Pascal's Triangle and Its Relatives

Background

A Quick Sketch of Some Group Theory

A subset H of a group G is a subgroup if it is itself a group, under the operation of G.

Consequently, H must contain the identity element, the inverse of each of its elements, and must be closed.

For example, it is easy to see that H1 = {0, 3} and H2 = {0, 2, 4} are subgroups of Z6.

Does Z6 have other subgroups?

Of course, {0} and Z6 itself are subgroups.

In fact every group G has the corresponding subgroups, the identity element and all of G.

Any other subgroups are called proper subgroups, so the question is does Z6 have any proper subgroups in addition to H1 and H2?

The answer, no, is fairly easy to see, especially using Lagrange's Theorem: if H is a subgroup of a finite group G, then the order of H must be a divisor of the order of G.

The divisors of 6 are 1, 2, 3, and 6, so proper subgroups must have order 2 or 3.

A subgroup of order 2 contains only one element in addition to the identity element. In Z6 the only element that is its own inverse is 3, so the only subgroup of order 2 is {0, 3}.

Because 2 and 4 are inverses of one another, a subgroup H of order 3 that is different from H2 must not contain 2 or 4. Suppose H contains 1. Closure implies H contains 1 + 1 = 2, impossible if H is different from H2. Suppose H contains 5. Closure implies H contains 5 + 5 which is congruent to 4 (mod 6), impossible if H is different from H2. This leaves the only possible elements of H as 0 and 3, but we have seen these form the group H1, of order 2.

Another easy consequence of Lagrange's theorem is that groups of prime order have no proper subgroups.

Return to Some group theory.