Exercise 3.3.2 (Proof of the Lattice Isomorphism Theorem)

Question

Prove all parts of the Lattice Isomorphism Theorem.

Richard Ganaye · Accepted Answer

Let $G$ be a group, and let $N$ be a normal subgroup of $G$. Let 
 $$\pi \
 \left\{
 \begin{array}{ccl}
  G &	o &G/N\
  x & \mapsto & xN
  \end{array}
  ight.
  $$
   be the natural projection. We know that $\pi$ is a surjective homomorphism.

\begin{itemize}
\item {\bf The map $H \mapsto H/N$ from the set of subgroups of $G$ containing $N$ to the set of subgroups of $G/N$ is bijective.}

In other words, for every subgroup $\overline{H}$ of $G/N$, there exists a unique subgroup $H$ of $G$ containing $N$ such that $\overline{H} = H/N$.
\begin{proof}
In the following diagram, the horizontal arrows are inclusion maps, and the vertical arrows are the natural projection.
\begin{center}
\begin{tikzpicture}
    
ode (A) at (4,3) {$N$};
    
ode (B) at (6,3) {$H$};
    
ode (C) at (4,1) {$\{\overline{1}\}$};
    
ode (D) at (6,1) {$H/N$};
    
ode (E) at (8,3) {$K$};
    
ode (F) at (8,1) {$K/N$};
    
ode (G) at (10,3){$G$};
    
ode (N) at (10,1){$G/N$};
    \draw[->] (A) edge (B)  ;
    \draw[->] (B) edge (E)  ;
    \draw[->] (E) edge (G)  ;
    \draw[->] (C) edge (D)  ;
    \draw[->] (D) edge (F)  ;
    \draw[->] (F) edge (N)  ;
    \draw[->] (A) edge (C);
    \draw[->] (B) edge   (D);
    \draw[->] (E) edge  (F);
    \draw[->] (G) edge  (N);
\end{tikzpicture}
\end{center}

Let $\mathscr G$ be the set of subgroups of $G$ containing $N$, and $\overline{\mathscr{G}}$ be the set of subgroups of $G/N$.

We define:
$$
\varpi
\left\{
\begin{array}{ccl}
\mathscr G & 	o & \overline{\mathscr{G}}\
H & \mapsto & \pi(H),
\end{array}
ight.
\qquad
\eta
\left\{
\begin{array}{ccl}
\overline{\mathscr{G}} & 	o & \mathscr G \
\overline{H} & \mapsto & \pi^{-1}(\overline{H}),
\end{array}
ight.
$$
where
\begin{align*}
&\pi(H) = \{\pi(x) \mid x \in H\} = \{\overline{x} \in G/N \mid \exists x \in H,\ \pi(x) = \overline{x} \},\
&\pi^{-1}(\overline{H}) = \{x \in G \mid \pi(x) \in \overline{H} \}.
\end{align*}

Note that $\pi(H) = \{xN \mid x \in H\} = H/N$, so $\varpi$ is the function defined by $\varpi(H) = H/N$.

If $\overline{H} \in \overline{\mathscr{G}}$ then $\pi(\pi^{-1}(\overline{H})) = \overline{H}$. This is a purely set-theoretic property due to the surjectivity of $\pi$:

\begin{align*}
\overline{x} \in \pi(\pi^{-1}(\overline{H})) &\iff \exists x \in \pi^{-1}(\overline{H}),\ \overline{x} = \pi(x)\
&\iff \exists x \in G,\ \pi(x) \in \overline{H} 	ext { and } \overline{x} = \pi(x)\
&\iff \overline{x} \in \overline{H},
\end{align*}
the surjectivity of $\pi$ justifying the implication ($\Leftarrow$) of the last line.

The equality $\pi(\pi^{-1}(\overline{H})) = \overline{H}$ shows that $\varpi \circ \eta= 1_{\overline{\mathscr G}}$.

Let us show, using $\pi(H) = H/N$, that for all $H\in \mathscr G $, $$(\eta \circ \varpi)(H) = \pi^{-1}(H/N) = H.$$
By definition of $\pi^{-1}(H/N)$, for all $x \in G$,
$$ x \in \pi^{-1}(H/N) \iff x N \in H/N.$$
Let's verify that:
$$x N \in H/N \iff x \in H.$$
Indeed, if $x \in H$, then $xN \in H/N$ by definition of $H/N$, and conversely, if $xN \in H/N$, where $x \in G$, then $xN = yN$ for some $y \in H$. Therefore, $x = xe \in xN = yN$, so there exists $h \in N$ such that $x = yh$. Thus, $y \in H$ and $h \in N \subseteq H$ show that $x \in H$, which completes the proof of the equivalence $x N \in H/N \iff x \in H$ for all $x \in G$.

Thus, $\eta \circ \varpi = 1_{\mathscr G}$.

We have proven that $\varpi$ is bijective, with reciprocal $\varpi^{-1} = \eta$.
\end{proof}

Now we write $\overline{H} = H/N$ the image $\pi(H)$ of any subgroup $H \leq G$, and $\overline{x} = xN$ for $x \in G$.

Suppose that $N \leq A \leq G$ and $N \leq B \leq G$, so that $A, B \in \mathscr{G}$.

\item {\bf (a) $A \leq B \iff \overline{A} \leq \overline{B}$.}
\begin{proof} Suppose that $A \leq B$. Let $\overline{x} \in \overline{A}$. Then $\overline{x} = x N$ for some $x \in A$. Since $A \subseteq B$, $x \in B$, thus $\overline{x} = x N \in \overline{B}$. This shows that $\overline{A} \leq \overline{B}$.

Conversely, suppose that $\overline{A} \leq \overline{B}$. Let $x \in A$. Then $xN \in \overline{A} \leq \overline{B}$, so $x N \in \overline{B} = B/N$. Therefore there is some $y \in B$ such that $xN = y N$. Then $x \in yN$, so there exists $n \in N$ such that $x = yN$. Since $y \in B$, and $n \in N \subseteq B$, we obtain $x \in B$. This shows that $A \leq B$. So (1) is proven.
\end{proof}
Note: In the theory of lattices, this show that the lattices $(\mathscr{G},\leq)$ and $(\overline{\mathscr{G}},\leq)$ are isomorphic. 

\item {\bf (b) If $A \leq B$, then $|B : A| = |\overline{B} : \overline{A}|$.}
\begin{proof} (We don't know if $B$ or $N$ is a finite subgroup, so we cannot use $|B/N| = |B|/|N|$.)

Suppose that $A\leq B$. Let $(a_i)_{i\in I}$ be a complete family of representatives of the left cosets of $A$ in $B$, so that $a_i \in B$ for all $i \in I$, and
\begin{align}
 &B = \bigcup_{i\in I} a_i A,\
 &\forall (i,j) \in I 	imes I, \ i 
e j \Rightarrow a_i A \cap a_j A = \varnothing.
 \end{align}
Let $\overline{x}$ be any element  of $G/N$, so that $\overline{x} = xN,\ x \in G$. If $\overline{x} \in \bigcup_{i\in I} \overline{a_i}\,  \overline{A}$, then  there is some index $i \in I$ and some $\overline{y} \in \overline{A}$ such that $\overline{x} =  \overline{a_i}\,  \overline{y}$. Since $\overline{a_i} \in \overline{B}$, and $\overline{y} \in \overline{A} \subseteq \overline{B}$ by item (a), we obtain $\overline{x} \in \overline{B}$.

Conversely, suppose that $\overline{x} \in \overline{B}$. Then $x N =zN$ for some $z \in B$, thus $x \in zN \subseteq B$, so $x \in B$. By (1), $x = a_i t$ for some $i\in I$ and $t \in A$. Then $ \overline{x} = \overline{a_i}\,  \overline{t} \in  \bigcup_{i\in I} \overline{a_i}\, \overline{A}$. This shows that
\begin{align}
\overline{B} = \bigcup_{i\in I} \overline{a_i}\,  \overline{A}.
\end{align}
Moreover, if $i 
e j$ ($i,j \in I$), we prove that $\overline{a_i}\,  \overline{A}\,  \cap\,  \overline{a_j}\,  \overline{A} = \varnothing$. If $\overline{x} \in \overline{a_i} \, \overline{A}$ and $\overline{x} \in \overline{a_j}\, \overline{A}$, then $\overline{x} = \overline{a_i}\,  \overline{z} = \overline{a_j}\,  \overline{t}$, where $z,t \in A$. Then $a_i z N = a_j t N$, thus $a_i z \in a_j t N \subseteq a_j A$. Therefore $a_i \in a_j Az^{-1} = a_j A $ (because $z \in A$), hence $a_i A = a_j A$. Then (2) implies $i = j$. So
\begin{align}
\forall (i,j) \in I 	imes I, \ i 
e j \Rightarrow \overline{a_i} \overline{A} \cap \overline{a_i} \overline{A} = \varnothing.
\end{align}
Then (3) and (4) show that $(\overline{a_i})_{i\in I}$ is a complete family of representatives of the cosets of $\overline{A}$ in $\overline{B}$. Hence
$$|\overline{B} : \overline{A} | = \mathrm{Card}(I) = |B : A |.$$
(The two are infinite if $I$ is a infinite set.)

\item {\bf (c) $\overline{\langle A, B angle} = \langle \overline{A}, \overline{B} angle$.}

(Here $\langle A, B angle = \langle A \cup B angle$ is the least upper bound (or join, supremum) $A \vee B$ of the two subgroups $A,B$ in the lattice of subgroups. Then the isomorphism $\varpi : A \mapsto \overline{A}$ of lattices preserves the least upper bound.)

More explicitly, by definition of $\langle A, B angle$, 
\begin{align}
&A \leq \langle A, B angle,\quad B \leq \langle A, B angle,\
&\forall C \in \mathscr{G},\ ( A \leq C 	ext{ and } B \leq C )\Rightarrow  \langle A, B angle \leq C.
\end{align}
Then by item (a),
\begin{align}
\overline{A} \leq \overline{\langle A, B angle},\quad \overline{B} \leq \overline{\langle A, B angle},
\end{align}
so $\overline{\langle A, B angle}$ is an upper bound of the pair $\{A,B\}$.

Moreover, suppose that some subgroup $\overline{C}$ is an upper bound of the pair $\{\overline{A}, \overline{B}\}$, so that $\overline{A} \leq \overline{C}$ and $\overline{B} \leq \overline{C}$. By item (a), $A \leq C$ and $B \leq C$, and by (6), $\langle A, B angle \leq C$, thus $\overline{\langle A, B angle} \leq \overline{C}$. This shows that
\begin{align}
\forall \overline{C} \in \overline{\mathscr{G}},\ (\overline{A} \leq \overline{C} 	ext{ and } \overline{B} \leq \overline{C}) \Rightarrow \overline{\langle A, B angle} \leq \overline{C}.
\end{align}
This shows that  $\overline{\langle A, B angle}$ is the least upper bound of the pair $\{\overline{A}, \overline{B}\}$. Since by definition of $\langle \overline{A}, \overline{B} angle$,
\begin{align}
&\overline{A} \leq \langle \overline{A}, \overline{B} angle,\quad \overline{B} \leq \langle \overline{A}, \overline{B} angle,\
&\forall \overline{} \in \overline{\mathscr{G}},\ ( \overline{A} \leq \overline{C} 	ext{ and } \overline{B} \leq \overline{C} )\Rightarrow  \langle \overline{A}, \overline{B} angle \leq \overline{C}.
\end{align}

There is some subgroup $C \in \mathscr{G}$ such that $\overline{C} = \langle \overline{A}, \overline{B} angle$. By (8) and (9), we obtain 
$$ \overline{\langle A, B angle} \leq \langle \overline{A}, \overline{B} angle.$$
Similarly, if $\overline{D} =  \overline{\langle A, B angle}$, then (7) and (10) show that
$$ \langle \overline{A}, \overline{B} angle \leq \overline{\langle A, B angle},$$
so
$$\overline{\langle A, B angle} = \langle \overline{A}, \overline{B} angle.$$
(This shows the unicity of the least upper bound of $\{A,B\}$.)
\item {\bf (d) $\overline{A\cap B} = \overline{A} \cap \overline{B}$.}

(Similarly, $A \cap B$ is the greatest lower bound of the pair $\{A,B\}$ in the lattice $\mathscr{G}$, and the isomorphism $\varpi : A \mapsto \overline{A}$ of lattices preserves the greatest lower bound.)

If $A,B \in \mathscr{G}$, the intersection $A \cap B \in \mathscr{G}$ satisfies
\begin{align}
& A \cap B \leq A,\quad A \cap B \leq B,\
&\forall C \in \mathscr{G},\ (C \leq A 	ext{ and } C \leq B) \Rightarrow C \leq A \cap B.
\end{align}
By item (a), (11) implies
\begin{align*}
\overline{A\cap B}  \leq \overline{A},\quad \overline{A\cap B}  \leq \overline{B},
\end{align*}
thus
$$\overline{A\cap B} \leq  \overline{A} \cap \overline{B}.$$

There is some subgroup $C \in \mathscr{G}$ such that $ \overline{C}  = \overline{A} \cap \overline{B}$. Then $\overline{C} \leq \overline{A}$ and $\overline{C} \leq \overline{B}$, therefore (by item (a)) $C \leq A$ and $C \leq B$, thus $C \leq A \cap B$, so $\overline{C} \leq \overline{A \cap B}$. This shows that
$$\overline{A} \cap \overline{B} \leq \overline{A \cap B}.$$
This proves $\overline{A\cap B} =  \overline{A} \cap \overline{B}$ and item (d).

\item {\bf (e) $A \unlhd G$ if and only if $\overline{A} \unlhd \overline{G}$.} 

Suppose that $A \unlhd G$. Let $\overline{a} \in \overline{A}$ and $\overline{g} \in \overline{G}$ (where $a \in A$ and $g \in G$). Then $ga g^{-1} \in A$, therefore
$$\overline{g} \overline{a} \overline{g}^{-1} = \overline{gag^{-1}} \in \overline{A},$$
so $\overline{A} \unlhd \overline{G}$.

Conversely, suppose that $\overline{A} \unlhd \overline{G}$. Let $a \in A$ and $g \in G$. Then $\overline{g}\, \overline{a} \, \overline{g}^{-1} \in \overline{A}$. Therefore $\overline{gag^{-1}} \in \overline{A}$, so $gag^{-1} N = b N$ for some $b \in A$, thus $gag^{-1} \in bN \subseteq A$, so $gag^{-1} \in A$. This proves $A \unlhd G$.
\end{proof}

\end{itemize}

Exercise 3.3.2 (Proof of the Lattice Isomorphism Theorem)

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Exercise 3.3.2 (Proof of the Lattice Isomorphism Theorem)

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