Note - The Correspondence Theorem

note group-theory

For a group $(G,\cdot )$, with normal subgroup $N⊴G$, there is a morphism $\pi :G\to G/N$ where $\pi (g)=gN$. We also have $\pi$ is a surjective morphism with kernel equal to $N$.

Question: How do subgroups of $G$, $G/N$ relate?

Let $H\le G$, $N\le H$.

$\pi (H)=\{\pi (h):h\in H\}=\{hN:h\in H\}$ $hN=\{hn:n\in N\}\subseteq H$ $N\le H$

$hN\in H/N\leftarrow N⊴H$ $H/N\subseteq G/N$ (← can see this)

Claim: $H/N\le G/N$ e.g. check stable under products:

$hN,h^{\prime }N\in H/N\subseteq G/N⟹$ $hN\cdot h^{\prime }N=(h{h}^{\prime })N\in H/N$ (← $\cdot$ is the product in $G/N$)

Recall: If $Q\le G/N$, then ${\pi }^{-1}Q=\{g\in G:\pi (g)\in Q\}$

Lemma: $N\le {\pi }^{-1}Q\le G$

Proof: …

Lemma: $Q={\pi }^{-1}Q/N$

Proof: If $gN\in Q$ then $gN=\pi (g)$ so $g\in {\pi }^{-1}Q$. Thus $gN\in {\pi }^{-1}Q/N$, so $Q\subseteq {\pi }^{-1}Q/N$. If $gN\in G/N$ and $gN\in {\pi }^{-1}Q/N$, then $gN=g^{\prime }N$ for some $g^{\prime }\in {\pi }^{-1}Q$. We have $g^{\prime }N=\pi (g^{\prime })\in Q$. Therefore, $gN=g^{\prime }N\in Q$, so ${\pi }^{-1}Q/N\subseteq Q$

The Correspondence Theorem

We have a one-to-one correspondence between subgroups of $G/N$ and subgroups of $G$ that contain $N$. $N\le H\le G\to H/N$ $N\le {\pi }^{-1}Q\le G\leftarrow Q\le G/N$

Proof

It remains to prove that if $N\le H\le G⟹{\pi }^{-1}(H/N)=H$.

Let $h\in H$, so $\pi (h)=hN\in H/N$. We have that $h\in {\pi }^{-1}(H/N)$. Let $g\in {\pi }^{-1}(H/N)⟹gN=\pi (g)\in H/N⟹\exists h\in H,g\in gN=hN⟹g\in hN⟹g-h\cdot n$ for some $n\in N⟹g\in H.\square$

Remark

Correspondences above preserve inclusions and normality. (Proofs omitted but are supposedly easy).

Question

If $f:G\to F$ is a morphism, then $f(G)\le F$. In particular, if $H\le G⟹f(H)\le F$.

If $N⊴G$, $\pi :G\to G/N$. Assume $H\le G$ does not contain $N$. This implies $\pi (H)\le G/N$. Which subgroup of $G/N$?

Answer:

$\pi (H)=HN/N$.

$HN-\{hn:h\in H,n\in N\}$ $HN\le G$ $N⊴H$

Proof

If $h\in H\subseteq G⟹\pi (h)=hN\in HN/N$, $h\in HN$ and $N⊴HN$. Therefore, $\pi (H)\subseteq HN/N$.

Conversely, if $gN\in HN/N$ ($gN=h\cdot nN=hN$) for some $h\in H,n\in N$, $gN=hN=\pi (h)⟹gN\in \pi (H).\square$

Example

$G=\mathbb{Z}$, $N=100\mathbb{Z}$, $H=40\mathbb{Z}$

What is the image of $H$ in $G/N={\mathbb{Z}}_{100}$?

Solution 1: $H=⟨40⟩\le \mathbb{Z}⟹$

$$\begin{array}{rl}\pi (H)& =⟨\pi (40)⟩,\\ & =⟨40⟩\le {\mathbb{Z}}_{100},\\ & =⟨\gcd (40,100)⟩,\\ & =⟨20⟩.\end{array}$$