Introduction to Central Products (5.1.12)

Dummit and Foote Abstract Algebra, section 5.1, exercise 12:

Let A and B be groups, and assume Z₁ ≤ Z(A) is isomorphic to Z₂ ≤ Z(B). Let this isomorphism be given by  x_i → y_i for all x_i∈Z₁. A central product of A and B is a quotient (A × B) / Z where Z = { (x_i,y_i^-1) | x_i∈Z₁ } and is denoted by A ∗ B - it is not unique due to the variability in choice of Z₁, Z₂, and the isomorphism φ between them. (Think of A ∗ B as the direct product of A and B "collapsed" by identifying each element of Z₁ with its corresponding element in Z₂.)
(a) Prove that the images of A and B in the quotient group A ∗ B are isomorphic to A and B, respectively, and that these images intersect in a central subgroup isomorphic to Z₁. Find | A ∗ B |.
(b) Let Z₄ ∗ D₈ be the central product that identifies x² and r² and let Z₄ ∗ Q₈ be the central product that identifies x² and -1. Prove that Z₄ ∗ D₈ ≅ Z₄ ∗ Q₈.

Proof: (a) Let ψ be the function mapping A to its image within the central product. It is a homomorphism by:

ψ(ac) = (ac, 1)Z = (a, 1)(c, 1)Z = (a, 1)Z*(c, 1)Z = ψ(a)ψ(c), and it is clearly surjective on its image. Prove that it is injective thus:

ψ(a) = ψ(c)
(a, 1)Z = (c, 1)Z
(c^-1a, 1) ∈ Z
(c^-1a, 1) = (x_i,y_i^-1), for some x_i∈Z₁,
c^-1a = x_i and y_i^-1 = 1, the latter of which implies x_i = 1 due to the assumed isomorphism, so
c^-1a = 1
a = c

The proof for the isomorphism of B with its image in the central product is parallel.

For the second part of (a), we can prove ψ(Z₁) = ψ(A)∩ψ(B). Prove that ψ(Z₁) ≤ ψ(A)∩ψ(B) by following:

ψ(Z₁) ≤ ψ(A) clearly, and
(x_a, y_a^-1)∈Z for any x_a∈Z₁ definitionally, so
(x_a, 1)Z = (1, y_a)Z = (1, y)Z for some y∈B, so ψ(Z₁) ≤ ψ(B).

Now prove ψ(A)∩ψ(B) ≤ ψ(Z₁):

We have (a, 1)Z = (1, b)Z, so
(a, 1)(x_i,y_i^-1) = (1, b)(x_j,y_j^-1), and in particular
ax_i = x_j, so a∈Z₁, which is to say every element of ψ(A) that is also in ψ(B) is within ψ(Z₁), so ψ(A)∩ψ(B) ≤ ψ(Z₁). We thus have Z₁ ≅ ψ(Z₁) = ψ(A)∩ψ(B).

Furthermore, we have | Z | = | Z₁ | simply by observing the obvious surjective maps from Z to Z₁ and from Z₁ to Z. Thus, | A ∗ B | = | (A × B) / Z | = | A |*| B | / | Z | = | A |*| B | / | Z₁ |.

(b) Note that | (x, 1) | = 4, | (1, r) | = 4, and | (1, s) | = 2 in Z₄ ∗ D₈. We have < (x, 1), (1, r) > is a group of order 4, 8, or 16. It is not 4 as (1, r)∉< (x, 1) > and it is not 16 as 1 ≠ (1, r)² = (x, 1)². Observing that Z₄ ∗ D₈ is of order 16 by the above order formula, we have that < (x, 1), (1, r), (1, s) > is equal either to < (x, 1), (1, r) > or Z₄ ∗ D₈; it must be the latter, as (1, s)∉< (x, 1), (1, r) >.

Note that | (x, 1) | = 4, | (1, i) | = 4, and | (1, j) | = 4 in Z₄ ∗ Q₈. We have < (x, 1), (1, i) > is a group of order 4, 8, or 16. It is not 4 as (1, i)∉< (x, 1) > and it is not 16 as 1 ≠ (1, i)² = (x, 1)². Observing that Z₄ ∗ Q₈ is of order 16 by the above order formula, we have that < (x, 1), (1, i), (1, j) > is equal either to < (x, 1), (1, i) > or Z₄ ∗ Q₈; it must be the latter, as (1, j)∉< (x, 1), (1, i) >.

 We have that the generators of < (x, 1), (1, r), (1, s) > all commute with each other, and (1, r)² = (x, 1)² so that Z₄ ∗ D₈ = { (x^a, n) | 0 ≤ a ≤ 3, n∈{1, r, s, rs} = D₄} (it helps to draw the element chart out), and similarly, the generators of < (x, 1), (1, i), (1, j) > all commute with each other, and (1, i)² = (x, 1)² so Z₄ ∗ Q₈ = { ((x^a, m) | 0 ≤ a ≤ 3, n∈{1, i, j, k} = Q₈ / < -1 > ≅ D₄ }. For each element of Z₄ ∗ Q₈ obtain the isomorphism to Z₄ ∗ D₈ by applying the isomorphism from Q₈ / < -1 > to D₄ to the second coordinate.

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