If H is a nonempty subset of the group G, H is a subgroup iff every a and b in H implies a/b in H. The forward direction is obvious, so assume the latter. Since H contains a/a, it includes the identity. Since H includes 1/a, we have inverses. Finally H contains ab, via a/(1/b), hence * does not map H to anything outside of H.
Let H be a substructure inside G. Shifted copies of H are called cosets of H, and they cover all of G without overlap. For example, let G be the xy plane and let H be the x axis. For every real number c, the line running parallel to the x axis, through y = c, is a coset of the x axis. These cosets cover the entire plane without overlap.
The quotient G/H is the collection of cosets. In this example, the quotient is really the y axis.
With this in mind, let's build cosets of the subgroup H inside the group G.
Given a subgroup H, the right coset of an element x is the set of elements H*x.
Left cosets are defined similarly, i.e. x*H.
(Note, this convention is reversed for left modules and left ideals,
an annoying inconsistency that causes confusion for anyone who works with both nonabelian groups and noncommutative rings.)
If a and b are in H, and ax = bx, then cancel x, and a = b. All the elements in a coset are distinct. There are |H| of them.
In fact the elements of any given coset, left or right, correspond 1-1 with the elements of H.
Thanks to the identity in H, x is always in the coset of x (reflexivity). Since we can always multiply by the inverse of an element in H, x is in the coset of y iff y is in the coset of x (symmetry). If z is in the coset of y is in the coset of x, z is in the coset of x (transitivity). In other words, y = ax and z = by, hence z = bax. We have an equivalence relation, and G can be partitioned into well defined cosets. These cosets have the same size, or cardinality, namely |H|.
Let H be a subgroup of G.
If x*H = H*x for every x in G, then H is a normal subgroup.
In other words, the left and right cosets coincide.
This does not mean x commutes with the elements of H,
it only means x*H and H*x produce the same set.
Of course, if H does commute with every x in G,
then H has to be normal.
Every subgroup of an abelian group is
An equivalent definition says H is normal iff x*y/x is in H for every x in G and every y in H. If left and right cosets are the same, then xy = zx for some z in H, hence xy/x = z, a member of H. Conversly, if xy/x is always some z in H, then xy is the same as zx, a member of the right coset. We can run the other direction by replacing x with x inverse. Multiply by x on the left, giving yx = xz. Each member of the right coset is a member of the left coset, and H is normal.
A simple group has no normal subgroups, other than 1 and itself. This may remind you of the definition of a prime number. If G has no proper subgroups at all, it has no normal subgroups, and is simple. Zp, for p prime, is an example, having no subgroups by Lagrange's theorem. An abelian group must be free of subgroups to be simple.