I know you've been waiting for this:
a nonabelian group that is easy to understand.
The dihedral group Dn is the rotations and reflections of a regular n-gon. Picture an n-gon with thickness, similar to a coin with edges. In fact dihedral comes from the latin, meaning two sides, like heads and tails. Assign numbers to the edges of the coin, so you know what is going on. Place the coin on the table with edge #1 facing front. Rotate the coin one notch clockwise, then turn it over. Edge #1 has been shifted to the left, then flipped over to the right. Edge #2 is directly in front of you. Now reverse the operations; flip the coin and then turn it clockwise. Edge #1 still faces front after the flip, then shifts left. This is not where it was before, hence the group is nonabelian.
The subgroup of pure rotations is Zn, and it has index 2, hence it is normal in the dihedral group. All other group elements are involutions, flipping the coin over through various axes.
For n = 1, the dihedral group is Z2, like a stick standing up, flipping end over end.
For n = 2, the coin has two edges, a front edge and a back edge, and some thickness, so we can flip it over. The group consists of vertical flips cross horizontal flips. These are independent of each other, hence the group is Z2Z2. This is not a cyclic group, not the same as Z4, but it is abelian.
When the coin has 3 or more edges, it looks like a traditional n-gon, and the group is nonabelian.
Let H be a subgroup of G. Intersect H with R and find a cycle K. If K is all of H then we are done. The cycles of R are subgroups of G. The elements of such a cycle are c+x, 2c+x, 3c+x, …, where c divides n. Apply j-x, then c+x, then j-x, and get -c+x. Each cycle is normal in G.
Now assume H contains some j-x. It also contains j-c-x. In other words, j implies a coset of K in R. If H also contains i-x, where i is in some other coset, compose the two flips and get i-j+x. This is a rotation that is not in K, and that is a contradiction. Thus H is K in R, along with a coset of K acting as reflections.
This subgroup is not usually normal in G. Conjugate j-x by 1-x and get 2-j-x. If j and 2-j are in the same coset of K, then K contains 2. The index of H in G is 2, and H is normal in G. For any smaller cycle, H (with its reflections present) is not normal in G.
Subgroups based on different cosets are isomorphic. Map c+x to c+x, and map j-x to j+v-x, where v shifts from one coset to the other. Verify this map respects the action of G, and is reversible, hence it builds an isomorphism.
All the subgroups have been determined. They are, up to isomorphism, the cycles of R, and the dihedral groups based on these cycles. And if R is odd you have to throw in Z2, the reflection, which cannot be realized as a cycle in R.