Exercise 2.5.17 ($\{x \in M \mid x^2 = 1\} \simeq Z_2 imes Z_2$)

Question

Use the lattice of subgroups of the modular group $M$ of order $16$ to show that the set $\{x \in M \mid x^2 = 1\}$ is a subgroup of $M$ isomorphic to the Klein $4$-group (cf. Exercise 14).

Richard Ganaye · Accepted Answer

\begin{proof} 
We obtained in Exercise 14 the lattice of subgroups of $M$:

\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) {$\langle u, v^4 angle$};

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}

The set $S = \{x \in M \mid x^2 = 1\}$ is the set of elements of order $1$ or $2$ in $M$. By this lattice,
$$S  = \{1, u, uv^4, v^4\} = \langle u, v^4 angle.$$
Since $S$ has no element of order $4$, it is not cyclic, and by Exercise 10,
$$S =  \langle u, v^4 angle \simeq Z_2 	imes Z_2$$
is isomorphic to the Klein $4$-group.

\end{proof}

Exercise 2.5.17 ($\{x \in M \mid x^2 = 1\} \simeq Z_2 \times Z_2$)

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Exercise 2.5.17 ($\{x \in M \mid x^2 = 1\} \simeq Z_2 \times Z_2$)

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