Supplementary Notes on Direct Products of Groups and the Fundamental Theorem of Finite Abelian Groups Definition. Let G ,G ,...G be groups. We can make the Cartesian product 1 2 n G × G × ··· × G into a group via coordinate-wise multiplication: 1 2 n (g ,g ,...,g ) · (g(cid:48),g(cid:48),...,g(cid:48) ) = (g g(cid:48),g g(cid:48),...,g g(cid:48) ). 1 2 n 1 2 n 1 1 2 2 n n It is absolutely routine to check that with the above operation, G ×G ×···×G 1 2 n is a group, called the (external) direct product of G ,G ,...G . 1 2 n In case G ,G ,...,G are abelian groups written additively, then it is more 1 2 n customary to write the direct product as G ⊕ G ⊕ ··· ⊕ G , and refer to this as 1 2 n an (external) direct sum. To “internalize” this notion, recall first that if G is a group with normal sub- groups H,K (cid:47) G, then the product HK is a subgroup of G. More generally, if we have normal subgroups H ,H ,...,H (cid:47)G, then the product H H ···H is also a 1 2 n 1 2 n subgroup of G. Sometimes this represents G as a direct product, as follows: Theorem 1. [Internal Direct Product Theorem] Let H ,H ,...,H be normal 1 2 n subgroups of G, and assume that (i) G = H H ···H , 1 2 n (ii) For each i = 1,2,...,n, we have H ∩ (H ···H H ···H ) = {e}. i 1 i−1 i+1 n ∼ Then G = H × H × ··· × H ; in fact the mapping 1 2 n φ : H × H × ··· × H −→ G 1 2 n given by φ(h ,h ,...,h ) = h h ···h is an isomorphism. When this happens, 1 2 n 1 2 n we shall slur the distinction between “internal” and “external” and simply write G = H × H × ··· × H , keeping in mind that H , H , ..., H are actually 1 2 n 1 2 n subgroups of G. Proof. We claim first that if h ∈ H and h ∈ H with i (cid:54)= j, then h h = h h . But i i j j i j j i this is easy: since both H and H are normal in G, we see that the “commutator” i j h h h−1h−1 ∈ H ∩H . But condition (ii) above clearly implies that H ∩H = {e}, i j i j i j i j and so it follows that h h = h h , as required. Now define i j j i φ : H × H × ··· × H −→ G 1 2 n by setting φ(h ,h ,...,h ) = h h ···h . Because of what we just proved above, φ 1 2 n 1 2 n is a homomorphism. By condition (i), φ is onto. Finally, if (h ,h ,...,h ) ∈ ker φ, 1 2 n then h h ···h = e implies that h−1 = h ···h h ···h ∈ H ∩(H ···H H ···H ) = 1 2 n i 1 i−1 i+1 n i 1 i−1 i+1 n {e}, i.e., h = e. Since i was arbitrary, we conclude that (h ,h ,...,h ) = i 1 2 n (e,e,...,e), and so φ is also injective. This proves the theorem. Please note that for abelian groups written additively, there is an obvious analogue of the above result. We shall now turn our attention to finite abelian groups. Basically we shall try to prove that every finite abelian group can be decomposed into cyclic groups. Without further stipulations, we can’t really hope for a uniqueness result, basically because of the Theorem 2. [Chinese Remainder Theorem] Let m and n be relatively prime ∼ integers and let Z , Z , Z be cyclic groups of the given orders. Then Z ×Z = m n mn m n Z . mn Proof. Let Z be generated by the element z. Then the elements z = zn, z = zm mn 1 2 have orders m and n, respectively. Set H = (cid:104)z (cid:105), H = (cid:104)z (cid:105) (cyclic groups of orders 1 1 2 2 m and n, respectively), and note that the hypotheses of the above theorem are met, ∼ i.e., Z = H × H . mn 1 2 Notice that the Chinese Remainder Theorem allows us to consolidate many direct product decompositions of cyclic groups into fewer such products, or to reconfigure direct products, as the following examples indicate: ∼ Z × Z = Z , 6 7 42 ∼ Z × Z × Z = Z , 2 5 7 70 ∼ ∼ Z × Z = Z × Z × Z = Z × Z . 6 35 2 3 35 2 105 Now let A be a finite abelian group, which we continue to write multiplicatively. Let |A| = pa1pa2 ···pak be the decomposition of the order of A into distinct prime 1 2 k factors. We set n = |A|, and set n n n n = , n = , ...,n = . 1 pa1 2 pa2 k pak 1 2 k Note that while the integers n ,n ,...,n are not pairwise relatively prime, they 1 2 k are collectively relatively prime, meaning that there is no integer greater than 1 than divides all of them. This implies, since Z is a principal ideal domain, that the ideal (n ,n ,...,n ) = (1) = Z, which further implies that there exist integers 1 2 k s ,s ,...,s ∈ Z, with s n + s n + ···s n = 1. 1 2 k 1 1 2 s k k Using the above, we can prove the following fundamental result. Theorem 3. [Primary Decomposition Theorem] Let A be a finite abelian group of order |A| = n = pa1pa2 ···pak, and let P be the p -Sylow subgroup of A, i = 1 2 k i i 1,2,...,k. Then A = P × P × ··· × P . 1 2 k Proof. We shall show that the hypotheses of Theorem 1 are satisfied. Note first that since A is abelian, then every subgroup of A is normal. In particular, the Sylow subgroups P , P , ..., P are all normal. Furthermore, if a ∈ A is an element 1 2 k whose order is a power of p , then it follows from Sylow’s Theorem (the “covering” i part) that a is contained in a p -Sylow subgroup. But there is only one p -Sylow i i subgroup since all are conjugate and any one is normal. Therefore, it follows that a ∈ P . i Next, let a ∈ A be an arbitrary element. Using the result above, we have s n + s n + ···s n = 1, for suitable s , s , ..., s ∈ Z, and so 1 1 2 s k k 1 2 k a = a1 = as1n1+s2ns+···sknk = as1n1as2n2 ···asknk. However, (asini)paii = (an)si = esi = e and so asini ∈ Pi, for i = 1,2,...,k. This already proves that A = P P ···P . That P ∩(P ···P P ···P ) = {e} should 1 2 k i 1 i−1 i+1 k be obvious. This proves the theorem. At this stage, we see that the decomposition of a finite abelian group into a direct product of cyclic groups can be accomplished once we show that any abelian p-group can be factored into a direct product of cyclic p-groups. Theorem 4. Let P be a finite abelian p-group of order pm. Then there exist ∼ powers e ,e ,...,e with e ≥ e ≥ ··· ≥ e such that P = Z ×Z ×···×Z . 1 2 r 1 2 r pe1 pe2 per Proof. We shall argue by induction on the order of P. First, let x ∈ P be an 1 element of maximal order pe1 in P. Clearly, e ≤ m. Set Z = (cid:104)x (cid:105). By induction, 1 1 1 the quotient group A/Z is a direct product of cyclic groups: 1 A/Z = (cid:104)y Z (cid:105) × (cid:104)y Z (cid:105) × ··· × (cid:104)y Z (cid:105), 1 2 1 3 1 r 1 where o(y Z ) = pe2,...,o(y Z ) = per, e ≥ e ≥ ··· ≥ e . Note that since x has 2 1 r 1 2 3 r 1 maximal order in A, it follows that e ≥ e . 1 2 Now we shall make some adjustments. Fix i, where 2 ≤ i ≤ r, and consider the cyclic subgroup (cid:104)y Z (cid:105) ≤ A/Z . Since o(y Z ) = pei we have ypei ∈ Z , say i 1 1 i 1 i 1 ypei = xki. Thus (xki)pe1−ei = (ypei)pe1−ei = ype1 = e and so pe1|k pe1−ei. Thus pei|k , i 1 1 i i i i say k = l pei, so ypei = xki = xlipei. Now the adjustment: we set x = y x−li, i i i 1 1 i i 1 so xpei = ypeix−lipei = e, and no smaller exponent kills x . Doing this for each i i 1 i i = 2,3,...,r produces elements x ,x ,...,x ∈ P such that o(x ) = pei, i = 2 3 r i 2,3,...,r. Set Z = (cid:104)x (cid:105), Z = (cid:104)x (cid:105),...,Z = (cid:104)x (cid:105). To finish the proof, we need 2 2 3 3 r r to show: (i) P = Z Z ···Z , 1 2 r (ii) Z ∩ (Z ···Z Z ···Z ) = {e}. i 1 i−1 i+1 r Let x ∈ P. From the product decomposition above for A/Z , we have powers 1 f ,f ,...,f with 2 3 r (y Z )f2(y Z )f3 ···(y Z )fr = xZ , 2 1 3 1 r 1 1 so it follows that yf2 ···yfr = xx(cid:48), for some x(cid:48) ∈ Z . Thus, since y = x xli, we get 2 r 1 i i 1 xf2xf3 ···xfr = xx(cid:48)(cid:48), 2 3 r for some x(cid:48)(cid:48) ∈ Z . Now let f be such that xf1 = (x(cid:48)(cid:48))−1, and conclude that 1 1 1 xf1xf2 ···xfr = x, proving part (i). 1 2 r To prove part (ii), note that it suffices to prove that if xf1xf2 ···xfr = e, 1 2 r then pei|f , for i = 1,2,...,r. Again, since x = y x−li, i = 2,3,...,r we get an i i i 1 equation of the form x(cid:48)yf2yf3 ···yfr = e, 2 3 r for some x(cid:48) ∈ Z , from which we conclude that 1 (y Z )f2(y Z )f3 ···(y Z )fr = eZ ; 2 1 3 1 r 1 1 this implies that pei|f , i = 2,3,...,r. But then it follows immediately that xf1 = e, i 1 and so pe1|f , as well. This completes the proof. 1 If we use the above theorem, together with the Chinese Remainder Theorem, we get the following ”existence” theorem. Theorem 5. [Existence of Invariant Factors] Let A be a finite abelian group, and let n = |A| be the order of A. Then there exists a decreasing sequence n ≥ n ≥ 1 2 ··· ≥ n , with n |n , n |n ,...,n |n , and cyclic subgroups Z , Z ,...,Z s s s−1 s−1 s−2 2 1 1 2 s of orders n ,n ,...,n , respectively with A = Z × Z × ··· × Z . The numbers 1 2 s 1 2 s n , n , ..., n are called invariant factors of A. 1 2 s It is pretty easy to see how the direct product decompositions of the Sylow subgroups in the Primary Decomposition Theorem can be synthesized into a direct product decomposition of the type indicated above. For example, if we had a group A = P × P × P × P , (direct decomposition into Sylow subgroups), with 1 2 3 4 ∼ ∼ ∼ ∼ P = Z × Z , P = Z × Z , P = Z × Z × Z × Z , and P = Z , (note here 1 8 4 2 9 3 3 125 5 5 5 4 11 that I’ve written Z for a cyclic group of order n; sorry about being somewhat n inconsistent in my usage), then we would have ∼ A = Z × Z × Z × Z × Z × Z × Z × Z × Z , 8 4 9 3 125 5 5 5 11 and so a reorganized direct product decomposition would look like: ∼ A = Z × Z × Z × Z , 8·9·125·11 4·3·5 5 5 and so the “invariant factors” are n = 8 · 9 · 125 · 11 = 99,000, n = 4 · 3 · 5 = 1 2 60, n = 5, n = 5. 3 4 To obtain a uniqueness result for the direct product decomposition, we shall first discuss the question of “cancellation” in a direct product decomposition. Basically, the question is this: If we have an isomorphism of direct product groups, ∼ G × H = G × H , 1 1 2 2 ∼ ∼ with H = H , is it true that G = G ? The naive student will quickly respond with 1 2 1 2 an affirmative answer; yet the answer is no, making the cancellation question rather subtle. As a counterexample to the cancellation problem, consider the infinite direct product of the infinite cyclic group Z: ∼ Z × (Z × ···) = Z × Z × (Z × ···); note that if we could cancel off the right hand factors, when we would end up with ∼ Z = Z ×Z, which is false (Z is cyclic, but Z ×Z is not). Therefore we must treat this question with some care. Before turning to the main cancellation result, let’s state and prove a very simple lemma, and recall another important result. Lemma. Suppose we have groups and normal subgroups: M (cid:47) A, N (cid:47) B. Then M × N (cid:47) A × B, and ∼ (A × B)/(M × N) = (A/M) × (B/N). Proof. Clearly the first statement above is true. As for the isomorphism, define φ : A × B → (A/M) × (B/N), by setting φ(a,b) = (aM,bN). Note that φ is a surjective homomorphism with kernel M × N. Theorem 6. [Noether Isomorphism Theorem] Let G be a group and let H,K ≤ G, with K (cid:47) G. Then ∼ HK/K = H/(H ∩ K). ∼ In particular, if we have H,K (cid:47) G and H ∩ K = {e}, then (H × K)/K = H. We turn now to the main result. I’ve pretty much followed the treatment given by R. Hirshon, in his paper Decomposition of groups, in the American Mathe- matical Monthly, vol. 76 (1969), pp. 1037-1039. Theorem 7. [Cancellation of Finite Groups] Assume that we have an isomorphism ∼ ∼ ∼ H × L = K × G, where L = G and are finite groups. Then H = K. Proof. We shall regard the above direct product as internal. Assume that φ : H × L → K × G is an isomorphism. Thus, if K(cid:48) = φ(H), G(cid:48) = φ(L), then we have K(cid:48) × G(cid:48) = K × G; the task, then, is to show that K ∼= K(cid:48). We now set X = K × G = K(cid:48) × G(cid:48); since we have regarded the direct products as internal, then K,K(cid:48),G,G(cid:48) are all normal subgroups of X, and G ∼= G(cid:48). We divide the proof into two steps. Step 1. If K(cid:48) ∩ G = {e}, or if K ∩ G(cid:48) = {e}, we shall show that K(cid:48) ∼= K. In the first case, we have K(cid:48)G = K(cid:48) × G ≤ X = K(cid:48) × G(cid:48). However, |G| = |G(cid:48)| and so the index of K(cid:48) in both K(cid:48) × G and K(cid:48) × G(cid:48) is the same. Therefore, K(cid:48) × G = K(cid:48) × G(cid:48) = K × G, and so K(cid:48) ∼= (K(cid:48) × G)/G = (K × G)/G ∼= K. The argument is similar in case K ∩ G(cid:48) = {e}. Step 2. Induction on |G|.) Set M = K∩G(cid:48), N = K(cid:48)∩G, and note that M,N (cid:47)X, and that M ∩ N = {e}. We have ∼ X/(M × N) = (K × G)/(M × N) = (K/M) × (G/N), and X/(M × N) = X/(N × M) = (K(cid:48) × G(cid:48))/(N × M) ∼= (K(cid:48)/N) × (G(cid:48)/M). Therefore (K/M) × (G/N) ∼= (K(cid:48)/N) × (G(cid:48)/M). But G ∼= G(cid:48) and so G × (K/M) × (G/N) ∼= G(cid:48) × (K(cid:48)/N) × (G(cid:48)/M). (*) We now work on the right hand side of (*): G(cid:48) × (K(cid:48)/N) × (G(cid:48)/M) ∼= ((G(cid:48) × K(cid:48))/N) × (G(cid:48)/M) = ((G × K)/N) × (G(cid:48)/M) ∼= (G/N) × K × (G(cid:48)/M) ∼= K × (G/N) × (G(cid:48)/M), In other words, G(cid:48) × (K(cid:48)/N) × (G(cid:48)/M) ∼= K × (G/N) × (G(cid:48)/M). (**a) In an entirely similar way, G × (K/M) × (G/N) ∼= K(cid:48) × (G(cid:48)/M) × (G/N). (**b) Now apply (*) to the left hand sides of (**a) and (**b) and infer that K × (G/N) × G(cid:48)/M) ∼= K(cid:48) × (G/N) × (G(cid:48)/M). By induction we may successively cancel (G(cid:48)/M) and then (G/N), resulting in the desired isomorphism: K ∼= K(cid:48). Corollary. [Uniqueness of Invariant Factors] Let A be a finite abelian group with invariant factors n ,n ,...,n , and m ,m ,...,m . Then s = t and m = 1 2 s 1 2 t i n , i = 1,2,...,s. i Proof. By assumption, we have ∼ ∼ Z × Z × ··· × Z = A = Z × Z × ··· × Z , n n n m m m 1 2 s 1 2 t where n |n , i = 1,2,...,s − 1, and m |m , j = 1,2,...,t − 1. Note that i+1 i j+1 j ∼ if n (cid:54)= m , say m < n , then since A = Z × Z × ··· × Z , we conclude 1 1 1 1 m m m 1 2 t that am1 = e for all a ∈ A. However, as A contains a subgroup isomorphic with Z , then A has an element of order n , a contradiction. Therefore, n = m . n 1 1 1 1 Now use induction on |A| and apply Theorem 7, to cancel off the Z factors n 1 and infer that the remaining invariant factors all agree. Example 1. The possible isomorphism types of abelian groups of order 16 are: Z , Z × Z , Z × Z , Z × Z × Z , Z × Z × Z × Z . 16 8 2 4 4 4 2 2 2 2 2 2 Example 2. The possible isomorphism types of abelian groups of order 100 are Z , Z × Z , Z × Z , Z × Z . 100 50 2 20 4 10 10