Exercise 3.3.6 (Modular group, the return)

Question

Let $M = \langle v,u \rangle$ be the modular group of order $16$ described in Exercise 14 of Section 2.5. Prove that $\langle v^4 \rangle$ is normal in $M$ and use the Lattice Isomorphism Theorem to draw the lattice of subgroups of $M /\langle v^4 \rangle$. Which group of order $8$ has the same lattice as this quotient? Use generators and relations for $M / \langle v^4 \rangle$ to decide the isomorphism type of this group.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

All is done in the solution of Exercise 2.5.14. I recall here the arguments.

\begin{proof} 
We have proved in Exercise 2.5.14 that the center of $M$ is $\langle v^2 angle$ (and not $\langle v^4 angle$). Since $\langle v^4 angle \leq \langle v^2 angle = Z(M)$, we know that

 $$\langle v^4 angle \unlhd M.$$

The lattice of subgroups of $M$ is given by

\bigskip
\begin{center}
\begin{tikzpicture}

ode (1)   at (-2,0) {$\langle 1angle$};

ode (2)   at (0,2) {$\langle v^4 angle$};

ode (3)   at (-2,2) {$\langle uv^4angle$};

ode (4)   at (-4,2) {$\langle u angle$};

ode (5)   at (2,4) {$\langle v^2 angle$};

ode (6)   at (0,4) {$\langle uv^2 angle$};

ode (7)   at (-2,4) {$\langle u angle$};

ode (8)   at (0,6) {$ \langle u,v^2angle $};

ode (9)   at (2,6) {$ \langle uv angle $};

ode (10)   at (4,6) {$ \langle vangle $};

ode (11)   at (2,8) {$\qquad \ \langle u,vangle = M$}; 

\draw (1) edge  (2) edge (3) edge (4) ;
\draw (7) edge  (2) edge (3) edge (4) ;
\draw (8) edge  (5) edge (6) edge (7) ;
\draw (2) edge  (5) edge (6)  ;
\draw (5) edge  (9) edge (10)  ;
\draw (11) edge  (8) edge (9) edge (10) ;
\end{tikzpicture}

\end{center}

By the Lattice Isomorphism Theorem, the lattice of subgroups of $M/ \langle v^4 angle$ is isomorphic to the lattice of subgroups of $M$ above $\langle v^4 angle$, which gives, using $\overline{v}^4 = \overline{1}$, 


\bigskip
\begin{center}
\begin{tikzpicture}

ode (2)   at (0,2) {$\langle \overline{1} angle$};

ode (5)   at (2,4) {$\langle \overline{v}^2 angle$};

ode (6)   at (0,4) {$\langle \overline{u}\overline{v}^2 angle$};

ode (7)   at (-2,4) {$\langle \overline{u} angle$};

ode (8)   at (0,6) {$ \langle \overline{u},\overline{v}^2angle $};

ode (9)   at (2,6) {$ \langle \overline{u}\overline{v} angle $};

ode (10)   at (4,6) {$ \langle \overline{v}angle $};

ode (11)   at (2,8) {$\qquad  \ \langle \overline{u},\overline{v}angle = M/\langle v^4 angle  $}; 

\draw (7) edge  (2)  ;
\draw (8) edge  (5) edge (6) edge (7) ;
\draw (2) edge  (5) edge (6)  ;
\draw (5) edge  (9) edge (10)  ;
\draw (11) edge  (8) edge (9) edge (10) ;
\end{tikzpicture}

\end{center}
We recognize in this diagram the lattice of $Z_2 	imes Z_4$.

I repeat the arguments to prove that  $M/ \langle v^4 angle \simeq Z_2 	imes Z_4$: Put $N = \langle v^4 angle$.

The cosets $\overline{u} = u  N$ and $\overline{v}= v N$ are generators of $M/N$ and since $\overline{v}^4 = \overline{1}$,
$$\overline{u}^2 = \overline{v}^4 = \overline{1},\qquad \overline{v}\, \overline{u} = \overline{u}\overline{v}.$$
Since $A = Z_2 	imes Z_4 = \langle a,b \mid a^2 = b^4 = 1,\ ab = ba angle$ (see Ex. 12), by van Dyck's Theorem, this exists a surjective homomorphism $\lambda: Z_2 	imes Z_4 	o M/N$ such that $\lambda(a) =u$ and $\lambda(b) = v$. Moreover, $|Z_2 	imes Z_4| =  |M/N| = 8$, therefore $\lambda$ is an isomorphism, so
$$M/\langle v^4 angle \simeq Z_2 	imes Z_4.$$
\end{proof}

Exercise 3.3.6 (Modular group, the return)

Answers

Comments

Exercise 3.3.6 (Modular group, the return)

Answers

Comments

Add answer