Exercise 2.5.14 (Lattice of subgroups of the modular group $M$ of order $16$)

Proof. To begin with, we take for granted that $M=⟨u,v\mid u^2=v^8=1,\mathit{vu}=u{v}^5⟩$ has order $16$ and has three subgroups of order $8$: $⟨u,v^2⟩$, $⟨v⟩$, and $⟨\mathit{uv}⟩$ and every proper subgroup is contained in one of these three.

Then $⟨v⟩$ and $⟨\mathit{uv}⟩$ are cyclic groups of order $8$, so $⟨v⟩\simeq Z_8$ and $⟨\mathit{uv}⟩\simeq Z_8$.

Since $u^2=e$, $|u|\le 2$. Moreover, if $u=1$, then $M=⟨v\mid v^4=1⟩\simeq Z_4$ has order $4$: this is false, so $|u|=2$. Moreover $⟨v⟩\simeq Z_8$, so $|v|=8$ and $|v^2|=4$. Note that

so $u$ and $v^2$ commute. Therefore

(If $Z_2=⟨a⟩$ and $Z_4=⟨b⟩$, then $f:Z_2\times Z_4\to ⟨u,v^2⟩$ defined by $f(a^i{b}^j)=u^i{v}^{2j}$ is an isomorphism.)

We know the lattice of the subgroups of $Z_2\times Z_4$( se Ex.12). We infer the lattice of subgroups of $⟨u,v^2⟩$ by replacing $a$ by $u$ and $b$ by $v^2$. Moreover, we know the lattice of the subgroups of cyclic groups. This gives

Then we glue together the three following lattices:

to obtain

This lattice is isomorphic to the lattice of subgroups of $Z_2\times Z_4$ given in Exercise 13. Nevertheless the groups $M$ and $Z_2\times Z_8$ are not isomorphic, because the former is not abelian, but the latter is abelian.

Now we prove the assertions of the statement, i.e. $M$ has order $16$ and contains three subgroups of order $8$: $H=⟨u,v^2⟩$, $K=⟨v⟩$, and $L=⟨\mathit{uv}⟩$ and every proper subgroup is contained in one of these three.

• By definition of $M$,

  $$\begin{array}{lll}\hfill u^2=v^8=1,\mathit{vu}=u{v}^5.& & \hfill \text{(1)}\end{array}$$

  By the relations (1), every element of $M$ is of the form $v^i{u}^j,0\le i<8,0\le j<2$, so $M$ has at most $16$ elements. We search how to multiply two such elements.

  We show by induction that $v^{k}u=u{v}^{5k}$. This is true for $k=0$. If we assume that $v^{k}u=u{v}^{5k}$ for some integer $k$, then

  $$v^{k+1}u=v{v}^{k}u=\mathit{vu}{v}^{5k}=u{v}^5{v}^{5k}=u{v}^{5(k+1)}.$$

  The induction is done, which proves that for all integers $k\ge 0$, $v^{k}u=u{v}^{5k}$. By multiplying this equality by $u$ on the left and on the right, we obtain $u{v}^k=v^{5k}u$, so $v^{-5k}u=u{v}^{-k}$. This shows that for all $k\in \mathbb{Z}$,

  $$\begin{array}{lll}\hfill v^{k}u=u{v}^{5k},u{v}^k=v^{5k}u.& & \hfill \text{(2)}\end{array}$$

  If $j=1$, then

  $$\begin{array}{llllll}\hfill (v^i{u}^j)(v^k{u}^l)& =v^{i}u{v}^k{u}^l& \hfill & & \hfill & \\ \hfill & =v^i{v}^{5k}u{u}^l& \hfill (\text{by (2)})& & \hfill & & \hfill \\ \hfill & =v^{i+5k}{u}^{1+l},& \hfill & & \hfill & \end{array}$$

  and if $j=0$,

  $$\begin{array}{llll}\hfill (v^i{u}^j)(v^k{u}^l)& =v^i{v}^k{u}^l& \hfill & \\ \hfill & =v^{i+k}{u}^l.& \hfill & \end{array}$$

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

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

  This suggests considering the semi-direct product $\mathbb{Z}∕8\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$, where

  $$\phi \{\begin{array}{ccc}\hfill \mathbb{Z}∕2\mathbb{Z}\hfill & \hfill \to \hfill & \mathrm{Aut}(\mathbb{Z}∕8\mathbb{Z})\simeq {(\mathbb{Z}∕8\mathbb{Z})}^{\text{∗}}\hfill \\ \hfill j\hfill & \hfill \mapsto \hfill & {\phi }_j\{\begin{array}{ccc}\hfill \mathbb{Z}∕8\mathbb{Z}\hfill & \hfill \to \hfill & \mathbb{Z}∕8\mathbb{Z}\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & 5^{j}x\hfill \end{array}\hfill \end{array}$$

  Since $g.c.d.(5,8)=1$, ${\phi }_j$ is an automorphism of $\mathbb{Z}∕8\mathbb{Z}$, and since $5^2\equiv 1(\mathit{mod}8)$, $\phi$ is a homomorphism. The law on the group $P=\mathbb{Z}∕8\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$ is given by

  $$(i,j)\cdot (k,l)=(i+5^{j}k,j+l).$$

  Put $U=(0,1)$ and $V=(1,0)$. Then $K=⟨U,V⟩$ and $U^2=V^8=1_P$. Moreover

  $$\begin{array}{lllllll}\hfill VU& =(1,0)(0,1)=(1,1),U{V}^5& \hfill =(0,1)(5,0)=(0+5^1\cdot 5,0+1)=(25,1)=(1,1),& & \hfill & & \hfill \end{array}$$

  so $VU=U{V}^5$. Since $U^2=V^8=1_K$ and $VU=U{V}^5$, where $P=⟨U,V⟩$, by van Dyck’s Theorem, there is a surjective homomorphism

  $$\chi :M=⟨u,v\mid u^2=v^8=1,\mathit{vu}=u{v}^5⟩\to P=⟨U,V⟩$$

  such that $\chi (u)=U=(0,1)$ and $\chi (v)=V=(1,0)$. Therefore $|M|\ge |P|=16$, and we know also that $|M|\le 16$, so $|M|=16$.

  Since $\chi :M\to P$ is a surjective homomorphism, where $|M|=|P|=16$, then $\chi$ is an isomorphism, so

  $$M=⟨u,v\mid u^2=v^8=1,\mathit{vu}=u{v}^5⟩\simeq \mathbb{Z}∕8\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}.$$

  In particular,

  $$|M|=16.$$

  Note that $\chi (u)=U=(0,1)$ and $\chi (v)=V=(1,0)$, where $|(1,0)|=8$ and $|(0,1)|=2$, hence $|v|=8$ and $|u|=2$.

• Since $|M|=16$,

  $$M=\{1,v,v^2,\dots ,v^7,\mathit{uv},u{v}^2,\dots ,u{v}^7\}=⟨v⟩\cup u\cdot ⟨v⟩,$$

  where all these elements are distinct.

  Moreover, $v$ has order $8$, thus

  $$\begin{array}{lll}\hfill K=⟨v⟩=\{1,v,v^2,v^3,v^4,v^5,v^6,v^7\}& & \hfill \end{array}$$

  has order $8$.

  In $\mathbb{Z}∕8\mathbb{Z}{⋊}_{\phi }\mathbb{Z}∕2\mathbb{Z}$, we obtain

  $$\begin{array}{llll}\hfill {(1,1)}^2& =(1,1)(1,1)=(1+5\cdot 1,1+1)=(6,0)& \hfill & \\ \hfill {(1,1)}^4& =(6,0)(6,0)=(12,0)=(4,0)\ne (0,0)& \hfill & \\ \hfill {(1,1)}^8& =(4,0)(4,0)=(8,0)=(0,0),& \hfill & \end{array}$$

  This sho<s that the order of $(1,1)=\chi (\mathit{uv})$ is $8$, thus $|\mathit{uv}|=8$. Therefore

  $$L=⟨\mathit{uv}⟩=\{1,\mathit{uv},v^2,u{v}^3,v^4,u{v}^5,v^6,u{v}^7\}$$

  has order $8$.

  Finally,

  $$H=⟨u,v^2⟩\supseteq \{1,v^2,v^4,v^6,u,u{v}^2,u{v}^4,u{v}^6\}=⟨u⟩\cdot ⟨v^2⟩.$$

  Since $u$ and $v^2$ commute, $⟨u⟩\cdot ⟨v^2⟩$ is a subgroup of $M$ which contains $u$ and $v^2$, thus $⟨u,v^2⟩\subseteq ⟨u⟩\cdot ⟨v^2⟩$, so $⟨u,v^2⟩=⟨u⟩\cdot ⟨v^2⟩$. This shows that

  $$H=\{1,v^2,v^4,v^6,u,u{v}^2,u{v}^4,u{v}^6\}$$

  has order $8$.

  Knowing that $|H|=|K|=|L|=8$, the arguments of the first part show that

  $$H=⟨u,v^2⟩\simeq Z_2\times Z_4,K=⟨v⟩\simeq Z_8,L=⟨\mathit{uv}⟩\simeq Z_8.$$

• It remains to show that every proper subgroup of $M$ is contained in one of these three, so that there is no other subgroup of $M$ than those given in the preceding figure.

  We first compute the center $Z$ of $M$. Let $x=v^i{u}^j$ $(0\le i<8,0\le j<2)$ be any element of $M$. Then $x\in Z$ if and only if for all integers $k,l$, $(v^i{u}^j)(v^k{u}^l)=(v^k{u}^l)(v^i{u}^j)$. By (3), this gives $v^{i+5^{j}k}{u}^{j+l}=v^{k+5^{l}i}{u}^{j+l}$. Therefore

  $$v^i{u}^j\in Z⟺\forall <!--\; FUNCTION\; APPLICATION\; -->k\in [[0,8[[,\forall <!--\; FUNCTION\; APPLICATION\; -->l\in [[0,2[[,i+5^{j}k\equiv k+5^{l}i(\mathit{mod}8).$$

  If $x=v^i{u}^j\in Z$, then $(5^l-1)i\equiv (5^j-1)k(\mathit{mod}8)$ for all $i,k$. For $l=1$ and $k=0$, this gives $4i\equiv 0(\mathit{mod}8)$, thus $i\equiv 0(\mathit{mod}2)$. For all $l\in \mathbb{Z}$, $5^l-1\equiv 0(\mathit{mod}4)$, therefore $(5^l-1)i\equiv 0(\mathit{mod}8)$, so $(5^j-1)k\equiv 0(\mathit{mod}8)$ for all $k$. In particular, for $k=1$, we obtain $5^j\equiv 1(\mathit{mod}8)$, where $j=0$ or $j=1$, thus $j=0$. This shows that $x\in \{1,v^2,v^4,v^6\}=⟨v^2⟩$, so $Z\le ⟨v^2⟩$.

  Conversely, $v^2$ commutes with $v$ and with $u$, because $v^{2}u=u{v}^{10}=u{v}^2$ where $M=⟨u,v⟩$, thus $v^2\in Z$, and $⟨v^2⟩\le Z$, so

  $$Z(M)=⟨v^2⟩.$$

  We have proved that

  $$\begin{array}{llll}\hfill M& =\{1,v,v^2,\dots ,v^7,\mathit{uv},u{v}^2,\dots ,u{v}^7\}=⟨u,v⟩,& \hfill & \\ \hfill H& =\{1,v^2,v^4,v^6,u,u{v}^2,u{v}^4,u{v}^6\}=⟨u,v^2⟩,& \hfill & \\ \hfill K& =\{1,v,v^2,v^3,v^4,v^5,v^6,v^7\}=⟨v⟩,& \hfill & \\ \hfill L& =\{1,\mathit{uv},v^2,u{v}^3,v^4,u{v}^5,v^6,u{v}^7\}=⟨\mathit{uv}⟩.& \hfill & \end{array}$$

  Therefore $M=H\cup K\cup L$. Thus 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 $M$. 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 $v^4\in S$, so that $⟨v^4⟩\le S$.

  If $v^4\notin S$, since $v^4={(v^2)}^2={(v^3)}^4={(v^5)}^4={(v^6)}^2={(v^7)}^4$, we may assert that $v,v^2,v^3,v^4,v^5,v^6,v^7$ don’t belong to $S$, so

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

  This implies that $|\mathit{SK}|=|S||K|=8|S|>16$. This is impossible because $\mathit{SK}\subseteq M$, where $|M|=16$. This contradiction shows that for every subgroup $S$ such that $|S|>2$,

  $$⟨v^4⟩\le S.$$

  Since $⟨v^4⟩\le ⟨v^2⟩=Z$, the subgroup $N=⟨v^4⟩$ is normal in $M$, so we may consider the quotient group $M∕N=M∕⟨v^4⟩$. The classes $\bar{u}=\mathit{uN}$ and $\bar{v}=\mathit{vN}$ are generators of $M∕N$ and since ${\bar{v}}^4=\bar{1}$,

  $${\bar{u}}^2={\bar{v}}^4=\bar{1},\bar{v}\bar{u}=\bar{u}.$$

  Since $A=Z_2\times Z_4=⟨a,b\mid a^2=b^4=1,\mathit{ab}=\mathit{ba}⟩$ (see Ex. 12), by van Dyck’s Theorem, this exists a surjective homomorphism $\lambda :Z_2\times Z_4\to M∕N$ such that $\lambda (a)=u$ and $\lambda (b)=v$, so

  $$M∕⟨v^4⟩\simeq Z_2\times Z_4.$$

  By Theorem 20 (the Fourth or Lattice Isomorphism Theorem), there is a bijection from the set of subgroups $S\le M$ which contain $⟨v^4⟩$ onto the set $\bar{S}=S∕N$ of $M∕N\simeq Z_2\times Z_4$. By Exercise 12, there are exactly $7$ proper subgroups of $Z_2\times Z_4$, therefore there are exactly $7$ proper subgroups of $M$ which contain $⟨v^4⟩$. We know already $7$ such subgroups by the first part, i.e.,

  $$⟨v^4⟩,⟨u,v^4⟩,⟨u{v}^2⟩,⟨v^2⟩,⟨u,v^2⟩,⟨\mathit{uv}⟩,⟨v⟩,$$

  and all are subgroups of $H$, $K$ or $L$.

  In conclusion, every subgroup of $M$ of order $4$ or $8$ contains $⟨v^4⟩$ and is contained in $H$, $K$ or $L$. This shows that every proper subgroup of $M$ is contained in $H$, $K$ or $L$. There are no other subgroups besides those already known in the lattice of the first part, which is complete.