Question

I just read on Wikipedia about elementary abelian groups which appear to be related to bit fields. I’d be grateful if someone could explain me this particular paragraph as I strive to fully master bit fields.

Answer 1

The group Z/2Z is the set {0,1} together with the binary operation + that works as follows:

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0

In that paragraph, the author refers to the group (Z/2Z)^n, which is just an ordered n-tuple of bits:

(b_1, b_2, ..., b_n)

where b_i = 0 or 1, and the binary operation + is taken coordinate-wise so that

(b_1, b_2, ..., b_n) + (d_1, d_2, ..., d_n) = (b_1+d_1, b_2+d_2, ..., b_n+d_n)

where b_i+d_i is done as in Z/2Z.

The partial order denoted <= that is discussed is the usual order on Z/2Z given by

0 <= 1

0 <= 0
1 <= 1

The last two are reflexive. This order is extended to (Z/2Z)^n coordinatewise, so that

(b_1, b_2, ..., b_n) <= (d_1, d_2, ..., d_n)

if and only if

b_i <= d_i for every i

For example, when n=2, we get the following relations:

(0,0) <= (0,0)
(0,0) <= (0,1)
(0,0) <= (1,0)
(0,0) <= (1,1)

(0,1) <= (0,1)
(0,1) <= (1,1)

(1,0) <= (1,0)
(1,0) <= (1,1)

(1,1) <= (1,1)

Notice that (1,0) and (0,1) are incomparable meaning that neither (0,1) <= (1,0) nor (1,0) <= (0,1).