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## The Lattice Isomorphism Theorem

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.2

Prove all parts of the Lattice Isomorphism Theorem.

Solution:

Let $G$ be a group and $N \leq G$ a normal subgroup, with the natural projection $\pi : G \rightarrow G/N$. Let $\mathcal{A}$ denote the set of all subgroups of $G$ containing $N$ and $\mathcal{B}$ the set of all subgroups of $G/N$, and define $\Phi : \mathcal{A} \rightarrow \mathcal{B}$ by $\Phi(A) = \pi[A]$. Let $A,B \in \mathcal{A}$. Then we have the following.

(1) $\Phi$ is a bijection.

(Injective) Suppose $\Phi(A) = \Phi(B)$; then $\pi[A] = \pi[B]$. Now let $x \in A$. We have $\pi(x) = \pi(y)$ for some $y \in B$, so that $y^{-1}x \in N$. Then $x \in yN \subseteq B$, and we have $x \in B$. Similarly, if $x \in B$ then $x \in A$. Thus $A = B$, and $\Phi$ is injective.

(Surjective) Let $B \in \mathcal{B}$; we saw previously that the preimage of a subgroup under a group homomorphism is a subgroup, so that there exists some $A \in \mathcal{A}$ such that $\Phi(A) = B$.

(2) $A \leq B$ if and only if $\Phi(A) \leq \Phi(B)$.

($\Rightarrow$) If $A \leq B$, then $\pi[A] \subseteq \pi[B]$, so $\pi[A] \leq \pi[B]$. Thus $\Phi(A) \leq \Phi(B)$.

($\Leftarrow$) Suppose $\Phi(A) \leq \Phi(B)$ and let $x \in A$. Now $\pi(x) = \pi(y)$ for some $y \in B$, so that $y^{-1}x \in N$. Then $x \in yN \subseteq B$, hence $A \leq B$.

(3) If $A \leq B$, then $[B:A] = [\Phi(B):\Phi(A)]$. If $A \leq B$, then by the previous part, we have partitions $B/A$ and $\Phi(B)/\Phi(A)$ and every equivalence class has the form $bA$ or $b\Phi(A)$, respectively. Define $\alpha : B/A \rightarrow \Phi(B)/\Phi(A)$ by $$\alpha(bA) = \pi(b)\Phi(A).$$ (Well defined) If $b_1A = b_2A$, then $b_2^{-1}b_1 \in A$, so $$\pi(b_2)^{-1}\pi(b_1) \in \pi[A] = \Phi(A),$$ hence $\pi(b_1)\Phi(A) = \pi(b_2)\Phi(A)$.

(Injective) Suppose $\alpha(b_1A) = \alpha(b_2A)$. Then $\pi(b_2)^{-1}\pi(b_1) \in \pi[A]$, and we have $\pi(b_2^{-1}b_1) = \pi(a)$ for some $a \in A$. Now $a^{-1}b_2^{-1}b_1 \in N$, so that $b_2^{-1}b_1 \in aN \subseteq A$. Thus $b_1A = b_2A$.

(Surjective) $\alpha$ is surjective because $\pi$ is surjective.

Thus $\alpha$ is a bijection and the conclusion follows.

(4) $\Phi(\langle A \cup B \rangle) = \langle \Phi(A) \cup \Phi(B) \rangle$. Let $x \in \Phi(\langle A \cup B \rangle)$. Then $x = \pi(y)$ for some $y \in \langle A \cup B \rangle$, so that $y = c_1c_2 \ldots c_k$ for some $c_i \in A \cup B$. Then $$x = \pi(c_1) \pi(c_2) \ldots \pi(c_k),$$ where $\pi(c_i) \in \Phi(A) \cup \Phi(B)$. Thus $x \in \langle \Phi(A) \cup \Phi(B) \rangle$. If $x \in \langle \Phi(A) \cup \Phi(B) \rangle$, then $x = d_1d_2 \ldots d_k$ for some $d_i \in \Phi(A) \cup \Phi(B)$. Now $\Phi(A) \cup \Phi(B) = \Phi(A \cup B)$, and $d_i = \pi(c_i)$ for each $i$. Thus $x = \pi(c_1c_2 \ldots c_k)$ for some $c_i \in A \cup B$, so that $x \in \Phi(\langle A \cup B \rangle)$.

(5) $\Phi(A \cap B) = \Phi(A) \cap \Phi(B)$. If $x \in \Phi(A \cap B)$, then $x = \pi(y)$ for some $y \in A \cap B$. Now $y \in A$, so $x \in \Phi(A)$, and similarly $x \in \Phi(B)$, so $x \in \Phi(A) \cap \Phi(B)$. If $x \in \Phi(A) \cap \Phi(B)$, then $x = \pi(y) = \pi(z)$ for some $y \in \Phi(A)$ and $z \in \Phi(B)$. Now $z^{-1}y \in N$, so that $y \in zN \subseteq B$, hence $y \in A \cap B$. Thus $x \in \Phi(A \cap B)$.

(6) $A \leq G$ is normal if and only if $\Phi(A) \leq G/N$ is normal. Suppose $A$ is normal, and let $x \in G/N$. Because $\pi$ is surjective, we have $x = \pi(y)$ for some $y \in G$. Now $$x \Phi(A) = \pi(y)\pi[A] = \pi[yA] = \pi[Ay] = \pi[A]\pi(y) = \Phi(A) x,$$ so that $\Phi(A)$ is normal. Now suppose $\Phi(A)$ is normal and let $x \in G$. We have $$\pi[xAx^{-1}] = \pi(x)\Phi[A]\pi(x)^{-1} = \Phi(A) \pi[A].$$ Thus if $xax^{-1} \in xAx^{-1}$, we have $b^{-1}xax^{-1} \in N$ for some $b \in A$, so $xax^{-1} \in bN \subseteq A$. Thus $A$ is normal.