## The Lattice Isomorphism Theorem

Prove all parts of the Lattice Isomorphism Theorem.

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.

• Gobi Ree  On November 28, 2011 at 1:57 am

error: 1,2. $\pi(x)=\pi(y)$ implies $xN=yN$, so that $x \in yN \subseteq B$. (not $xy^{-1} \in N$ but $y^{-1}x \in N$)
3. $ab$ should be $bA$. And the same errors are in there. For example, $b_1 A = b_2 A$ implies $(b_2)^{-1}b_1 \in A$
5. Similarly, $z^{-1}y \in N$.
6. $\Phi(A) \pi[A]$ should be $\Phi(A)=\pi[A]$. Also the same error: From $\pi[xAx^{-1}]=\pi[A]$, we get $xax^{-1}N=bN$ for some $b \in A$, so that $xax^{-1} \in bN \subseteq A$. (not $xax^{-1}b \in N$ but $b^{-1}xax^{-1}$ \in N\$

my additional explanation: 1.(surjective) The surjectivity of $\pi$ is important. Let $\bar{A}$ be any subgroup of $G/N$. Define $A:= \pi^{-1}(\bar{A})$. Then $\pi(A)=\pi(\pi^{-1}(\bar{A})=\bar{A}$ since $\pi$ is surjective. Obviously $A \supseteq N$ so $A \in \mathcal{A}$.
6. Just using the equivalent condition is also straightforward. $A$ is normal iff $gag^{-1} \in A$ for all $g \in G$ iff $(gag^{-1})N=gNaN(gN)^{-1} \in A/N$ for all $gN \in G/N$ iff $A/N$ is normal in $G/N$.

question: In 4, is it true that $\Phi(A) \cup \Phi(B) = \Phi(A \cup B)$? Isn’t the proof necessary? Actually, I think that we only need $\Phi(A) \cup \Phi(B) \subseteq \Phi(A \cup B)$, which is quite straightforward.

ps. Is there an easy way to type your image-filed-latex-code?

• nbloomf  On November 29, 2011 at 12:28 am

I was halfway though fixing these errors when I realized I wrote this proof using the ‘right’ congruence induced by $N$ rather than the usual ‘left’ congruence, hence the consistent backwardness. I went ahead and changed the proof to use the left coset notation. (Given any subset $A \subseteq G$, we can define two relations on $G$ using $A$, one ‘left’ and one ‘right’, by $x \lambda y$ iff $y^{-1}x \in A$ and $x \rho y$ iff $xy^{-1} \in A$. These are equivalences if and only if $A$ is a subgroup (a pleasant fact to check) and are congruences (i.e. the naturally induced operator on classes is well defined) if and only if $A$ is normal. The two relations are dual, so it doesn’t matter which one we use so long as we’re consistent.)

As far as easy typing, I find that the best way to get faster at typing latex is to type lots of latex. đź™‚

• Gobi Ree  On November 28, 2011 at 2:03 am

ps2. Is there any way to edit my comments? Or is there anywhere to check the result of my comments? My elaborate comment above has some typos, and I’ll be glad if you fix them.

• nbloomf  On November 29, 2011 at 12:15 am

Unfortunately there is no way to preview comments as far as I know- although that would be an incredibly useful feature.