Supplementary Notes on Direct Products of Groups and the Fundamental Theorem of Finite Abelian Groups

Definition. Let G1,G2,...Gn be groups. We can make the Cartesian product G1 × G2 × · · · × Gn into a group via coordinate-wise multiplication: 0 0 0 0 0 0 (g1, g2, . . . , gn) · (g1, g2, . . . , gn) = (g1g1, g2g2, . . . , gngn).

It is absolutely routine to check that with the above operation, G1 × G2 × · · · × Gn is a group, called the (external) direct product of G1,G2,...Gn. In case G1,G2,...,Gn are abelian groups written additively, then it is more customary to write the direct product as G1 ⊕ G2 ⊕ · · · ⊕ Gn, and refer to this as an (external) direct sum. To “internalize” this notion, recall first that if G is a group with normal sub- groups H,K/G, then the product HK is a subgroup of G. More generally, if we have normal subgroups H1,H2,...,Hn /G, then the product H1H2 ··· Hn is also a subgroup of G. Sometimes this represents G as a direct product, as follows:

Theorem 1. [Internal Direct Product Theorem] Let H1,H2,...,Hn be normal subgroups of G, and assume that

(i) G = H1H2 ··· Hn,

(ii) For each i = 1, 2, . . . , n, we have Hi ∩ (H1 ··· Hi−1Hi+1 ··· Hn) = {e}. ∼ Then G = H1 × H2 × · · · × Hn; in fact the mapping

φ : H1 × H2 × · · · × Hn −→ G given by φ(h1, h2, . . . , hn) = h1h2 ··· hn is an isomorphism. When this happens, we shall slur the distinction between “internal” and “external” and simply write G = H1 × H2 × · · · × Hn, keeping in mind that H1,H2,...,Hn are actually subgroups of G.

Proof. We claim first that if hi ∈ Hi and hj ∈ Hj with i 6= j, then hihj = hjhi. But this is easy: since both Hi and Hj are normal in G, we see that the “commutator” −1 −1 hihjhi hj ∈ Hi ∩ Hj. But condition (ii) above clearly implies that Hi ∩ Hj = {e}, and so it follows that hihj = hjhi, as required. Now define

φ : H1 × H2 × · · · × Hn −→ G by setting φ(h1, h2, . . . , hn) = h1h2 ··· hn. Because of what we just proved above, φ is a homomorphism. By condition (i), φ is onto. Finally, if (h1, h2, . . . , hn) ∈ ker φ, −1 then h1h2 ··· hn = e implies that hi = h1 ··· hi−1hi+1 ··· hn ∈ Hi∩(H1 ··· Hi−1Hi+1 ··· Hn) = {e}, i.e., hi = e. Since i was arbitrary, we conclude that (h1, h2, . . . , hn) = (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 Zm,Zn,Zmn be cyclic groups of the given orders. Then Zm ×Zn = Zmn.

n m Proof. Let Zmn be generated by the element z. Then the elements z1 = z , z2 = z have orders m and n, respectively. Set H1 = hz1i,H2 = hz2i (cyclic groups of orders m and n, respectively), and note that the hypotheses of the above theorem are met, ∼ i.e., Zmn = H1 × H2. 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: ∼ Z6 × Z7 = Z42, ∼ Z2 × Z5 × Z7 = Z70, ∼ ∼ Z6 × Z35 = Z2 × Z3 × Z35 = Z2 × Z105.

Now let A be a finite abelian group, which we continue to write multiplicatively. a1 a2 ak Let |A| = p1 p2 ··· pk be the decomposition of the order of A into distinct prime factors. We set n = |A|, and set n n n n1 = a1 , n2 = a2 , . . . , nk = ak . p1 p2 pk

Note that while the integers n1, n2, . . . , nk are not pairwise relatively prime, they 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 (n1, n2, . . . , nk) = (1) = Z, which further implies that there exist integers s1, s2, . . . , sk ∈ Z, with s1n1 + s2ns + ··· sknk = 1. Using the above, we can prove the following fundamental result.

Theorem 3. [Primary Decomposition Theorem] Let A be a finite abelian group a1 a2 ak of order |A| = n = p1 p2 ··· pk , and let Pi be the pi-Sylow subgroup of A, i = 1, 2, . . . , k. Then A = P1 × P2 × · · · × Pk. 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 P1,P2,...,Pk are all normal. Furthermore, if a ∈ A is an element whose order is a power of pi, then it follows from Sylow’s Theorem (the “covering” part) that a is contained in a pi-Sylow subgroup. But there is only one pi-Sylow subgroup since all are conjugate and any one is normal. Therefore, it follows that a ∈ Pi. Next, let a ∈ A be an arbitrary element. Using the result above, we have s1n1 + s2ns + ··· sknk = 1, for suitable s1, s2, . . . , sk ∈ Z, and so a = a1 = as1n1+s2ns+···sknk = as1n1 as2n2 ··· asknk .

s n pai n s s s n However, (a i i ) i = (a ) i = e i = e and so a i i ∈ Pi, for i = 1, 2, . . . , k. This already proves that A = P1P2 ··· Pk. That Pi ∩(P1 ··· Pi−1Pi+1 ··· Pk) = {e} should 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 e1, e2, . . . , er with e1 ≥ e2 ≥ · · · ≥ er such that P = Zpe1 × Zpe2 × · · · × Zper .

Proof. We shall argue by induction on the order of P . First, let x1 ∈ P be an e element of maximal order p 1 in P . Clearly, e1 ≤ m. Set Z1 = hx1i. By induction, the quotient group A/Z1 is a direct product of cyclic groups:

A/Z1 = hy2Z1i × hy3Z1i × · · · × hyrZ1i,

e e where o(y2Z1) = p 2 , . . . , o(yrZ1) = p r , e2 ≥ e3 ≥ · · · ≥ er. Note that since x1 has maximal order in A, it follows that e1 ≥ e2. Now we shall make some adjustments. Fix i, where 2 ≤ i ≤ r, and consider ei ei p the cyclic subgroup hyiZ1i ≤ A/Z1. Since o(yiZ1) = p we have yi ∈ Z1, say e e −e e e −e e p i ki ki p 1 i p i p 1 i p 1 e e −e e y = x . Thus (x ) = (y ) = y = e and so p 1 |kip 1 i . Thus p i |ki, i 1 1 e i e i e p i ki lip i −li say ki = lip i , so y = x = x . Now the adjustment: we set xi = yix , e e e i 1 1 1 p i p i −lip i so xi = yi x1 = e, and no smaller exponent kills xi. Doing this for each e i = 2, 3, . . . , r produces elements x2, x3, . . . , xr ∈ P such that o(xi) = p i , i = 2, 3, . . . , r. Set Z2 = hx2i,Z3 = hx3i,...,Zr = hxri. To finish the proof, we need to show:

(i) P = Z1Z2 ··· Zr,

(ii) Zi ∩ (Z1 ··· Zi−1Zi+1 ··· Zr) = {e}.

Let x ∈ P . From the product decomposition above for A/Z1, we have powers f2, f3, . . . , fr with f2 f3 fr (y2Z1) (y3Z1) ··· (yrZ1) = xZ1,

f2 fr 0 0 li so it follows that y2 ··· yr = xx , for some x ∈ Z1. Thus, since yi = xix1 , we get

f2 f3 fr 00 x2 x3 ··· xr = xx ,

00 f1 00 −1 for some x ∈ Z1. Now let f1 be such that x1 = (x ) , and conclude that f1 f2 fr x1 x2 ··· xr = x, proving part (i). To prove part (ii), note that it suffices to prove that if

f1 f2 fr x1 x2 ··· xr = e,

ei −li then p |fi, for i = 1, 2, . . . , r. Again, since xi = yix1 , i = 2, 3, . . . , r we get an equation of the form 0 f2 f3 fr x y2 y3 ··· yr = e, 0 for some x ∈ Z1, from which we conclude that

f2 f3 fr (y2Z1) (y3Z1) ··· (yrZ1) = eZ1;

ei f1 this implies that p |fi, i = 2, 3, . . . , r. But then it follows immediately that x1 = e, e and so p 1 |f1, as well. This completes the proof.

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 n1 ≥ n2 ≥ · · · ≥ ns, with ns|ns−1, ns−1|ns−2, . . . , n2|n1, and cyclic subgroups Z1,Z2,...,Zs of orders n1, n2, . . . , ns, respectively with A = Z1 × Z2 × · · · × Zs. The numbers n1, n2, . . . , ns are called invariant factors of A. 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 = P1 × P2 × P3 × P4, (direct decomposition into Sylow subgroups), with ∼ ∼ ∼ ∼ P1 = Z8 × Z4, P2 = Z9 × Z3, P3 = Z125 × Z5 × Z5 × Z5, and P4 = Z11, (note here that I’ve written Zn for a cyclic group of order n; sorry about being somewhat inconsistent in my usage), then we would have

∼ A = Z8 × Z4 × Z9 × Z3 × Z125 × Z5 × Z5 × Z5 × Z11, and so a reorganized direct product decomposition would look like:

∼ A = Z8·9·125·11 × Z4·3·5 × Z5 × Z5, and so the “invariant factors” are n1 = 8 · 9 · 125 · 11 = 99, 000, n2 = 4 · 3 · 5 = 60, n3 = 5, n4 = 5. 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, ∼ G1 × H1 = G2 × H2, ∼ ∼ with H1 = H2, is it true that G1 = G2? The naive student will quickly respond with 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 / A, N / B. Then M × N/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/G. Then HK/K ∼= H/(H ∩ K). In particular, if we have H,K/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 K0 = φ(H),G0 = φ(L), then we have K0 × G0 = K × G; the task, then, is to show that K ∼= K0. We now set X = K × G = K0 × G0; since we have regarded the direct products as internal, then K,K0, G, G0 are all normal subgroups of X, and G ∼= G0. We divide the proof into two steps. Step 1. If K0 ∩ G = {e}, or if K ∩ G0 = {e}, we shall show that K0 ∼= K. In the first case, we have K0G = K0 × G ≤ X = K0 × G0. However, |G| = |G0| and so the index of K0 in both K0 × G and K0 × G0 is the same. Therefore, K0 × G = K0 × G0 = K × G, and so K0 ∼= (K0 × G)/G = (K × G)/G ∼= K. The argument is similar in case K ∩ G0 = {e}. Step 2. Induction on |G|.) Set M = K ∩G0,N = K0 ∩G, and note that M,N/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) = (K0 × G0)/(N × M) ∼= (K0/N) × (G0/M). Therefore (K/M) × (G/N) ∼= (K0/N) × (G0/M). But G ∼= G0 and so G × (K/M) × (G/N) ∼= G0 × (K0/N) × (G0/M). (*) We now work on the right hand side of (*): G0 × (K0/N) × (G0/M) ∼= ((G0 × K0)/N) × (G0/M) = ((G × K)/N) × (G0/M) ∼= (G/N) × K × (G0/M) ∼= K × (G/N) × (G0/M), In other words, G0 × (K0/N) × (G0/M) ∼= K × (G/N) × (G0/M). (**a) In an entirely similar way, G × (K/M) × (G/N) ∼= K0 × (G0/M) × (G/N). (**b) Now apply (*) to the left hand sides of (**a) and (**b) and infer that K × (G/N) × G0/M) ∼= K0 × (G/N) × (G0/M). By induction we may successively cancel (G0/M) and then (G/N), resulting in the desired isomorphism: K ∼= K0.

Corollary. [Uniqueness of Invariant Factors] Let A be a finite abelian group with invariant factors n1, n2, . . . , ns, and m1, m2, . . . , mt. Then s = t and mi = ni, i = 1, 2, . . . , s.

Proof. By assumption, we have

∼ ∼ Zn1 × Zn2 × · · · × Zns = A = Zm1 × Zm2 × · · · × Zmt , where ni+1|ni, i = 1, 2, . . . , s − 1, and mj+1|mj, j = 1, 2, . . . , t − 1. Note that ∼ if n1 6= m1, say m1 < n1, then since A = Zm1 × Zm2 × · · · × Zmt , we conclude that am1 = e for all a ∈ A. However, as A contains a subgroup isomorphic with

Zn1 , then A has an element of order n1, a contradiction. Therefore, n1 = m1.

Now use induction on |A| and apply Theorem 7, to cancel off the Zn1 factors and infer that the remaining invariant factors all agree.

Example 1. The possible isomorphism types of abelian groups of order 16 are:

Z16,Z8 × Z2,Z4 × Z4,Z4 × Z2 × Z2,Z2 × Z2 × Z2 × Z2.

Example 2. The possible isomorphism types of abelian groups of order 100 are

Z100,Z50 × Z2,Z20 × Z4,Z10 × Z10.