DIRECT PRODUCTS

DIRECT PRODUCTS

Given two sets, S and T , we can form their Cartesian product, S × T = {(s, t)|s S, t T }, whose elements are ordered pairs. For example, the product of the real line, R, with itself is the plane, R × R = R2. We now show how to define the product of any two groups; the underlying set of the product is the Cartesian product of the underlying sets of the original groups.

The group (G × H, ·) is called the direct product of the groups (G, Ž ) and (H , ).

Proof. All the group axioms follow from the axioms for (G, Ž ) and (H , ). The identity of G × H is (eG, eH ), and the inverse of (g, h) is (g−1, h−1).

This construction can be iterated any finite number of times to obtain the direct product of n groups.

Sometimes the direct product of two groups G and H is called the direct sum and is denoted by G H . (The direct sum of a finite number of groups is the same as the direct product. It is possible to define a direct sum and direct product of an infinite number of groups; these are different. An element of the direct product is obtained by taking one element from each group, while an element of the direct sum is obtained by taking one element from each group, but with only a finite number different from the identity.)

Example 4.31. Write down the table for the direct product of C2 with itself.

Solution. Let C2 = {e, g}, so that C2 × C2 = {(e, e), (e, g), (g, e), (g, g)}. Its table is given in Table 4.5. We see that this group C2 × C2 is isomorphic to the

since gcd(m, n) = 1, r is divisible by mn. Hence Kerf = {e}, and the image of f is isomorphic to Cmn. However, |Cmn| = mn and |Cm × Cn| = |Cm| · |Cn| = mn; hence Imf = Cm × Cn, and f is an isomorphism.

The following is an easy consequence of this result.

If m and n are not coprime, then Cmn is never isomorphic to Cm × Cn. For

(h, k) in Cm × Cn is always less than mn.

Direct products can be used to classify all finite abelian groups. It can be shown that any finite abelian group is isomorphic to a direct product of cyclic groups. For example, see Nicholson [11] or Baumslag and Chandler [25]. The results above can be used to sort out those products of cyclic groups that are isomorphic to each other. For example, there are three nonisomorphic abelian

Theorem 4.34. If (G, ·) is a finite group for which every element g G satisfies g2 = e, then |G| = 2n for some n 0, and G is isomorphic to the n-fold direct product C2n = C2 × C2 × · · · × C2.

Proof. Every element in G has order 1 or 2, and the identity is the only element of order 1. Therefore, every element of G is its own inverse. The group G is abelian because for any g, h G, gh = (gh)−1 = h−1g−1 = hg.

Choose the elements a1, a2, . . . , an G so that ai = e and ai cannot be written as a product of powers of a1, . . . , ai−1. Furthermore, choose n maximal, so that every element can be written in terms of the elements ai . If C2 is generated by g, we show that the function

which is a contradiction. Therefore, all the exponents are even, and f is injective.

Example 4.35. Describe all the group morphisms from C10 to C2 × C5. Which of these are isomorphisms?

Solution. Since C10 is a cyclic group, generated by g, for example, a morphism from C10 is determined by the image of g. Let h and k be generators of C2 and C5, respectively, and consider the function fr,s : C10 → C2 × C5 which maps g to the element (hr , ks ) C2 × C5. Then, if fr,s is a morphism, fr,s (gn) = (hrn, ksn) for 0 n 9. However, this would also be true for all integers n, because if gn = gm, then 10|n − m. Hence 2|n − m and 5|n − m and hrn = hrm and ksn = ksm.

We now verify that fr,s is a morphism for any r and s. We have

fr,s (ga gb) = fr,s (ga+b) = (h(a+b)r , k(a+b)s ) = (har , kas )(hbr , kbs )

= fr,s (ga )fr,s (gb).

Therefore, there are ten morphisms, fr,s , from C10 to C2 × C5 corresponding to the ten elements (hr , ks ) of C2 × C5.

Now

Kerfr,s = {gn|(hrn, ksn) = (e, e)} = {gn|rn ≡ 0 mod 2 and sn ≡ 0 mod 5}.

Kerfr,s contains more than one element, fr,s is not an injection and cannot be an isomorphism. By the morphism theorem,

|C10|/|Kerfr,s | = |Imfr,s |,

and if Kerfr,s = {e}, then | Im fr,s | = 10, so fr,s is surjective also. Therefore, the isomorphisms are f1,1, f1,2, f1,3, and f1,4.

GROUPS OF LOW ORDER

We find all possible isomorphism classes of groups with eight or fewer elements.

Lemma 4.36. Suppose that a and b are elements of coprime orders r and s, respectively, in an abelian group. Then ab has order rs.

Proof. Let A and B denote the subgroups generated by a and b, respectively. Since ab = ba, we have (ab)rs = ars brs = (ar )s (bs )r = es er = e. Suppose that

|A ∩ B| = 1. It follows that ak = e and b−k = e, so r divides k and s divides k. Hence rs divides k by Theorem 11, Appendix 2 (again because r and s are coprime), as required.

With this we can describe the groups of order eight or less.

Order 1. Every trivial group is isomorphic to {e}.

Order 2. By Corollary 4.11, every group of order 2 is cyclic. Order 3. By Corollary 4.11, every group of order 3 is cyclic. Order 4. Each element has order 1, 2, or 4.

Case (i). If there is an element of order 4, the group is cyclic.

Case (ii). If not, every element has order 1 or 2 and, by Theorem 4.34, the group is isomorphic to C2 × C2.

Order 5. By Corollary 4.11, every group of order 5 is cyclic. Order 6. Each element has order 1, 2, 3, or 6.

Case (i). If there is an element of order 6, the group is cyclic.

Case (ii). If not, the elements have orders 1, 2, or 3. By Theorem 4.34, all the elements in a group of order 6 cannot have orders 1 and 2. Hence there is an element, say a, of order 3. The subgroup H = {e, a, a2} has index 2, and if b / H , the underlying set of the group is then H H b = {e, a, a2, b, ab, a2b}. By Proposition 4.20, H is normal, and the quotient group of H is cyclic of order 2. Hence

is normal, bab−1 H . We cannot have bab−1 = e, because a = e. If bab−1 = a, then ba = ab, and we can prove that the entire group is abelian. This cannot happen because by Lemma 4.36, ab would have order 6. Therefore, bab−1 = a2, and the group is generated by a and b with relations a3 = b2 = e and ba = a2b. This group is isomorphic to D3 and S3.

Order 7. Every group of order 7 is cyclic.

Order 8. Each element has order 1, 2, 4, or 8.

Case (i). If there is an element of order 8, the group is cyclic.

Case (ii). If all elements have order 1 or 2, the group is isomorphic to C2 × C2 × C2 by Theorem 4.34.

Case (iii). Otherwise, there is an element of order 4, say a. The subgroup H = {e, a, a2, a3} is of index 2 and therefore normal. If b / H , the underlying set of the group is H H b = {e, a, a2, a3, b, ab, a2b, a3b}. Now b2 H , but b2 cannot have order 4; otherwise, b would have order 8. Therefore, b2 = e or a2. As H is normal, bab−1 H and has the same order as a because (bab−1)k = bak b−1.

Case (iiia). If bab−1 = a, then ba = ab, and the whole group can be proved to be abelian. If b2 = e, each element can be written uniquely in the form ar bs , where 0 r 3 and 0 s 1. Hence the group is isomorphic to C4 × C2 by mapping ar bs to (ar , bs ). If b2 = a2, let c = ab, so that c2 = a2b2 = a4 = e. Each element of the group can now be written uniquely in the form ar cs , where 0 r 3, 0 s 1, and the group is still isomorphic to C4 × C2.

Case (iiic). If bab−1 = a3 and b2 = a2, then the group is isomorphic to the quaternion group Q, described in Exercise 3.47. The isomorphism maps ar bs to ir j s .

Any group with eight or fewer elements is isomorphic to exactly one group in Table 4.6.