A pathconnected compact abelian matrix group H is a torus (i.e. it's isomorphic to U(1) to the n for some n).
Optional: Compact connected abelian groups are tori
Compact connected abelian groups are tori
This video is optional.
Proof
Since H is abelian, we have exp X_1 exp X_2 equals exp (X_1 + X_2) Therefore the exponential map is a group homomorphism exp from little h to H where we think of little h as R n with addition.
Here's how the proof will go:

We'll prove that exp is surjective.

The first isomorphism theorem for groups will then show that big H equals little h over ker of exp.

Then we will identify ker of exp inside little h with Z n inside R n.
Why is exp surjective? Because H is pathconnected, we know from an earlier exercise that H is generated by any neighbourhood of the identity matrix. In particular, H is generated by the image of an exponential chart. We can always add generators to a generating set to get a bigger generating set, so H is generated by exp of little h. But exp of little h is a subgroup, as we saw in the last video, so the subgroup generated by exp of little h is exp of little h itself, therefore big H equals exp of little h, and the exponential map is surjective.
We now identify the kernel of the exponential map.
Take k linearly independent vectors v_1 up to v_k and take all possible integer linear combinations of them to get a subgroup Lambda equals the set of vectors in R n of the form sum from i=1 to k of a_i v_i where the a_i are integers. A subgroup of this form is called a
Z n inside R n is a lattice, taking v_i to be the ith standard basis vector.
ker exp inside little h is a lattice of rank n (where n equals the dimension of little h).
Once we have established this, we are done. To see why, consider the basis of little h given by the v_is. This identifies little h with R n in such a way that the kernel of exp is identified with Z n. Therefore big H, which equals little h over ker exp, is isomorphic to R n over Z n, which equals (R over Z) to the n, which equals U(1) to the n.
Note that U(1) equals R over Z because we have a surjective homomorphism from R to U(1), sending theta to e to the 2 pi i theta, with kernel Z, so the first isomorphism theorem tells us U(1) equals R over Z.
The fact that the rank must be n is easy to see once you know that the kernel of exp is a lattice. If the rank were strictly less than n then we would get H isomorphic to (R over Z) to the k times R to the (n minus k), so H would fail to be compact because R to the (n minus k) is not compact unless k = n.
We will need to assume the following lemma:
A discrete subgroup of R n is a lattice.
The proof of this lemma would take us too far away from the main business of the course. You can find it in Arnold's book "Mathematical Methods of Classical Mechanics" (on page 276 in my version of the book, where he is proving the (Arnold)Liouville theorem on integrable systems).
We will just define discrete subgroups and prove that the kernel of exp is discrete.
A subgroup Gamma inside R n is
In other words, the points of Gamma are sufficiently spaced out that you can separate them by balls.
An example of something that is not discrete would be the rationals inside R: any ball (interval) contains infinitely many rational numbers.
By constrast, the integers Z inside R are discrete: an interval of length epsilon less than 1 containing an integer contains only one integer.
the kernel of exp is a discrete subgroup of little h.
The origin 0 in little h is in the kernel of exp, so let's start by finding a ball B inside little h such that B intersect ker exp equals the singleton set 0. We can take B to be the domain of an exponential chart.
Remember that we have the exponential map exp from little h to big H and that we can find neighbourhoods U inside little h of the origin and V inside big H of the identity such that exp restricted to U from U to V is bijective. In particular, exp restricted to U is injective, so ker exp intersects U only at 0 (i.e. 0 is the only possible preimage of the identity in this neighbourhood).
For a general X in ker exp, we will take B to be the set of points X + Y where Y is in U. This works because B intersect ker exp is the set of points X + Y where Y is in U and exp of (X + Y) is the identity but exp of (X + Y) equals exp X exp Y, which equals exp Y because H is abelian and X is in ker exp, so we deduce that Y is in ker exp. Since Y is in U, this implies Y = 0. Therefore B intersects ker exp only at X
Summary
This final lemma shows that the kernel of exp is a discrete subgroup of little h, so by the previous (unproven) lemma, it is a lattice (which we saw has rank n). This implies that big H equals little h over ker exp is isomorphic to U(1) to the n as required.