Abstract Algebra – How to Use the Lattice Isomorphism Theorem

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I am having trouble understanding some of the wordings of the Lattice isomorphism theorem (Also known as 4th isomorphism theorem) in group theory. I quote here the theorem as in Dummit and Foote

Let $G$ be a group and let $N$ be a normal subgroup of $G$. Then there is a bijection from the set of subgroups of $A$ of $G$ which contain $N$ onto the set of subgroups $ \overline{A}=A/N$ of $G/N$. In particular every subgroup of $\overline{G}$ is of the form $A/N$ for some subgroup $A$ of $G$ containing $N$.

I'm a bit lost on the part where it says "Then there is bijection from the set of ……"

  1. What exactly do they mean there is a bijection?. Do they mean this in the sense there is literally a bijection or are they using this word loosely?
  2. How can one use this particular property to prove something useful? 🙂
    My Professor used to say that this is the most natural of the isomorphism theorems but I just don't see/get it.

Can someone be kind enough to explain this better?. If you can include a typical example where this theorem is used can clear things up a bit.

Thanks

Edit

Question: Let $N$ be a order $7$ normal subgroup of $G$. Suppose that $G/N \cong D_{10}$ the dihedral group of order $10$. I want to prove that $G$ has a normal subgroup of order $35$.

Here is how I proceed. Since the given dihedral group has a normal subgroup (The cyclic subgroup of order $5$), $G/N$ must also have such a subgroup $M$. Now by the lattice isomorphism theorem, $M=A/N$ for some subgroup $A$ of $G$ containing $N$. Can I immediately conclude that $A$ is normal in $G$ here without referring to the index of $A$ in $G$?. (It turns out that $A$ in this instance has index $2$ so normal in $G$).

Best Answer

Let $S$ be the set of all subgroups $A$ of $G$ such that $N\subseteq A$. Let $T$ be the set of all subgroups of $G/N$. The claim is that there is a bijection between the sets $S$ and $T$. The use of the word bijection is the usual one: there exists a function $\psi:S\to T$ which is both injective and surjective.

Side remark: The theorem is much stronger. Each of the sets $S$ and $T$ above are in fact lattices and the bijection is a lattice isomorphism. Moreover, the isomorphism preserves and reflects normality.

Applications:

1) A normal subgroup $N\subseteq G$ is maximal (i.e., no subgroup $H$ exists with $N\subseteq H \subseteq G$) iff the index of $N$ in $G$ is a prime number.

2) Let $N$ be normal in $G$. Then $G/N$ is simple iff there exists no normal subgroups $K$ of $G$ with $N\subset K \subset G$.

A similar result holds for rings, giving the classical and very important application

3) If $I$ is an ideal in a commutative ring $R$, then $R/I$ is a field iff $I$ is maximal in $R$. This gives a very efficient way to building fields.

Answer to you later edit: Yes, normality is both preserved and reflected by the bijection. That means that if $\psi(A)=H$, then $A$ is normal in $G$ iff $H$ is normal in $G/N$. Most other properties of groups though will not be preserved or reflected in this way.

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