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Home , Cayley table, Direct product

4 Direct products

4.1 Definition and Examples Direct products are a straightforward way to define new groups from old. The simplest example is the Z2 × Z2. Writing Z2 = {0, 1}, we see that

Z2 × Z2 = {(0, 0), (0, 1), (1, 0), (1, 1)} has four elements. This can be given a structure in an obvious way. Simply define

(p, q) + (r, s) := (p + r, q + s) where p + r and q + s are computed in Z2. This is clearly a on Z2 × Z2. It is straightforward to check that this operation defines a group. Indeed, we obtain the following Cayley table: + (0, 0) (0, 1) (1, 0) (1, 1) (0, 0) (0, 0) (0, 1) (1, 0) (1, 1) (0, 1) (0, 1) (0, 0) (1, 1) (1, 0) (1, 0) (1, 0) (1, 1) (0, 0) (0, 1) (1, 1) (1, 1) (1, 0) (0, 1) (0, 0) ∼ This should look familiar: it is exactly the Cayley table for the Klein 4-group: Z2 × Z2 = V. This construction works in general.

Theorem 4.1. Let G1,..., Gn be multiplicative groups. The binary operation

(a1,..., an) · (b1,..., bn) := (a1b1,..., anbn) induces a group structure on the set G1 × · · · × Gn.

Definition 4.2. The group G1 × · · · × Gn is the direct product of the groups G1,..., Gn.

If the groups G1,..., Gn are additive, then the operation will also be written additively. It is immediate that a direct product has cardinality equal to the product of the cardinalities of G1,..., Gn. Proof of Theorem. Simply check the group axioms.

Closure Since each aibi ∈ Gi this is immediate.

Associativity Each Gi is associative: it is a short calculation to see that the associativity condition on the whole direct product is equivalent to each individual Gi being associative.

Identity If ei is the identity in Gi, then (e1,..., en) is the identity in the direct product.

−1 −1 −1 Inverse (a1,..., an) = (a1 ,..., an ).

The proof uses nothing beyond the fact that each factor Gi is a group in its own right—this should not be surprising as we have no other information to work with! In particular you should note that the individual groups Gi do not interact with each other.

1 Examples

1. Z2 × Z3 = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)} under (+2, +3). Observe that if we choose a = (1, 1), then the group can be written (in the above ) as {e, 4a, 2a, 3a, a, 5a}. Thus Z2 × Z3 is generated by a and is therefore cyclic. Being a of order 6, we necessarily ∼ have Z2 × Z3 = Z6. 2. The of vector spaces W = U ⊕ V is a more general example. Indeed in linear algebra it is typical to use direct sum notation rather than Cartesian products. For example the direct sum of n copies of the real line R is the familiar

n Rn = M R = R ⊕ · · · ⊕ R i=1

4.2 Orders of elements in direct products Z = 12 Z In 12 the element 10 has order 6 gcd(10,12) , while in 8 the element 2 has order 4. What is the order of the element (10, 2) ∈ Z12 × Z8? Simply compute the cyclic group generated by this element: n o h(10, 2)i = (10, 2), (8, 4), (6, 6), (4, 0), (2, 2), (0, 4), (10, 6), (8, 0), (6, 2), (4, 4), (2, 6), (0, 0)

The order is therefore 12. If you consider what happens to each of the entries of each pair, it should be obvious that the order is the least common multiple of the orders of the originals: 12 = lcm(6, 4). In general we have the following.

Theorem 4.3. Suppose that ai ∈ Gi has order ri for each i. Then (a1,..., an) ∈ G1 × · · · × Gn has order lcm(r1,..., rn). Proof. We simply compute the order:

k k k k (a1,..., an) = (a1,..., an) = (e1, e2,..., en) ⇐⇒ ∀i, ai = ei ⇐⇒ ∀i, ri | k

The order is the minimal positive integer k satisfying the above, which is precisely

k = lcm(r1,..., rn)

1 Example What is the order of (1, 4, 3, 2) ∈ Z4 × Z7 × Z5 × Z20? ∈ Z n Recalling that the order of x n is gcd(x,n) we see that the above elements have orders 4, 7, 5 and 10 respectively. Thus the order of (1, 4, 3, 2) is lcm(4, 7, 5, 10) = 140.

4.3 Finite(ly generated) abelian groups

We have already seen that all finite cyclic groups are isomorphic to some Zn. As we shall see in a moment, all finite abelian groups are isomorphic to direct products of these. ∼ We have already seen that Z2 × Z3 = Z6 but that Z2 × Z2  Z4. What is the pattern here? Your gut should suspect that it has something to do with the fact that 2 and 3 are relatively prime.

1 Note that (1, 4, 3, 2) is really a quadruple of equivalence classes ([1]4, [4]7, [3]5, [2]20).

2 Theorem 4.4. Zm × Zn is cyclic ⇐⇒ gcd(m, n) = 1, specifically, ∼ Zm × Zn = Zmn ⇐⇒ gcd(m, n) = 1 More generally, ∼ Zm1 × · · · × Zmk = Zm1···mk ⇐⇒ gcd(mi, mj) = 1, ∀i 6= j

r1 rk Moreover if n = p1 ··· pk is the unique prime factorization of an integer n, then ∼ Zn = Z r1 × · · · × Z rk p1 pk Proof. This is merely a corollary of Theorem 4.3. For brevity we just prove the first part: the remainder follows by induction.

If gcd(m, n) = 1, then the element (1, 1) ∈ Zm × Zn has order mn lcm(m, n) = gcd(m, n)

Hence (1, 1) is a generator of Zm × Zn, which is then cyclic. Conversely, suppose gcd(m, n) = d ≥ 2. Then the maximum order of an element (p, q) ∈ Zm × Zn is mn mn lcm(m, n) = = < mn gcd(m, n) d

It follows that Zm × Zn is not cyclic.

Example Is Z5 × Z12 × Z43 cyclic? The Theorem says yes, since no pairs of the numbers 5, 12, 43 have any common factors. It is ghastly to write out, but there are 15 different ways (up to reordering) of expressing this group! ∼ ∼ ∼ ∼ Z2580 = Z3 × Z860 = Z4 × Z645 = Z5 × Z516 = Z43 × Z60 ∼ ∼ ∼ = Z12 × Z215 = Z15 × Z172 = Z20 × Z129 ∼ ∼ ∼ = Z3 × Z4 × Z215 = Z3 × Z5 × Z172 = Z3 × Z20 × Z43 ∼ ∼ ∼ = Z4 × Z5 × Z129 = Z4 × Z15 × Z43 = Z5 × Z12 × Z43 ∼ = Z3 × Z4 × Z5 × Z43 Direct products of cyclic groups have a universal application here. As the next Theorem shows, every finitely generated is isomorphic to a direct product of cyclic groups. Theorem 4.5 (Fundamental Theorem of finitely generated Abelian groups). Every finitely generated abelian group is isomorphic to a group of the form

Z r1 × · · · × Zprn × Z × · · · × Z p1 n where the pi are (not necessarily distinct) primes, the ri are positive integers and there are finitely many factors of Z. The proof is far too difficult for this course. Its purpose here is to allow us to classify finite abelian groups up to . Recall the optional section on generating sets to remind yourself of what finitely generated means: the above direct product is only a finite group if it has no factors of Z.

3 Example Find, up to isomorphism, all abelian groups of order 450. First note that 450 = 2 · 32 · 52. Now apply the fundamental theorem to see that the complete list is ∼ 1. Z450 = Z2 × Z32 × Z52

2. Z2 × Z3 × Z3 × Z52

3. Z2 × Z32 × Z5 × Z5

4. Z2 × Z3 × Z3 × Z5 × Z5

2 Theorem 4.6. If m is a square free integer (@k ∈ Z≥2 such that k | m) then there is only one abelian group of order m (up to isomorphism).

Proof. By the Fundamental Theorem such a group G must be isomorphic to some Z r1 × · · · × Zprn p1 n r1 rn with m = p ··· pn . But m being square free implies that every exponent si is equal to 1 and all the 1 ∼ primes pi are distinct. By Theorem 4.4 we have G = Zp1···pn = Zm.

All groups of small order As examples of the above we list all the groups of orders 1 through 15 and the abelian groups of order 16 up to isomorphism. The Fundamental Theorem gives us all abelian groups. In particular observe where Theorem 4.6 applies. There are three groups we haven’t previously encountered: the Q8, the A4, and the Dicyclic group Dic3. We will consider the alternating group properly later, the others you can look up if you’re interested. There are nine non- abelian groups of order 16 up to isomorphism, six of which we have no notation for in this class! You may be suspicious from looking at the table that there are no non-abelian groups of any odd order. This is not so, but you need to go to order 21 before you find one.

Order Abelian Non-Abelian 1 Z1 2 Z2 3 Z3 ∼ 4 Z4, V = Z2 × Z2 5 Z5 ∼ ∼ 6 Z6 = Z2 × Z3 D3 = S3 7 Z7 ∼ 8 Z8, Z2 × Z4, Z2 × Z2 × Z2 D4, Q8 = Dic2 9 Z9, Z3 × Z3 ∼ 10 Z10 = Z2 × Z5 D5 11 Z11 ∼ ∼ 12 Z12 = Z3 × Z4, Z2 × Z6 = Z2 × Z2 × Z3 D6, A4, Dic3 13 Z13 ∼ 14 Z14 = Z2 × Z7 D7 ∼ 15 Z15 = Z3 × Z5 16 Z16, Z4 × Z4, Z2 × Z8, Z2 × Z2 × Z4, Z2 × Z2 × Z2 × Z2 Many

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