**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.12**

Solution:

(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 $c \equiv 0 \pmod 2$ and $b \equiv 0 \pmod 4$. Now if $c \equiv 0 \pmod 4$, then $a \equiv 0 \pmod 4$ and $(t,u) = (1,1) \in Z$. If $c \not\equiv 0 \pmod 4$, then $a \equiv 0 \pmod 2$ and $(t,u) = (x^2,-1) \in Z$.

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

## Mathis Duguin

6 Aug 2021In showing that alpha-bar extends to a homomorphism alpha, you do not verify the relation alpha((1,i))^2=alpha((1,j))^2 and in fact this isn't true, yet Q_8=

, so this would need to be verified for the extension to work. With your definition of alpha we get alpha((1,-1))=(1,r^2) but also alpha(1,-1)=(x^2,1) so that alpha isn't well defined.