Basic properties of the central product of groups

Let A and B be groups. Suppose Z_1 \leq Z(A) and Z_2 \leq Z(B) are subgroups and that there exists an isomorphism \varphi : Z_1 \rightarrow Z_2. Define Z \leq A \times B by Z = \{ (x, \varphi(x)^{-1}) \ |\ x \in Z_1 \}. Note that Z is normal in A \times B since Z \leq Z(A \times B), using a previous theorem. We denote by A \ast_\varphi B the quotient (A \times B)/ Z. (In particular, A \ast_\varphi B depends on \varphi.) Think of A \ast_\varphi B as the direct product A \times B “collapsed” by identifying each element x \in Z_1 with its \varphi-image in Z_2.

  • Prove that the images of A and B in A \ast_\varphi B are isomorphic to A and B, respectively, and that these images intersect in a central subgroup isomorphic to Z_1. Find |A \ast_\varphi B|.
  • Let Z_4 = \langle x \rangle. Let D_8 = \langle r,s \rangle and Q_8 = \langle i,j \rangle as usual. Let Z_4 \ast_\varphi D_8 be the central product of Z_4 and D_8 which identifies x_2 and r^2. (I.e. Z_1 = \langle x^2 \rangle, Z_2 = \langle r^2 \rangle, and \varphi(x^2) = r^2.) Let Z_4 \ast_\psi Q_8 be the central product of Z_4 and Q_8 which identifies x^2 and -1. (I.e. Z_1 = \langle x^2, Z_2 = \langle -1 \rangle, and \psi(x^2) = -1.) Prove that Z_4 \ast_\varphi D_8 \cong Z_4 \ast_\psi Q_8.

  1. Let \pi : A \times B \rightarrow (A \times B)/Z denote the canonical projection, and identify A with A \times 1 and B with 1 \times B in A \times B.

    Let (a_1,1), (a_2,1) \in A \times 1, and suppose \pi((a_1,1)) = \pi((a_2,1)). Then (a_2a_1^{-1},1) \in Z, so that a_2a_1^{-1} = \pi(1) = 1. Thus a_1 = a_2, and hence \pi|_A is injective. Similarly, \pi|_B is injective.

    Note that the restriction \pi|_{Z_1} is also injective. Suppose (x,y)Z \in \pi[A] \cap \pi[B]; then for some a \in A and b \in B we have (x,y)Z = (a,1)Z = (1,b)Z. Thus (a,b^{-1}) \in Z; by definition, then, a \in Z_1 and b = \varphi(a). Note that for all (z,w)Z \in A \ast_\varphi B we have (z,w)Z(a,1)Z = (za,w)Z = (az,w)Z (a,1)Z(z,w)Z, so that \pi[A] \cap \pi[B] is in the center of A \ast_\varphi B. Moreover, (x,y)Z \in \mathsf{im}\ \pi|_{Z_1}. Conversely, if (z,1) ]in Z_1 \times 1, we have (z,1)Z = (z,1)(z^{-1}, \varphi(z))Z = (1,\varphi(z))Z, so that \mathsf{im}\ \pi|_{Z_1} \subseteq \pi[A] \cap \pi[B]. Then \mathsf{im}\ \pi|_{Z_1} = \pi[A] \cap \pi[B], and by the First Isomorphism Theorem, Z_1 \cong \pi[A] \cap \pi[B]. In addition, \pi[A] \cap \pi[B] is a central subgroup of A \ast_\varphi B.

    Finally, by Lagrange write |A| = n|Z_1| and |B| = m|Z_2|. Note that |Z_1| = |Z_2|. Now |A \ast_\varphi B| = |A \times B|/|Z| = nm|Z_1|^2/|Z_1| = nm|Z_1|. We may also write this equation in the form |A \ast_\varphi B| = |A| \cdot [B : Z_2] = |B| \cdot [A : Z_1].

  2. Define \overline{\alpha} : \{(x,1), (1,i), (1,j)\} \rightarrow Z_4 \times D_8 as follows: \overline{\alpha}((x,1)) = (x,1), \overline{\alpha}((1,i)) = (1,r), and \overline{\alpha}((1,j)) = (x,s). Because \{(x,1),(1,i),(1,j)\} generates Z_4 \times Q_8 and these images satisfy the relations \overline{\alpha}((x,1))^4 = \overline{\alpha}((1,i))^4 = \overline{\alpha}((1,j))^4 = 1, \overline{\alpha}((x,1))\overline{\alpha}((1,i)) = \overline{\alpha}((1,i))\overline{\alpha}((x,1)), \overline{\alpha}((x,1))\overline{\alpha}((1,j)) = \overline{\alpha}((1,j))\overline{\alpha}((x,1)), and \overline{\alpha}((1,i))\overline{\alpha}((1,j)) = \overline{\alpha}((1,j))\overline{\alpha}((1,i))^3, \overline{\alpha} extends to a homomorphism \alpha : Z_4 \times Q_8 \rightarrow Z_4 \times D_8.

    If we let \pi denote the natural projection, \pi \circ \alpha is a mapping Z_4 \times Q_8 \rightarrow Z_4 \ast_\varphi D_8. We claim that Z \leq Z_4 \times Q_8 is contained in the kernel of this mapping; it suffices to show this for the (unique) nonidentity element of Z, namely (x^2,-1). In fact, (\pi \circ \alpha)(x^2,-1) = \pi(\alpha(x^2, i^2)) = \pi(\alpha((x,1)^2(1,i)^2) = \pi(x^2,r^2) = 1. Thus Z \leq \mathsf{ker}(\pi \circ \alpha). By the remarks on page 100 in the text, there exists a unique group homomorphism \theta : Z_4 \ast_\psi Q_8 \rightarrow Z_4 \ast_\varphi D_8 such that \theta \circ \pi = \pi \circ \alpha, and in particular, \theta(tZ) = (\pi \circ \alpha)(t).

    We now show that \mathsf{ker}\ \theta = 1. Suppose \theta((t,u)Z) = 1. Then \pi(\alpha(t,u)) = 1; note that (t,u) = (x^a,i^bj^c) for some integers a,b,c, and that \alpha(t,u) = (x^{a+c},r^bs^c). Then (x^{a+c},r^bs^c) \in \{ (1,1), (x^2,r^2) \}.

    If (x^{a+c},r^bs^c) = (1,1), then c \equiv 0 mod 2 and b \equiv 0 mod 4. Now if c \equiv 0 mod 4, then a \equiv 0 mod 4 and (t,u) = (1,1) \in Z. If c \not\equiv 0 mod 4, then a \equiv 0 mod 2 and (t,u) = (x^2,-1) \in Z.

    If (x^{a+c},r^bs^c) = (x^2,r^2), then c \equiv 0 mod 2 and b \equiv 2 mod 4. If c \equiv 0 mod 4, then a \equiv 2 mod 4 and (t,u) = (x^2,-1) \in Z. If c \not\equiv 0 mod 4, then a \equiv 0 mod 4 and (t,u) = (1,1) \in Z.

    Thus \mathsf{ker}\ \theta = 1, hence \theta is injective.

    Now by part (a), we see that both Z_4 \ast_\varphi D_8 and Z_4 \ast_\psi Q_8 have order 16. Thus \theta is an isomorphism.

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