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Volume C   Chapter 2
Cosets and Natural Subgroups
It is possible to combine a single object with a subset of objects in a group. Simply combine the single object with every object in the subset to get a new subset.

Example1:
Let S be a subset of the additive integers J, defined by   S = {1, 3, 6}. Then:

2 + S = {2+1, 2+3, 2+6} = {3, 5, 8},  ,   -4 + S = {-4+1, -4+3, -4+6} = {-3, -1, 2}

Example2:
Now let S be a subset of the set N   ⋅   of natural numbers with the binary operation of multiplication. Then

5S = {5(1), 5(3), 5(6)} = {5, 15, 30}

Example3:
Let S be a subset of some group G, defined by S = {a, b, c, d}. Then for any object   k in G,

kS = {ka, kb, kc, kd}

The previous example shows that all objects in   kS   have the form   ks &nbps;. So, for an object x to be in   kS   x must equal something with this form. The following expresses this idea as a definition:

[1] Let S be a subset of some group   G   and   k   any object in G. Then

for any x in G,   x is in kS   if and only if   x = ks,  for some s in S.

More useful situations occur if the subset is actually a subgroup. If H denotes a subgroup of G, then   kH   is called a left coset of H. In these discussions   k   is called the coefficient before the subgroup H.

If the coefficient before the subgroup is any object in the subgroup then the coset is equal to the subgroup.
Notation: hH = H, if h is in H.    In particular eH = H.

With no restrictions on a subgroup:

[2] (Left cosets forming partition) The collection of all left cosets of a subgroup of a group form a partition of the group.

Let x run through all the objects in group G. Let H be any subgroup. Then xH runs through all the left cosets of H. They form a disjoint covering of G. However, for different values of x, the left cosets produced may or may not be different. For example, let h, k be two different objects in H. Then hH =kH. because:

hH = H = kH.

Let H be any subgroup of a group G.

   (a) The left cosets cover G.
For any object x in G, x is in the left coset xH because x = xe and identity e is in H.

   (b) Any two left cosets are disjoint or equal.
Let cH and dH be any two left cosets of H. If they are disjoint then proof is finished.
Suppose that they are not disjoint. Then their intersection cH ∩ dH is not empty. Some object, say w, is in both left cosets.

As pointed out above, in a non-commutative group a left coset and a right coset of the same subgroup may or may not be the same.

[?] (Normal subgroups) A subgroup of a group is anormal subgroup if every left coset of the subgroup is equal to the corresponding right coset.
Notation: Let H be a subgroup of a group G. H is normal if and only if for every object x in G, xH = Hx. Example1. Let G be the non-commutative group of six objects. Let H = {e, a, a2}. Then bH = {b, a2b, ab} = Hb. The reader can verify that abH = Hab and a2b = H a2b. (Why is it not necessary to test cosets with multiplication with e,a, a2? Therefore xH = Hx and H is normal. Exmple2. This time let H = {e, b}. The reader can verify that aH ≠ Ha. So H is not a normal subset.