Exercise 3.2.2 (Lattice of subgroups of $S_3$)

Question

Prove that the lattice of subgroups of $S_3$ in Section 2.5 is correct (i.e., prove that it contains all subgroups of $S_3$ and that their pairwise joins and intersections are correctly drawn).

Richard Ganaye · Accepted Answer

\begin{proof} 
We must justify the following lattice:

\bigskip
\begin{center}
\begin{tikzpicture}
    
ode (n0) at (3,0) {$\langle 1 angle$};
    
ode (n1) at (1.5,1) {$\langle (2\ 3) angle$};
    
ode (n2) at (0,1) {$\langle  (1\ 3) angle$};
    
ode (n3) at (-1.5,1) {$\langle (1\ 2)angle$};
    
ode (n4) at (4.5,2) {$\langle  (1\ 2\ 3)angle$};
    
ode (n5) at (3,3) {$\langle  S_3angle$};
     \draw (n0) edge (n1) edge (n2) edge (n3) edge (n4)  ;
     \draw (n5) edge (n1) edge (n2) edge (n3) edge (n4)  ;
   \end{tikzpicture}
\end{center}

The order of the elements of $S_3$ are
$$
\begin{array}{c|c}
\sigma & |\sigma|\
\hline
() & 1\
(1\ 2) & 2\
(1\ 3) & 2\
(2\ 3) & 2\
(1\ 2 \ 3) & 3\
(1\ 3 \ 2) & 3
\end{array}
$$

By Lagrange's Theorem, a proper non trivial subgroup of $S_3$ has order $2$ or $3$. Since $2$ and $3$ are prime numbers, such a subgroup is cyclic. So a subgroup of order $2$ is generated by one of the three elements of order $2$, and these subgroups are distinct. 

A subgroup of order $3$ is generated by a $3$-cycle. Since $(1\ 3\ 2) = (1\ 2\ 3)^{-1}$, there is only one subgroup of order $3$, which is $\langle (1\ 2\ 3) angle$. Therefore there is no subgroup other than the $6$ subgroups of the preceding lattice.

\bigskip
Since a subgroup of order $2$ cannot be a subgroup of $\langle  (1\ 2\ 3)angle$ of order $3$, there are no other inclusion than those given, so the lattice is complete.

\end{proof}

Exercise 3.2.2 (Lattice of subgroups of $S_3$)

Answers

Comments

Exercise 3.2.2 (Lattice of subgroups of $S_3$)

Answers

Comments

Add answer