Direct products Direct products, visually IfA andB are groups, there is a natural group structure on the set One way to think of the direct product of two cyclic groups, sayZ/n Z/m: A B= (a,b) a A,b B . × × { | ∈ ∈ } Imagine a slot machine with two wheels, one with spaces numbered 0 through n 1, and the other with spaces numbered 0 throughm 1. − − Definition The actions are: spin both of wheels. Each action can be labeled by where we The direct product of groupsA andB consists of the set A B, and the group end up on each wheel, say (i,j). × 2 operation is done componentwise: if (a,b),(c,d) A B, then An example for a more general case: the element (r , 4) inD 4 Z/6. ∈ × × (a,b) (c,d)=(ac,bd). ∗ e 0 We callA andB the factors of the direct product. f 5 1 Note that the binary operations onA andB could be different. One might be and r3 r3f rf r ∗ the other +. r2f 4 2 For example, inD 3 Z 4: 2 × r 3 (r 2, 1) (fr, 3) = (r 2fr, 1 + 3) = (rf, 0). ∗ These elements do not commute: Key idea (fr, 3) (r 2, 1) = (fr 3, 3 + 1) = (f, 0). The direct product of two groups joins them so they act independently of each other. ∗ Sec 3.4 Direct products Abstract Algebra I 3 / 10 Sec 3.4 Direct products Abstract Algebra I 4 / 10 Cayley diagrams of direct products Cayley diagrams of direct products Remark Lete A be the identity ofA ande B the identity ofB. Just because a group is not written with doesn’t mean it isn’t secretly a direct × Given a Cayley diagram ofA with generators a1,...,a k , and a Cayley diagram ofB product. Example:V 4 (rectangle puzzle) is reallyC 2 C 2 (two light switches). with generators b1,...,b �, we can create a Cayley diagram forA B as follows: × × Here are some examples of direct products: Vertex set: (a,b) a A,b B . { | ∈ ∈ } Generators: (a1,e b),...,(a k ,e b) and (ea,b 1),...,(e a,b �). It is helpful to arrange the vertices in a rectangular grid. For example, here is a Cayley diagram for the groupZ/4 Z/3: × Exercise: List all subgroups ofZ/4 Z/3. × (0, 0) (1, 0) (2, 0) (3, 0) Hint: There are six. C3 C 3 C3 C 2 C2 C 2 C 2 C2 C 2 C 2 C 2 × × × × × × × Surprisingly, the groupC 3 C 2 is isomorphic to the cyclic groupC 6! The Cayley (0, 1) (1, 1) (2, 1) (3, 1) × 2 3 diagram forC 6 using generators r and r is the same as the Cayley diagram for C3 C 2 above. × (0, 2) (1, 2) (2, 2) (3, 2) Sec 3.4 Direct products Abstract Algebra I 5 / 10 Sec 3.4 Direct products Abstract Algebra I 6 / 10 Subgroups of direct products Direct products, visually We can construct the Cayley diagram of a direct product using the following Proposition “inflation” method. IfH A, andK B, thenH K is a subgroup ofA B. ≤ ≤ × × Inflation algorithm ForZ/4 Z/3, all subgroups had this form. However, this is false in general! To make a Cayley diagram ofA B from the Cayley diagrams ofA andB: × × 1. Begin with the Cayley diagram forA. Example when this is not true: Consider the groupZ/2 Z/2, which is alsoV 4. × 2. Inflate each node ofA, and place in each node a copy of the Cayley diagram for SinceZ/2 has two subgroups, the following four sets are subgroups ofZ/2 Z/2: × B. Z2 Z 2, 0 0 ,Z 2 0 = (1, 0) , 0 Z 2 = (0, 1) . × { } × { } × { } � � { } × � � 3. Remove the (inflated) nodes ofA while using the arrows ofA to connect corresponding nodes from each copy ofB. That is, remove theA diagram but One subgroup ofZ/2 Z/2 is missing from this list: (1, 1) = (0, 0), (1, 1) . × � � { } treat its arrows as a blueprint for how to connect corresponding nodes in the Practice copies ofB. What are the subgroups ofZ/2 Z/2 Z/2? 100 × × 101 000 Here is a Cayley diagram, writing the elements of the 001 110 product as abc rather than (a,b,c). 111 010 Hint: There are 16 subgroups! Check with Group Explorer. 011 Cyclic groupZ 2 each node contains direct product a copy ofZ/4 groupZ/4 Z/2 × Sec 3.4 Direct products Abstract Algebra I 7 / 10 Sec 3.4 Direct products Abstract Algebra I 8 / 10 Properties of direct products Multiplication tables of direct products Recall the following important definition A subgroupH G is normal if xH= Hx for allx G. We denote this byH�G. ≤ ∈ Direct products can also be visualized using multiplication tables. For example, we construct the table for the direct product Z/4 Z/2: Observation × IfA andB are not trivial, the direct productA B has at least 4 normal subgroups: × 0 1 0 1 0 1 0 1 (0,0)(0,1)(1,0)(1,1)(2,0)(2,1)(3,0)(3,1) eA eB ,A e B , e A B,A B. 0 1 2 3 { } × { } ×{ } { } × × 1 0 1 0 1 0 1 0 (0,1)(0,0)(1,1)(1,0)(2,1)(2,0)(3,1)(3,0) Proof: 0 1 2 3 0 1 0 1 0 1 0 1 (1,0)(1,1)(2,0)(2,1)(3,0)(3,1)(0,0)(0,1) 1 2 3 0 1 2 3 0 1 0 1 0 1 0 1 0 (1,1)(1,0)(2,1)(2,0)(3,1)(3,0)(0,1)(0,0) 2 3 0 1 0 1 0 1 0 1 0 1 (2,0)(2,1)(3,0)(3,1)(0,0)(0,1)(1,0)(1,1) Sometimes we “abuse notation”: writeA�A B to meanA e B �A B and 2 3 0 1 × ×{ } × 3 0 1 2 1 0 1 0 1 0 1 0 (2,1)(2,0)(3,1)(3,0)(0,1)(0,0)(1,1)(1,0) writeB�A B to mean e A B�A B. 0 0 0 0 (3,0)(3,1)(0,0)(0,1)(1,0)(1,1)(2,0)(2,1) × { } × × 3 1 0 1 1 1 2 1 (Technically,A andB are not even subsets ofA B.) 1 0 1 0 1 0 1 0 (3,1)(3,0)(0,1)(0,0)(1,1)(1,0)(2,1)(2,0) × multipl. table inflate each cell to contain a copy join the little tables and element names Observation for /4 of the multiplication table of /2 to form the table for /4 /2 Z Z Z ×Z In a Cayley diagram forA B, any arrow fromA commutes with any arrow fromB. × Algebraically, this is saying that (a,e b) (e a,b)=(a,b)=(e a,b) (a,e b) for all ∗ ∗ elementsa A,b B. ∈ ∈ Sec 3.4 Direct products Abstract Algebra I 9 / 10 Sec 3.4 Direct products Abstract Algebra I 10 / 10.