Subgroup

Let G be a Group and let S be a nonempty subset of G. The fact that G is a group will give us some of the things we need.

S is a subgroup if it satisfies two requirements:

• S is closed to the operation of G.
• S has inverses, with respect to the operation of G, for every element in S.

As a check of your understanding, can you think of one or more ways to make S “fail” for either of the two rules bulleted above?

A simple example:

• Let G be a group, G={-1, 0, +1} closed to multiplication(*).
• Let S be S={-1, +1}, which is also closed to multiplication(*).

Proving it…

The first step to determine if a set is a subgroup is to prove

Closure by building a multiplication table with the group operation so you can confirm that the operation on any two elements of your set results in an element of the set.

Associativity is a guarantee, since we had it for G. Recall that it means any elements in the group can be put through the change of parenthesis without changing the answer, and since this was true for all elements present in G, it has to be true for the elements in S.

We have to have the Identity Element. We had it for G, or else G would not have been a group. We just have to make sure that we brought it over with us when we created the set S. For both G and S, e=+1.

Our last step is to take every member of S and determine its Inverse Element and then make certain the inverse Element is present in S. The template below shows the calculation:

For +1:

(+1)(+1)=+1, so +1 is the inverse element, and it in S={+1,-1}.

For -1:

(-1)(-1)=+1, so -1 is the inverse element, and it is in S={+1,-1}.