2$ and $\phi (s)=\tau$. Moreover, by (1), $|D_8|=|⟨{\sigma }^2,\tau ⟩|=8$, therefore $\phi$ is an isomorphism. Hence it it sufficient to replace $r$ by ${\sigma }^2$ and $s$ by $\tau$ in the lattice of subgroups of $D_8$ to obtain the corresponding lattice of the subgroups of $⟨{\sigma }^2,\tau ⟩$ (cf. Example 4, slightly modified).

This gives the left unknown subgroups of the lattice: $⟨{\sigma }^4,\tau {\sigma }^2⟩$, $⟨\tau {\sigma }^6⟩$, and $⟨\tau {\sigma }^4⟩$.

(b)

Lattice of the subgroups of $⟨\sigma ⟩$.

Since ${\sigma }^8=1$ and $|⟨\sigma ⟩|=8$ (by (1)), we obtain $⟨\sigma ⟩\simeq Z_8\simeq \mathbb{Z}∕8\mathbb{Z}$. Starting from the lattice of $\mathbb{Z}∕8\mathbb{Z}$ given in Example 1, we obtain the lattice of subgroups of $⟨\sigma ⟩$, by the isomorphism which maps $\bar{1}$ on $\sigma$.

This gives the unknown subgroup $⟨{\sigma }^2⟩$.

(c)

Lattice of the subgroups of $⟨{\sigma }^2,\mathit{\tau \sigma }⟩$.

We recall the presentation of $Q_8$ given by

$$Q_8\simeq ⟨a,b\mid a^2=b^2,a^{-1}\mathit{ba}=b^{-1}\}$$

(cf. Ex. 6.3.7, or Ex. 1.5.3).

Consider the elements $I={\sigma }^2$, $J=\mathit{\tau \sigma }$ (and $K=\mathit{IJ}={\sigma }^2\mathit{\tau \sigma }=\tau {\sigma }^7$). Then, using the relations of the presentation of $Q{D}_{16}$,

$$\begin{array}{llll}\hfill I^2& ={\sigma }^4,& \hfill & \\ \hfill J^2& =(\mathit{\tau \sigma })(\mathit{\tau \sigma })=\tau (\tau {\sigma }^3)\sigma ={\sigma }^4=I^2,& \hfill & \\ \hfill J^{-1}& ={\sigma }^{-1}{\tau }^{-1}={\sigma }^{-1}\tau =\tau {\sigma }^{-3},& \hfill & \\ \hfill I^{-1}\mathit{JI}& ={\sigma }^{-2}\mathit{\tau \sigma }{\sigma }^2=\tau {\sigma }^{-3}=J^{-1}.& \hfill & \end{array}$$

Since $I^2=J^2$ and $I^{-1}\mathit{JI}=J^{-1}$, using anew the van Dyck’s Theorem, there is a surjective homomorphism $\chi :Q_8\to ⟨{\sigma }^2,\mathit{\tau \sigma }⟩$ such that $\chi (a)=I={\sigma }^2,\chi (b)=J=\mathit{\tau \sigma }$. Since $|Q_8|=|⟨{\sigma }^2,\mathit{\tau \sigma }⟩|=8$, $\chi$ is an isomorphism. Using $\chi$, we obtain the sublattice of the subgroups of $⟨{\sigma }^2,\mathit{\tau \sigma }⟩$ (cf. Example 5).

This gives the two remaining unknown subgroups: $⟨\mathit{\tau \sigma }⟩$ and $⟨\tau {\sigma }^7⟩=⟨\tau {\sigma }^{-1}⟩$.

Since $|⟨{\sigma }^2,\tau ⟩|=|⟨\sigma ⟩|=|⟨{\sigma }^2,\mathit{\tau \sigma }⟩|=8$, by Lagrange’s Theorem, these three subgroups are maximal. Putting all sublattices together, we obtain the complete lattice of subgroups of $Q{D}_{16}$.

(2°)

Now we explain why $|Q{D}_{16}|=16$ (this is not obvious: it may happen some collapsing, as in Ex. 1.2.18) and $|⟨{\sigma }^2,\tau ⟩|=|⟨\sigma ⟩|=|⟨{\sigma }^2,\mathit{\tau \sigma }⟩|=8$.

(a)

By definition of $Q{D}_{16}$, $$\begin{array}{lll}\hfill {\sigma }^8={\tau }^2=1,\mathit{\sigma \tau }=\tau {\sigma }^3.& & \hfill \text{(2)}\end{array}$$

By the relations (2), every element of $Q{D}_{16}$ is of the form ${\sigma }^i{\tau }^j,0\le i<8,0\le j<2$, so $Q{D}_{16}$ has at most $16$ elements. We search how to multiply two such elements.

We show by induction that ${\sigma }^k\tau =\tau {\sigma }^{3k}$. This is true for $k=0$. If we assume that ${\sigma }^k\tau =\tau {\sigma }^{3k}$ for some integer $k\ge 0$, then

$${\sigma }^{k+1}\tau =\sigma ({\sigma }^k\tau )=\mathit{\sigma \tau }{\sigma }^{3k}=\tau {\sigma }^3{\sigma }^{3k}=\tau {\sigma }^{3(k+1)}.$$

The induction is done, which shows that for every $k\in \mathbb{N}$,

$$\begin{array}{lll}\hfill {\sigma }^k\tau =\tau {\sigma }^{3k}.& & \hfill \text{(3)}\end{array}$$

Then $\tau {\sigma }^{-3k}={\sigma }^{-k}\tau$, so (3) remains true for all $k\in \mathbb{Z}$.

Moreover, multiplying (3) by $\tau$ both left and right, we obtain

$$\begin{array}{lll}\hfill \tau {\sigma }^k={\sigma }^{3k}\tau (k\in \mathbb{Z}).& & \hfill \text{(4)}\end{array}$$

If $j=1$, then

$$\begin{array}{llll}\hfill ({\sigma }^i{\tau }^j)({\sigma }^k{\tau }^l)& ={\sigma }^i\tau {\sigma }^k{\tau }^l& \hfill & \\ \hfill & ={\sigma }^i{\sigma }^{3k}\tau {\tau }^l& \hfill & \\ \hfill & ={\sigma }^{i+3k}{\tau }^{1+l},& \hfill & \end{array}$$

and if $j=0$, then

$$\begin{array}{llll}\hfill ({\sigma }^i{\tau }^j)({\sigma }^k{\tau }^l)& ={\sigma }^i{\sigma }^k{\tau }^l& \hfill & \\ \hfill & ={\sigma }^{i+k}{\tau }^l,& \hfill & \end{array}$$

In both cases, we obtain for all integers $i,j,k,l$,

$$({\sigma }^i{\tau }^j)({\sigma }^k{\tau }^l)={\sigma }^{i+3^{j}k}{\tau }^{j+l}.$$

Consider now the group $\mathbb{Z}∕8\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ with the law

$$\begin{array}{lll}\hfill (i,j)\cdot (k,l)=(i+3^{j}k,j+l).& & \hfill \text{(5)}\end{array}$$

(This makes sense because $3^2=1$ in $\mathbb{Z}∕8\mathbb{Z}$).

Those who know semi-direct products (see Chapter 5) recognize the group $G=\mathbb{Z}∕8Z{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$, where

$$\phi \{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}(\mathbb{Z}∕8\mathbb{Z})\simeq {(\mathbb{Z}∕8\mathbb{Z})}^{\text{∗}}\hfill \\ \hfill j\hfill & \hfill \mapsto \hfill & {\chi }_j\{\begin{array}{ccc}\hfill \mathbb{Z}∕8\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕8\mathbb{Z}\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & 3^{j}x\hfill \end{array}\hfill \end{array}$$

is a homomorphism, which maps $0$ on ${\chi }_0={\mathrm{Id}}_{\mathbb{Z}∕8\mathbb{Z}}$ and $1$ on ${\chi }_1:x\mapsto 3x$, because $3\in {(\mathbb{Z}∕8\mathbb{Z})}^{\text{∗}}$ and $3^2=1$ in ${(\mathbb{Z}∕8\mathbb{Z})}^{\text{∗}}$.

(The others may verify the axioms of groups for the law (4), and in particular the associativity.)

Put $s=(1,0)$ and $t=(0,1)$ and $e=(0,0)=1_G$. Then $s^8=e,t^2=e$, and using (4)

$$\begin{array}{llll}\hfill \mathit{st}& =(1,0)(0,1)=(1+3^0\cdot 0,0+1)=(1,1),& \hfill & \\ \hfill t{s}^3& =(0,1)(3,0)=(0+3^1\cdot 3,1)=(9,1)=(1,1)=\mathit{st}.& \hfill & \end{array}$$

Since $s^8=t^2=e$ and $\mathit{st}=t{s}^3$, there exists by van Dyck’s Theorem a homomorphim $\xi :Q{D}_{16}\mapsto G$ such that $\xi (\sigma )=s,\xi (\tau )=t$. Therefore $16=|G|\le |Q{D}_{16}|$, where $|Q{D}_{16}|\le 16$, so $|Q{D}_{16}|=16$, and $\xi$ is an isomorphism. In conclusion

$$Q{D}_{16}\simeq \mathbb{Z}∕8Z{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z},$$

and in particular $|Q{D}_{16}|=16$.

(b)

By part (a), $$\begin{array}{llll}\hfill Q{D}_{16}& =\{1,\sigma ,{\sigma }^2,\dots ,{\sigma }^7,\tau ,\mathit{\sigma \tau },\dots ,{\sigma }^7\tau \}& \hfill & \\ \hfill & =\{1,\sigma ,{\sigma }^2,\dots ,{\sigma }^7,\tau ,\mathit{\tau \sigma },\dots ,\tau {\sigma }^7\},& \hfill & \end{array}$$

(because $\mathit{\tau H}=\mathit{H\tau }(=G-H)$, where $H=⟨\sigma ⟩$),

and the elements $1,\sigma ,{\sigma }^2,\dots ,{\sigma }^7,\tau ,\mathit{\tau \sigma },\dots ,\tau {\sigma }^7$ are distinct.

• Since ${\sigma }^8=1$, $⟨\sigma ⟩=\{1,\sigma ,{\sigma }^2,\dots ,{\sigma }^7\}$, so we obtain $|⟨\sigma ⟩|=8$.

• Now

  $$⟨{\sigma }^2,\tau ⟩\supseteq \{1,{\sigma }^2,{\sigma }^4,{\sigma }^6,\tau ,\tau {\sigma }^2,\tau {\sigma }^4,\tau {\sigma }^6\},$$

  thus $|⟨{\sigma }^2,\tau ⟩|\ge 8$.

  Since ${({\sigma }^2)}^4={\tau }^2=1$ and $\tau {\sigma }^2={({\sigma }^2)}^3\tau$, by van Dyck’s Theorem, there is a surjective homomorphism $\phi :D_8=⟨r,s\mid r^4=s^2=1,\mathit{sr}=r^{3}s⟩\to ⟨{\sigma }^2,\tau ⟩$ such that $\phi (r)={\sigma }^2$ and $\phi (s)=\tau$. Therefore $8=|D_8|\ge |⟨{\sigma }^2,\tau ⟩|$, so

  $$|⟨{\sigma }^2,\tau ⟩|=8.$$

• Finally

  $$⟨{\sigma }^2,\mathit{\tau \sigma }⟩\supseteq \{1,{\sigma }^2,{\sigma }^4,{\sigma }^6,\mathit{\tau \sigma },\tau {\sigma }^3,\tau {\sigma }^5,\tau {\sigma }^7\},$$

  thus $|⟨{\sigma }^2,\mathit{\tau \sigma }⟩|\ge 8.$

  Put $I={\sigma }^2$, $J=\mathit{\tau \sigma }$. Since $I^2=J^2$ and $I^{-1}\mathit{JI}=J^{-1}$, by van Dyck’s Theorem, there is a surjective homomorphism $\chi :Q_8\to ⟨{\sigma }^2,\mathit{\tau \sigma }⟩$ such that $\chi (a)=I={\sigma }^2,\chi (b)=J=\mathit{\tau \sigma }$. Therefore $8=|Q_8|\ge |⟨{\sigma }^2,\mathit{\tau \sigma }⟩|$, so

  $$|⟨{\sigma }^2,\mathit{\tau \sigma }⟩|=8.$$

This shows (1), and (1) being assumed, we have proved in the first part that the three maximal groups satisfy

$$⟨{\sigma }^2,\tau ⟩\simeq D_8,⟨\sigma ⟩\simeq Z_8,⟨{\sigma }^2,\mathit{\tau \sigma }⟩\simeq Q_8.$$

(c)

Finally, we prove that every proper subgroup of $$Q{D}_{16}=\{1,\sigma ,{\sigma }^2,\dots ,{\sigma }^7,\tau ,\mathit{\tau \sigma },\dots ,\tau {\sigma }^7\},$$

is contained in one of the subgroups $H,K,L$, where

$$\begin{array}{llll}\hfill H& =⟨{\sigma }^2,\tau ⟩& \hfill & \\ \hfill & =\{1,{\sigma }^2,{\sigma }^4,{\sigma }^6,\tau ,\tau {\sigma }^2,\tau {\sigma }^4,\tau {\sigma }^6\},& \hfill & \\ \hfill K& =⟨\sigma ⟩& \hfill & \\ \hfill & =\{1,\sigma ,{\sigma }^2,{\sigma }^3,{\sigma }^4,{\sigma }^5,{\sigma }^6,{\sigma }^7\},& \hfill & \\ \hfill L& =⟨{\sigma }^2,\mathit{\tau \sigma }⟩& \hfill & \\ \hfill & =\{1,{\sigma }^2,{\sigma }^4,{\sigma }^6,\mathit{\tau \sigma },\tau {\sigma }^3,\tau {\sigma }^5,\tau {\sigma }^7\}.& \hfill & \end{array}$$

We compute first the center $Z$ of $Q{D}_{16}$. Let $x={\sigma }^i{\tau }^j$, $(0\le i<8,0\le j<2)$. Then $x\in Z$ if and only if for all integers $k,l$, $({\sigma }^i{\tau }^j)({\sigma }^k{\tau }^l)=({\sigma }^k{\tau }^l)({\sigma }^i{\tau }^j)$, that is ${\sigma }^{i+3^{j}k}{\tau }^{j+l}={\sigma }^{k+3^{l}i}{\tau }^{l+j}$ (see equality (5) and above). Therefore

$${\sigma }^i{\tau }^j\in Z⟺\forall <!--\; FUNCTION\; APPLICATION\; -->k\in [[0,8[[,\forall <!--\; FUNCTION\; APPLICATION\; -->l\in [[0,2[[,i+3^{j}k\equiv k+3^{j}i(\mathit{mod}8).$$

If ${\sigma }^i{\tau }^j\in Z$, then $(3^l-1)i\equiv (3^j-1)k(\mathit{mod}8)$ for all $i,k$. For $l=1$ and $k=0$, this gives $2i\equiv 0(\mathit{mod}8)$, thus $i\equiv 0(\mathit{mod}4)$, where $0\le i<8$, therefore $i=0$ or $i=4$.

In both cases, since $3^l-1$ is always even, $0\equiv (3^j-1)k(\mathit{mod}8)$ for all integers $k$. In particular, for $k=1$, $3^j-1\equiv 0(\mathit{mod}8)$, where $j\in \{0,1\}$, thus $j=0$. This gives $x\in \{1,{\sigma }^4\}$.

Conversely, $1\in Z$, and since ${\sigma }^4\sigma ={\sigma }^4\sigma$, and $\tau {\sigma }^4={\sigma }^{12}\tau ={\sigma }^4\tau$, where $\sigma ,\tau$ are generators of $Q{D}_{16}$, this shows that ${\sigma }^4\in Z$. Hence

$$\begin{array}{llll}\hfill & Z(Q{D}_{16})=⟨{\sigma }^4⟩.& \hfill & \end{array}$$

Since $H\cup K\cup L=Q{D}_{16}$, every cyclic subgroup $⟨h⟩$ satisfies $h\in H$ or $h\in K$ or $H\in L$, where $H,K,L$ are subgroups, so $⟨h⟩\le H$, or $⟨h⟩\le K$ or $⟨h⟩\le L$.

Now let $S$ be any proper subgroup of $Q{D}_{16}$. Then $S$ has order $1,2,4$ or $8$.

• If $|S|\le 2$, then $S$ is cyclic, so $S\le H$ or $S\le K$ or $S\le L$ by the preceding argument.

• We suppose now that $|S|>2$, i.e., $|S|=4$ or $|S|=8$.

  We prove first that ${\sigma }^4\in S$, so that $Z\le S$.

  If ${\sigma }^4\notin S$, since ${\sigma }^4={({\sigma }^2)}^2={({\sigma }^3)}^4={({\sigma }^5)}^4={({\sigma }^6)}^2={({\sigma }^7)}^4$, we know that $\sigma ,{\sigma }^2,{\sigma }^3,{\sigma }^4,{\sigma }^5,{\sigma }^6,{\sigma }^7$ don’t belong to $S$. Put $T=⟨\sigma ⟩$. Then

  $$S\cap T=\{1\}.$$

  Since $S\cap T=\{1\}$, the map $f:S\times T\to \mathit{ST}$ defined by $f(h,k)=\mathit{hk}$ is bijective. Therefore $|\mathit{ST}|=|S||T|=8|S|>16$ . This is impossible, because $\mathit{ST}\subseteq Q{D}_{16}$, so $\mathit{ST}$ has at most $16$ elements. This contradiction shows that ${\sigma }^4\in S$, and since $Z=\{1,{\sigma }^4\}$

  $$Z\le S.$$

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

  $$\overline{{\sigma }^4}={\bar{\tau }}^2=\bar{1},\bar{\tau }\bar{\sigma }={\bar{\sigma }}^3\bar{\tau }.$$

  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

  $$Q{D}_{16}∕Z\simeq D_8.$$

  By Theorem 20 (The Fourth or Lattice Isomorphism Theorem), there is a bijection from the set of subgroups $S\le Q{D}_{16}$ which contain $Z$ onto the set of subgroups $\bar{S}=S∕Z$ of $Q{D}_{16}∕Z\simeq D_8$. There are exactly $9$ proper subgroups of $D_8$ (see Example 4), therefore there are exactly $9$ proper subgroups $S$ of $Q{D}_{16}$ which contain $⟨{\sigma }^4⟩$. We know already $9$ such subgroups (see the lattices of subgroups of $H,K,L$ in the first part), i.e.,

  $$⟨{\sigma }^4⟩,⟨{\sigma }^4,\tau {\sigma }^2⟩,⟨{\sigma }^4,\tau ⟩,⟨{\sigma }^2⟩,⟨\tau ,\sigma ⟩,⟨\tau {\sigma }^7⟩,⟨{\sigma }^2,\tau ⟩,⟨\sigma ⟩,⟨{\sigma }^2,\mathit{\tau \sigma }⟩,$$

  and all are subgroups of $H$, $K$ or $L$ (Note that the lattice of the subgroups of $D_8$ and the lattice of intermediate groups between $⟨{\sigma }^4⟩$ and $Q{D}_{16}$ are isomorphic.)

  In conclusion, every subgroup of $Q{D}_{16}$ of order $4$ or $8$ contains $⟨{\sigma }^4⟩$ and is contained in $H$, $K$ or $L$. There are no other subgroups besides those of the lattice of part (a), which is complete.