The center of a group is the set of elements that commute with every other element in the group. In other words, if we have a group G, the center Z(G) is the subset of G such that for any element g in Z(G) and any element h in G, gh = hg.

In this case, we are given a group H = {[1, p, q, 0, 1, r, 0, 0, 1]: p, q, r ∈ R}, where R represents the set of real numbers. We want to determine the center of this group and show that it is isomorphic to another group.

To find the center of H, we need to find the elements that commute with every other element in H. Let's consider an arbitrary element g = [1, p, q, 0, 1, r, 0, 0, 1] in H and another element h = [1, x, y, 0, 1, z, 0, 0, 1] in H.

Using matrix multiplication, we can compute the product gh as follows:
gh = [1, p, q, 0, 1, r, 0, 0, 1] [1, x, y, 0, 1, z, 0, 0, 1]
= [1 + px, p + xy, q + y, 0, 1 + x, r + xz, 0, 0, 1].

Similarly, we can compute the product hg as follows:
hg = [1, x, y, 0, 1, z, 0, 0, 1] [1, p, q, 0, 1, r, 0, 0, 1]
= [1 + xp, x + py, y + q, 0, 1 + p, z + pr, 0, 0, 1].

For gh = hg to hold for any values of p, q, r, x, y, and z, the following conditions must be satisfied:

1 + px = 1 + xp,
p + xy = x + py,
q + y = y + q,
1 + x = 1 + p,
r + xz = z + pr.

Simplifying these equations, we get:
px = xp,
xy = py,
x = p,
xz = z.

From the first equation, we can conclude that x = 0, since otherwise, p would need to commute with all real numbers, which is not possible. Therefore, p must be equal to 0 as well.

From the second equation, we have y = 0, and from the third equation, we have q = 0.

Finally, from the fourth equation, we have r = z.

Thus, the center of H is the set of all elements of the form [1, 0, 0, 0, 1, r, 0, 0, 1], where r is a real number. This set is isomorphic to the group R, the set of real numbers under addition.

To show the isomorphism, we can define a map φ from the center of H to R as follows:

φ([1, 0, 0, 0