TITLE: Properties about central product
QUESTION [6 upvotes]: I'm proving some basic properties of central product of 2 groups but it seems too complex to understand (even just the definition).So I hope some one can help me clear this up and solve this for me. Really appreciated.

Let $A$ and $B$ be groups. Assume $Z(A)$ contains a subgroup $Z_{1}$ and $Z(B)$ contains a subgroup $Z_{2}$ with $Z_{1}$ is isomorphic with $Z_{2}$ by the map $x_{i} \mapsto y_{i}$ for all $x_{i} \in Z_{1}$. A central product of $A$ and $B$ is a quotient
$(A \times B) / Z$ where $Z = \{(x_{i}, y_{i}^{-1})\mid x_{i} \in Z_{1}\}$ and is denoted by $A*B$. Prove that 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 $Z_{1}$. Find $|A*B|$

REPLY [13 votes]: This exercise is just working with the definition of direct product and isomorphism. You should try it again.
Verifications
Let $\pi : A \times B \to (A\times B)/Z: (a,b) \mapsto (a,b)Z$, $\epsilon_1:A \to A\times B:a \mapsto (a,1_B)$, and $\epsilon_2:B \to A \times B : b \mapsto (1_A,b)$.
The kernel of $A \xrightarrow{\epsilon_1} A\times B \xrightarrow{ \pi} (A\times B)/Z$ is $\{ a : (a,1_B) \in Z \}$. Since $Z = \{ (x_i,y_i^{-1}) : x_i \in Z_1 \}$ we just need to find the $x_i$ with $y_i = 1_B$. But $x_i \mapsto y_i$ is an isomorphism, so $x_i = 1_A$. Thus the kernel is $\{ 1_A \}$ and $A \xrightarrow{\epsilon_1} A\times B \xrightarrow{ \pi} (A\times B)/Z$ is an injection of $A$ into the central product.
Similarly $B \xrightarrow{\epsilon_2} A\times B \xrightarrow{ \pi} (A\times B)/Z$ is an injection of $B$ into the central product.
Now suppose $(a,1_B) Z = (1_A,b) Z$ in $(A\times B)/Z$. Then $(1_A,b)^{-1}(a,1_B) \in Z$, but we can calculate in direct products easily:  $(1_A,b)^{-1}(a,1_B) =  (1_A,b^{-1})(a,1_B)=(a,b^{-1})$. Now $(a,b^{-1}) \in Z$ iff $a=x_i$ and $b=y_i$ for some $x_i \in Z_1$. Hence $(a,1_B)$ is in the image of $Z_1 \xrightarrow{\epsilon_1} A\times B \xrightarrow{ \pi} (A\times B)/Z$ and $(1_A,b)$ is in the image of $Z_2 \xrightarrow{\epsilon_2} A\times B \xrightarrow{ \pi} (A\times B)/Z$. Since these maps are injections, we get that each is an isomorphism on its image, so that the intersection is isomorphic to $Z_1$ and to $Z_2$.
Finding the order is now standard: $|(A\times B)/Z| = | A\times B| / |Z| = |A| \times |B|/ |Z_1|$.
Examples
Sometimes it is nice to understand a definition more intuitively. $D_8$ is a neat group: very small center (so not very abelian) but also very small derived subgroup (so nearly abelian). Can we make a bigger group with these properties? Can we find $G$ with $Z(G)=[G,G]$ of order 2?
Well an idea that almost works is $D_8 \times D_8$. We still get $Z(G) = [G,G]$, but unfortunately we get $Z(G) \cong C_2 \times C_2$ has order 4. Why do we need two copies of $C_2$ in the center? Can't we get rid of one?
In general, “getting rid of” something in a group means quotienting out by a subgroup that says it is the identity. The nice thing about the center, is that every subgroup is normal, so we get to choose which subgroup to quotient out by.
Why don't we just quotient out by the center of one of the $D_8$s and leave the other? That should leave us with $Z(G)$ order 2 right? Let's check: $$(D_8 \times D_8) / ( Z(D_8) \times 1 ) \cong (D_8/Z(D_8)) \times D_8 \cong V_4 \times D_8$$ does have $[G,G] = 1 \times Z(D_8)$ of order 2, but it has center of order 8, oh no!
We can't quotient out by $1 \times Z(D_8)$ either, lest the same thing happen with left and right switched. If only there was another subgroup of order 2 contained in $Z(D_8) \times Z(D_8)$. What's that you say? There is?! Yes, a diagonal subgroup works very nicely $Z=\{ (x,x) : x \in Z(D_8) \}$. Now check that $G=(D_8 \times D_8) / Z$ satisfies $[G,G]=Z(G)$ has order 2, but the group itself has grown for order 8, to order 32, yay!
In fact we can do this again, and form the central product of our order 32 group with another $D_8$ to get an order 128 group with derived subgroup and center of order 2. These are called extraspecial groups.
Textbook presentations
Ranked in order I suspect they are useful. In particular, Suzuki's book is wonderful, and Doerk–Hawkes is brief and exhaustive.

Suzuki's Group Theory: 2.4.15 (page 137, vol1) easy definition, 2.4.16 internal external equivalence, 2.4.17 more precise external, 2.4 Example (page 139, vol1) the Kronecker product AND non-uniqueness of central product decomposition, 4.4.16 (page 68, vol2) classification of the groups with $\Phi(G)$ of order $p$ using central products, 4.4.18 classification of extraspecial groups.
Doerk–Hawkes's Finite Soluble Groups: A.19.3 (page 74) is internal, A.19.5 is external, A.19.6 is sylows, A.19.7 is quotients, A.19.8 is upper and lower central series. Then A.20 is extraspecial groups, A.20.5 (page 79) being the  standard characterization.
Gorenstein's Finite Groups: Theorem 2.5.3 (page 29) internal central product, theorem 3.7.2 (page 102) external central product equivalence, Theorem 5.5.2 (page 204) characterizes extraspecial groups.
Huppert's Endliche Gruppen: Satz I.9.10 (page 49) und Aufgabe I.44a. Extraspecial groups are Satz III.13.8 page 355.
Robinson's Course in the Theory of Groups: 5.3.8 page 141. Just rephrasing previous theorems in new language.

I verified Hall, Rose, Rose, Scott, and Wehrfritz don't discuss them.