TITLE: Explanation/How to use the Lattice isomorphism theorem
QUESTION [8 upvotes]: 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 ......"

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?
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 $G$).

REPLY [5 votes]: 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 $G/N$.