I had studied group theory a year ago, but still could not understand the proof involving The Correspondence theorem.

let $G$ be a group and let $N⊴G$, where $N⊴G$ indicates that $N$ is a normal subgroup of $G$. Then there is a bijection from the set of subgroups $A$ of $G$ that contain $N$ onto the set of subgroups $\overline A $=$A/N$ of $G/N.$

Can anyone explain the steps to me in:

$(1.)$ a very easy-going manner so that I understand importance of each step.

$(2.)$I don't require the proof but only how the step follow each other .

I'll be thankful if there is anyone kind enough to do this....

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    $\begingroup$ This is not the 4th isom theorem. $\endgroup$ Sep 28 '14 at 12:56
  • $\begingroup$ @MartinBrandenburg thanks I've edited the mistake.. $\endgroup$
    – coool
    Sep 28 '14 at 13:21

Ok, Here we go...

Let $Sub(G,N)$ denote the subgroups of $G$ that contains $N$, and $Sub(G/N)$ denote the subgroups of $G/N$

$\textbf{STEP 1-}$ Consider $\Phi$ : $Sub(G,N) \to$ $Sub(G/N)$, which sends $H$ in $Sub(G,N)$ to $H/N$ in $Sub(G/N)$. we would like to show this is a bijection.

$\textbf{Injectivity-}$ Let $H_1$ and $H_2$ $\in$ Sub(G,N) such that $H_1 \neq H_2$, then there exist an element $x \in H_1$but not in $H_2$ (WLG), thus $H_1/N \neq H_2/N$, and hence injective.

$\textbf{Surjectivity-}$ Let $A$ be any subgroup of $G/N$, then $A$ will have cosets as its elements, so consider $S$={$g \in G: gN \in A$}. Then by one step subgroup test, you can check $S$ is a subgroup of $G$ and obviously $N \subseteq S$ , then consider $\Phi(S)$. It is $A$. And hence $\Phi$ is also onto.

And thus $\Phi$ is a bijection.

There is also correspondence between Normal subgroups of $G/N$ and normal subgroups of $G$ containing $N$, for that you must prooced by showing that $\Phi$ preserves normality, i.e. if $H_1$ , $H_2$ $\in$ $Sub(G,N)$ such that $H_1 \unlhd H_2$ then $\Phi(H_1)\unlhd \Phi(H_2)$. If you are further Intrested in this, try it out, and if can't get it, comment here, I will edit and upload in the same answer.

Thanks. Hope it helps.

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    $\begingroup$ Although intutively I'll go with lhf's answer but your stepwise proof is mathematically very explanatory .Thanks...+1 for it!! $\endgroup$
    – coool
    Oct 2 '14 at 13:10

Perhaps it is clearer if you consider the (apparently) more general setting:

Let $\phi: G \to G'$ be a homomorphism of groups. Then $\phi$ induces two maps on the sets $ \Gamma$ and $\Gamma'$ of subgroups of $G$ and $G'$ respectively.

$\phi_* : \Gamma \to \Gamma'$ given by $\phi_*(H)=\phi(H)$

$\phi^* : \Gamma' \to \Gamma$ given by $\phi^*(H')=\phi^{-1}(H')$

These maps preserve inclusions but they are not the inverse of each other because $\phi_*(H)$ is always a subgroup of the image of $\phi$ and $\phi^*(H')$ always contains the kernel of $\phi$.

However, when restricted to the set of subgroups of $G$ that contain the kernel of $\phi$ and the set of subgroups of $G'$ that are contained in the image of $\phi$, then these maps are the inverse of each other. This is the correspondence theorem.

More generally, $\phi_*\phi^*(H') = H' \cap \mbox{im}\ \phi$ and $\phi^*\phi_*(H) = \langle H, \ker \phi \rangle$, the smallest subgroup of $G$ that contains $H$ and $\ker \phi$; this happens to be $ H \ker \phi$.

  • $\begingroup$ thanks for answer ,but can you please make it more clearer to me this:"These maps preserve inclusions" $\endgroup$
    – coool
    Oct 2 '14 at 12:39
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    $\begingroup$ @coool, $H_1 \subseteq H_2 \implies \phi_*(H_1) \subseteq \phi_*(H_2)$ etc. $\endgroup$
    – lhf
    Oct 2 '14 at 12:40
  • $\begingroup$ I liked your approach .but just I last thing make me clear with is that what do you mean by: ϕ induces two maps .In general can you please explain what does a function inducing another map mean: $\endgroup$
    – coool
    Oct 2 '14 at 12:45
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    $\begingroup$ @coool, it just means that you can define $\phi_*$ and $\phi^*$ using $\phi$. $\endgroup$
    – lhf
    Oct 2 '14 at 12:47
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    $\begingroup$ @spectraa, see the last sentence in the answer. $\endgroup$
    – lhf
    Oct 8 '14 at 14:19

We need to find some function that will map the subgroups of $G$ containing $N$ to the subgroups of $G/N$. Since we are comparing $G$ and $G/N$, there is a homomorphism that should immediately jump to mind.

Consider the quotient map $$\pi: G \to G/N\\g \mapsto gN$$ which is well-defined as $N$ is normal.

When considering normal subgroups and quotient groups, this is the first homomorphism you should think of. It is the most natural way of getting from $G$ to $G/N$.

The idea of the proof is to show that this map will take the subgroups of $G$ containing $N$ to the subgroups of $G/N$.

The subtle point that will make this work is that homomorphisms and their inverse maps preserve subsets - that is if $\theta$ is a homomorphism and $H\subset G$, then $\theta(H) \subset \theta(G)$.

Hence, if $A$ is a subgroup containing $N$, then not only will its image will be a subgroup of $G/N$, but also the inverse map $\pi^{-1}$ taking $A/N$ back into $G$ must contain $\pi^{-1}(e) = N$.

We will make this concrete by describing an explicit bijection.

Suppose $A<G$ is a subgroup containing $N$. Then the image $\pi(A)$ of $A$ under this map will be a subgroup of $G/N$.

Define a function (not a homomorphism): $$f: \{\text{subgroups of $G$ containing $N$}\}\to \{\text{subgroups of $G/N$}\}\\A \mapsto\pi(A)$$

We wish to show that this map is a bijection. The easiest way to do this is to show that it has an inverse map $f^{-1}$.

Define a map:$$g:\{\text{subgroups of $G/N$}\}\to \{\text{subgroups of $G$ containing $N$}\}\\A/N\mapsto\pi^{-1}(A/N)$$where $\pi^{-1}(A/N)=\{g \in G:\pi(g) \in A/N\}$.

We want to show that this map is the inverse of $f$, but we first need to check that it's well-defined and that it does actually take subgroups of $G/N$ to subgroups of $G$ containing $N$

If $A/N=B/N$, then $$\begin{align}g(A/N) &= \pi^{-1}(A/N) \\&=\{g \in G:\pi(g) \in A/N\}\\&=\{g \in G:\pi(g) \in B/N\} \text{ since $A/N = B/N$}\\&=g(B/N)\end{align}$$as required.

Now suppose $\overline A$ is a subgroup of $G/N$, and consider $A=g(\overline A)$.

We want to show that $A$ is a subgroup of $G$ containing $N$.

We have $g(\overline A) = \{g \in G:gN \in \overline A\}$.

Since $\pi(n) = e$ $\forall n \in N$, certainly $N \subset g(\overline A)$.

Since $e \in N$, certainly $e \in g(\overline A)$.

And if $g, h \in g(\overline A)$, then $gN, hN \in \overline A$ so $ghN \in \overline A$ as $\overline A$ is a group, and hence $gh \in g(\overline A)$.

So $g(\overline A)$ is a subgroup of $G$ containing $N$ as required.

We now finish by showing that $f$ and $g$ are inverses of each other, and hence that $f$ is the required bijection. This is more or less immediate because the definition of $g$ is the inverse map of $f$. Most of the work was in showing that $g$ is well defined.

Let $A$ be a subgroup of $G$ containing $N$. Then $$\begin{align}g\circ f(A) &= g\big(\{aN:a \in A\}\big)\\&=\{g \in G : gN = aN \text{ for some } a \in A\}\\&\supset A \end{align}$$

So far we only have enough to tell us that $A \subset g\circ f(A)$. For equality, we need the fact that $N \subset A$.

If $b \notin A$ and $bN = aN$ for some $a \in A$, then $ba^{-1}N = N$, so $ba^{-1} \in N$.

But $N \subset A$, so $ba^{-1} \in A$ and hence $b \in A$ since $A$ is a group - a contradiction.

Similarly $f \circ g(\overline A) = \overline A$


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