Explanation/How to use the Lattice isomorphism theorem

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.



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 $G$).

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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.


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.

Imagine making a giant chart of all the subgroups in the group $G$. I believe Dummit and Foote actually has a few illustrated for you. If not, the subgroups of, say, $\mathbb{Z}/12\mathbb{Z}$ will do nicely. Now, in your diagram, draw an arrow from subgroup $H$ to subgroup $K$ exactly when $H \subset K$. It might keep things tidier if you only draw an arrow to the “next largest” subgroup. The drawing you make helps you understand the subgroup structure of your group. You see all the subgroups and how each is contained in the other.

Now, find a normal subgroup. If you’re working with $\mathbb{Z}/12\mathbb{Z}$, all of them are normal since the group is abelian. Since its normal, you can quotient and obtain a brand new group $G/N$. You can then go through and make this “subgroup chart” for this group. If you did things correctly, you can now see what the lattice isomorphism theorem is saying. The picture of the subgroup structure of $G/N$ should look exactly like all the subgroups of $G$ sitting “above” $N$. Each subgroup $K \supset N$ should correspond to a unique subgroup in $G/N$.