Direct Products and Finite Abelian Groups

Study Sheet

Direct Products and Finite Abelian Groups

Building groups from smaller ones, and classifying every finite abelian group

Direct Products

Concept
The direct product

For groups GG and HH, the direct product G×HG \times H is the set of pairs {(g,h):g∈G, h∈H}\{(g,h) : g \in G,\ h \in H\} under the componentwise operation (g1,h1)(g2,h2)=(g1g2, h1h2)(g_1,h_1)(g_2,h_2) = (g_1g_2,\ h_1h_2). It is a group, with identity (eG,eH)(e_G,e_H), inverse (g,h)−1=(g−1,h−1)(g,h)^{-1} = (g^{-1},h^{-1}), and order ∣G×H∣=∣G∣ ∣H∣|G\times H| = |G|\,|H|. The order of an element is

ord⁡(g,h)=lcm⁡(ord⁡(g), ord⁡(h)).\operatorname{ord}(g,h) = \operatorname{lcm}\big(\operatorname{ord}(g),\, \operatorname{ord}(h)\big).

Every axiom for G×HG \times H reduces to the same axiom for GG and for HH, checked in each coordinate at once — associativity, the identity, and inverses all come along for free, so forming a direct product never requires checking anything new. The construction says a group of order ∣G∣∣H∣|G||H| can be built from two smaller, already-understood groups running side by side, like two independent clocks whose hands are read together. The order formula is the engine behind the rest of this topic: (g,h)(g,h) returns to (eG,eH)(e_G,e_H) exactly when BOTH coordinates have returned home, so the wait is the first time that is simultaneously a multiple of ord⁡(g)\operatorname{ord}(g) and of ord⁡(h)\operatorname{ord}(h) — the least common multiple, never the sum and never the plain product.

Example
The order of a specific element

Find the order of (3,2)(3,2) in Z4×Z6\mathbb Z_4 \times \mathbb Z_6. By the order formula in Zn\mathbb Z_n (Topic 3), ord⁡(3)=4/gcd⁡(4,3)=4\operatorname{ord}(3) = 4/\gcd(4,3) = 4 in Z4\mathbb Z_4, and ord⁡(2)=6/gcd⁡(6,2)=3\operatorname{ord}(2) = 6/\gcd(6,2) = 3 in Z6\mathbb Z_6. So ord⁡(3,2)=lcm⁡(4,3)=12\operatorname{ord}(3,2) = \operatorname{lcm}(4,3) = 12. Check directly: 12⋅(3,2)=(36,24)≡(0,0)(mod(4,6))12\cdot(3,2) = (36,24) \equiv (0,0) \pmod{(4,6)}, and no smaller positive multiple of (3,2)(3,2) can be (0,0)(0,0), since that would need a common multiple of 44 and 33 smaller than 1212.

Example
The table of Z2×Z2\mathbb Z_2\times\mathbb Z_2
Z_2×Z_2+000110110000011011010100111010101100011111100100
Z_2×Z_2: a symmetric table in which every element is its own inverse.

Build the Cayley table of Z2×Z2={(0,0),(0,1),(1,0),(1,1)}\mathbb Z_2\times\mathbb Z_2 = \{(0,0),(0,1),(1,0),(1,1)\} under componentwise addition mod 22. Every entry reduces coordinate by coordinate, e.g. (1,0)+(1,1)=(0,1)(1,0)+(1,1) = (0,1). The table is symmetric, so the group is abelian, and the diagonal shows every element is its own inverse: (0,1)+(0,1)=(0,0)(0,1)+(0,1)=(0,0), and likewise for (1,0)(1,0) and (1,1)(1,1). This is the Klein four-group, the unique group of order 44 with no element of order 44.

Example
How many elements of Z4×Z4\mathbb Z_4\times\mathbb Z_4 have order 22?

An element (a,b)(a,b) has order dividing 22 exactly when both aa and bb have order dividing 22 in Z4\mathbb Z_4 — that means a,b∈{0,2}a,b \in \{0,2\}. That gives 2×2=42\times2 = 4 pairs, and removing the identity (0,0)(0,0) leaves exactly 33 elements of order exactly 22: (0,2)(0,2), (2,0)(2,0), (2,2)(2,2). (Every other pair, with at least one coordinate equal to 11 or 33, has order 44 by the order formula, since lcm⁡\operatorname{lcm} with a 44 present is always 44.)

Example
A direct product of a non-abelian group

The construction needs neither factor to be abelian. In S3×Z2S_3 \times \mathbb Z_2, take g=((1 2 3), 1)g = \big((1\,2\,3),\,1\big). Since ord⁡((1 2 3))=3\operatorname{ord}((1\,2\,3)) = 3 in S3S_3 (Topic 4) and ord⁡(1)=2\operatorname{ord}(1) = 2 in Z2\mathbb Z_2, ord⁡(g)=lcm⁡(3,2)=6\operatorname{ord}(g) = \operatorname{lcm}(3,2) = 6. Check: g6=((1 2 3)6, 6⋅1)=(e,0)g^6 = \big((1\,2\,3)^6,\ 6\cdot1\big) = (e,0), the identity of S3×Z2S_3\times\mathbb Z_2, and no smaller positive power works, since that would need a common multiple of 33 and 22 below 66. S3×Z2S_3 \times \mathbb Z_2 itself is not abelian, since S3S_3 is not: a direct product inherits non-abelian-ness from either factor.

Side note
Where it goes wrong

The order of (g,h)(g,h) is lcm⁡(ord⁡(g),ord⁡(h))\operatorname{lcm}(\operatorname{ord}(g),\operatorname{ord}(h)) — never the sum, and never the plain product, of the two orders. Take (1,1)(1,1) in Z4×Z6\mathbb Z_4\times\mathbb Z_6: ord⁡(1)=4\operatorname{ord}(1)=4 in Z4\mathbb Z_4 and ord⁡(1)=6\operatorname{ord}(1)=6 in Z6\mathbb Z_6, so the true order is lcm⁡(4,6)=12\operatorname{lcm}(4,6)=12 — NOT the sum 4+6=104+6=10, and NOT the product 4⋅6=244\cdot6=24. Stepping through the multiples of (1,1)(1,1) by hand reaches (0,0)(0,0) for the first time after exactly 1212 steps, ruling out the other two guesses entirely.

Try it
Try it: order in a direct product

Find the order of (2,3)(2,3) in Z6×Z9\mathbb Z_6 \times \mathbb Z_9.

Show the answer

In Z6\mathbb Z_6, ord⁡(2)=6/gcd⁡(6,2)=3\operatorname{ord}(2) = 6/\gcd(6,2) = 3; in Z9\mathbb Z_9, ord⁡(3)=9/gcd⁡(9,3)=3\operatorname{ord}(3) = 9/\gcd(9,3) = 3. So ord⁡(2,3)=lcm⁡(3,3)=3\operatorname{ord}(2,3) = \operatorname{lcm}(3,3) = \mathbf{3}. Check: 3⋅(2,3)=(6,9)≡(0,0)(mod(6,9))3\cdot(2,3) = (6,9) \equiv (0,0) \pmod{(6,9)}.

When Is a Product Cyclic?

Concept
The cyclic product criterion

Zm×Zn\mathbb Z_m \times \mathbb Z_n is cyclic — and then Zm×Zn≅Zmn\mathbb Z_m\times\mathbb Z_n \cong \mathbb Z_{mn} — if and only if gcd⁡(m,n)=1\gcd(m,n)=1. More generally, for finite groups GG and HH, G×HG\times H is cyclic exactly when GG and HH are both cyclic and gcd⁡(∣G∣,∣H∣)=1\gcd(|G|,|H|)=1.

A cyclic group of order NN needs some element of order exactly NN. By the order formula (Section 1), the largest possible order in G×HG\times H is lcm⁡(∣G∣,∣H∣)\operatorname{lcm}(|G|,|H|), achieved only when some element of GG has order ∣G∣|G| and some element of HH has order ∣H∣|H| — which needs both factors cyclic. And lcm⁡(m,n)\operatorname{lcm}(m,n) equals the full order mnmn precisely when gcd⁡(m,n)=1\gcd(m,n)=1; otherwise lcm⁡(m,n)=mn/gcd⁡(m,n)<mn\operatorname{lcm}(m,n) = mn/\gcd(m,n) < mn, and no element can reach the whole group. This single gcd test is the direct-product version of the Chinese Remainder Theorem (Number Theory), now phrased as a statement about groups rather than about congruences.

Example
Z3×Z5≅Z15\mathbb Z_3 \times \mathbb Z_5 \cong \mathbb Z_{15}
001122031420011223041021021324
Stepping by (1,1) around Z_3×Z_5 visits all 15 pairs before returning to (0,0).

gcd⁡(3,5)=1\gcd(3,5)=1, so by the criterion Z3×Z5\mathbb Z_3\times\mathbb Z_5 is cyclic of order 1515. The generator is (1,1)(1,1): ord⁡(1,1)=lcm⁡(3,5)=15\operatorname{ord}(1,1) = \operatorname{lcm}(3,5) = 15, matching the full order of the group, so (1,1)(1,1) generates everything and Z3×Z5≅Z15\mathbb Z_3\times\mathbb Z_5 \cong \mathbb Z_{15}.

Example
Z4×Z6\mathbb Z_4\times\mathbb Z_6 is not cyclic

gcd⁡(4,6)=2≠1\gcd(4,6)=2\ne1, so the criterion says Z4×Z6\mathbb Z_4\times\mathbb Z_6 (order 2424) cannot be cyclic. Indeed, the largest possible element order is lcm⁡(4,6)=12\operatorname{lcm}(4,6)=12, achieved at (1,1)(1,1) — well short of 2424 — and no other pair can do better, since no coordinate order can exceed 44 or 66. With no element of order 2424, nothing generates the whole group.

Example
Is Z10×Z15\mathbb Z_{10}\times\mathbb Z_{15} cyclic?

gcd⁡(10,15)=5≠1\gcd(10,15)=5\ne1, so no. The group has order 150150, but the largest element order is lcm⁡(10,15)=30\operatorname{lcm}(10,15)=30: for instance (1,1)(1,1) has order 3030, and no pair of coordinate-orders dividing 1010 and 1515 respectively can beat that, since lcm⁡(a,b)≤lcm⁡(10,15)\operatorname{lcm}(a,b)\le\operatorname{lcm}(10,15) whenever a∣10a\mid10 and b∣15b\mid15. A group of order 150150 with largest element order only 3030 is far from cyclic.

Example
A square of a cyclic group is never cyclic

For n≥2n\ge2, Zn×Zn\mathbb Z_n\times\mathbb Z_n is never cyclic, since gcd⁡(n,n)=n≠1\gcd(n,n)=n\ne1. Concretely in Z5×Z5\mathbb Z_5\times\mathbb Z_5 (order 2525), every nonzero coordinate has order 55 by Lagrange (Topic 5: 55 is prime, so every non-identity order divides 55 and must equal 55), so the lcm⁡\operatorname{lcm} of two orders from {1,5}\{1,5\} never exceeds 55 — far short of the 2525 a generator would need.

Side note
Where it goes wrong

Both factors being cyclic is not enough on its own; the orders must also be coprime. Z2\mathbb Z_2 and Z4\mathbb Z_4 are each cyclic, yet Z2×Z4\mathbb Z_2\times\mathbb Z_4 (order 88) is not: gcd⁡(2,4)=2\gcd(2,4)=2, and the largest element order is lcm⁡(2,4)=4\operatorname{lcm}(2,4)=4, reached at (1,1)(1,1) — no element of order 88 exists. Looking cyclic piece by piece is not the same as being cyclic as a whole.

Try it
Try it: a coprime product

Is Z7×Z9\mathbb Z_7\times\mathbb Z_9 cyclic? If so, what group is it isomorphic to?

Show the answer

gcd⁡(7,9)=1\gcd(7,9)=1, so yes: by the criterion, Z7×Z9\mathbb Z_7\times\mathbb Z_9 is cyclic, generated by (1,1)(1,1) since ord⁡(1,1)=lcm⁡(7,9)=63\operatorname{ord}(1,1)=\operatorname{lcm}(7,9)=63, the full order of the group. So Z7×Z9≅Z63\mathbb Z_7\times\mathbb Z_9 \cong \mathbf{\mathbb Z_{63}}.

The Fundamental Theorem of Finite Abelian Groups

Concept
The Fundamental Theorem of Finite Abelian Groups

Every finite abelian group GG is isomorphic to a direct product of cyclic groups of prime-power order,

G≅Zp1a1×Zp2a2×⋯×Zpkak,G \cong \mathbb Z_{p_1^{a_1}} \times \mathbb Z_{p_2^{a_2}} \times \cdots \times \mathbb Z_{p_k^{a_k}},

and this list of prime powers (the elementary divisors of GG) is unique up to the order in which the factors are written.

This is the group-theory analogue of unique prime factorization: just as every integer breaks uniquely into prime powers, every finite abelian group breaks uniquely into cyclic prime-power pieces, its atoms. Existence comes in two stages, both proved in "Why It Works": first GG splits along its distinct primes into primary components, by the same coprimality idea as Section 2; then each prime-power piece, taken alone, further splits into cyclic factors. Uniqueness means there is exactly one correct list of prime powers for a given GG, not several equally valid ones — which is what makes "the isomorphism type of GG" a well-defined, checkable piece of data rather than a matter of how you happened to write it.

Example
Decomposing Z12\mathbb Z_{12}

12=4⋅312 = 4\cdot3 with gcd⁡(4,3)=1\gcd(4,3)=1, so by Section 2, Z12≅Z4×Z3\mathbb Z_{12}\cong\mathbb Z_4\times\mathbb Z_3. Both factors already have prime-power order (4=224=2^2, 3=313=3^1), so this IS the elementary divisor decomposition: the elementary divisors of Z12\mathbb Z_{12} are 44 and 33.

Example
Decomposing Z360\mathbb Z_{360}

360=23⋅32⋅5=8⋅9⋅5360 = 2^3\cdot3^2\cdot5 = 8\cdot9\cdot5, and 8,9,58,9,5 are pairwise coprime, so Z360≅Z8×Z9×Z5\mathbb Z_{360}\cong\mathbb Z_8\times\mathbb Z_9\times\mathbb Z_5 — three cyclic factors of prime-power order, with 8⋅9⋅5=3608\cdot9\cdot5=360 confirming nothing was lost.

Example
Identifying a group of order 88 from its element orders

An abelian group GG of order 88 has an element of order 44 but none of order 88. Which group is GG? There are three abelian groups of order 88 up to isomorphism (Section 4): Z8\mathbb Z_8, Z4×Z2\mathbb Z_4\times\mathbb Z_2, and Z2×Z2×Z2\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2. Z8\mathbb Z_8 has an element of order 88, which GG does not, ruling it out; Z2×Z2×Z2\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2 has every non-identity element of order 22, so it has no element of order 44 either, ruling it out too. Only Z4×Z2\mathbb Z_4\times\mathbb Z_2 fits both conditions: it has (1,0)(1,0) of order 44 and nothing of order 88. So G≅Z4×Z2G \cong \mathbb Z_4\times\mathbb Z_2.

Example
Regrouping Z6×Z10\mathbb Z_6\times\mathbb Z_{10} by prime

Neither factor of Z6×Z10\mathbb Z_6\times\mathbb Z_{10} already has prime-power order, so split each one first: Z6≅Z2×Z3\mathbb Z_6\cong\mathbb Z_2\times\mathbb Z_3 and Z10≅Z2×Z5\mathbb Z_{10}\cong\mathbb Z_2\times\mathbb Z_5. So Z6×Z10≅Z2×Z3×Z2×Z5\mathbb Z_6\times\mathbb Z_{10} \cong \mathbb Z_2\times\mathbb Z_3\times\mathbb Z_2\times\mathbb Z_5, with elementary divisors 2,2,3,52,2,3,5 — note TWO separate factors of 22, since the theorem allows repeats among different primes' powers as long as each one individually has prime-power order. The order checks: 2⋅2⋅3⋅5=60=6⋅102\cdot2\cdot3\cdot5 = 60 = 6\cdot10.

Side note
Where it goes wrong

A direct-product split into two coprime-looking pieces is not automatically the elementary divisor decomposition. It is tempting to write Z12\mathbb Z_{12} as Z2×Z6\mathbb Z_2\times\mathbb Z_6, since 2⋅6=122\cdot6=12 — but Z2×Z6\mathbb Z_2\times\mathbb Z_6 is NOT isomorphic to Z12\mathbb Z_{12} at all: gcd⁡(2,6)=2≠1\gcd(2,6)=2\ne1, so by Section 2 it is not even cyclic, with largest element order only lcm⁡(2,6)=6\operatorname{lcm}(2,6)=6, while Z12\mathbb Z_{12} plainly has an element of order 1212. The theorem's factors must each have prime-power order precisely to rule out non-coprime splits like this one; 12=4⋅312=4\cdot3 is the correct split, not 12=2⋅612=2\cdot6.

Try it
Try it: elementary divisors of Z200\mathbb Z_{200}

Write Z200\mathbb Z_{200} as a product of cyclic groups of prime-power order.

Show the answer

200=23⋅52=8⋅25200 = 2^3\cdot5^2 = 8\cdot25, and gcd⁡(8,25)=1\gcd(8,25)=1, so Z200≅Z8×Z25\mathbb Z_{200} \cong \mathbf{\mathbb Z_8\times\mathbb Z_{25}}.

Counting Abelian Groups of a Given Order

Concept
Counting by partitions

If n=p1a1p2a2⋯pkakn = p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}, the number of abelian groups of order nn, up to isomorphism, is

p(a1) p(a2)⋯p(ak),p(a_1)\,p(a_2)\cdots p(a_k),

where p(a)p(a) is the number of integer partitions of aa — the number of ways to write aa as a sum of positive integers, order not mattering.

By the Fundamental Theorem, an abelian group of prime-power order pap^a is a direct product of cyclic factors Zpλ1×⋯×Zpλr\mathbb Z_{p^{\lambda_1}}\times\cdots\times\mathbb Z_{p^{\lambda_r}} with λ1≥⋯≥λr≥1\lambda_1 \ge \cdots \ge \lambda_r \ge 1 and λ1+⋯+λr=a\lambda_1+\cdots+\lambda_r = a — exactly a partition of aa, and distinct partitions give non-isomorphic groups, by the uniqueness half of the theorem. A general order nn factors into primes, its primary components (Section 3) are built independently, one partition choice per prime, and different primes never interact — so the total count across all primes multiplies rather than adds. This turns "how many groups of order nn" from a search into ordinary partition-counting.

Example
Abelian groups of order 88

8=238=2^3, and the partitions of 33 are 3, 2+1, 1+1+13,\ 2+1,\ 1+1+1 — so p(3)=3p(3)=3, and there are exactly 33 abelian groups of order 88: Z8\mathbb Z_8, Z4×Z2\mathbb Z_4\times\mathbb Z_2, and Z2×Z2×Z2\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2, matching Section 3's example.

Example
Abelian groups of order 1616

16=2416=2^4, and 44 has p(4)=5p(4)=5 partitions: 4, 3+1, 2+2, 2+1+1, 1+1+1+14,\ 3+1,\ 2+2,\ 2+1+1,\ 1+1+1+1. So there are 55 abelian groups of order 1616: Z16\mathbb Z_{16}, Z8×Z2\mathbb Z_8\times\mathbb Z_2, Z4×Z4\mathbb Z_4\times\mathbb Z_4, Z4×Z2×Z2\mathbb Z_4\times\mathbb Z_2\times\mathbb Z_2, and Z2×Z2×Z2×Z2\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2.

Example
How many abelian groups of order 360360?

360=23⋅32⋅51360 = 2^3\cdot3^2\cdot5^1 (Section 3). Each prime contributes its own partition count: p(3)=3p(3)=3 for the 232^3 part, p(2)=2p(2)=2 for the 323^2 part (22, 1+11+1), and p(1)=1p(1)=1 for the 515^1 part. By the counting formula, there are 3⋅2⋅1=63\cdot2\cdot1=6 abelian groups of order 360360, up to isomorphism.

Example
A million groups? Order 10610^6

106=26⋅5610^6 = 2^6\cdot5^6. Both exponents are 66, and p(6)=11p(6)=11 — there are 1111 partitions of 66. So the number of abelian groups of order 10610^6 is 11⋅11=12111\cdot11 = 121: each prime's partition choice is independent of the other's, so the count squares rather than merely doubling.

Side note
Where it goes wrong

The count uses partitions of the EXPONENTS, not divisors of nn itself. For n=12=22⋅3n=12=2^2\cdot3, the number of divisors of 1212 is 66 (1,2,3,4,6,121,2,3,4,6,12), but the number of abelian groups of order 1212 is p(2)⋅p(1)=2⋅1=2p(2)\cdot p(1) = 2\cdot1 = 2: just Z4×Z3≅Z12\mathbb Z_4\times\mathbb Z_3 \cong \mathbb Z_{12} and Z2×Z2×Z3\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_3. Counting divisors answers a different question — how many ways 1212 factors as a⋅ba\cdot b — not how many non-isomorphic groups of order 1212 exist.

Try it
Try it: counting groups of order 100100

How many abelian groups of order 100100 are there, up to isomorphism?

Show the answer

100=22⋅52100=2^2\cdot5^2, and p(2)=2p(2)=2 for each prime, so the count is 2⋅2=42\cdot2=\mathbf{4}: Z4×Z25\mathbb Z_4\times\mathbb Z_{25}, Z4×Z5×Z5\mathbb Z_4\times\mathbb Z_5\times\mathbb Z_5, Z2×Z2×Z25\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_{25}, and Z2×Z2×Z5×Z5\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_5\times\mathbb Z_5.

Elementary Divisors and Invariant Factors

Concept
Two canonical forms

A finite abelian group has two standard descriptions. The elementary divisor form is the Fundamental Theorem's decomposition into cyclic groups of prime-power order (Section 3). The invariant factor form writes G≅Zd1×Zd2×⋯×ZdsG \cong \mathbb Z_{d_1}\times\mathbb Z_{d_2}\times\cdots\times\mathbb Z_{d_s} with d1∣d2∣⋯∣dsd_1 \mid d_2 \mid \cdots \mid d_s (each divides the next) — also unique. To convert elementary divisors to invariant factors: group them by prime in decreasing order, pad any short prime-list with 11s so every prime has the same number of entries, then multiply straight across each column; the column products, in increasing order, are the invariant factors.

The two forms carry the same information, read two different ways — much like a number written as its prime factorization versus as its list of divisors. The largest invariant factor dsd_s is always the EXPONENT of the group: the largest possible element order, since it is built from the largest power of every prime at once (ties back to the order formula of Section 1). Invariant factors are often the more useful form for comparing two groups at a glance, since G≅HG\cong H exactly when their invariant factor lists match term for term — no regrouping by prime is needed once you already have them.

Example
From elementary divisors to invariant factors
3669432
Elementary divisors grouped by prime (9,3 and 4,2) multiply across to invariant factors 6 and 36.

A group has elementary divisors 4,9,2,34, 9, 2, 3 (order 4⋅9⋅2⋅3=2164\cdot9\cdot2\cdot3=216). Group by prime, largest first: prime 22 gives 4,24, 2; prime 33 gives 9,39, 3. Multiply across each column: 4⋅9=364\cdot9=36, then 2⋅3=62\cdot3=6. The invariant factors, in increasing order, are 66 and 3636, and indeed 6∣366\mid36. So G≅Z6×Z36G\cong\mathbb Z_6\times\mathbb Z_{36}, with 6⋅36=2166\cdot36=216 confirming the order.

Example
From invariant factors to elementary divisors

A group has invariant factors 22 and 6060 (with 2∣602\mid60), so G≅Z2×Z60G\cong\mathbb Z_2\times\mathbb Z_{60}, order 120120. Factor each invariant factor into prime powers: 2=212=2^1, and 60=22⋅3⋅5=4⋅3⋅560=2^2\cdot3\cdot5=4\cdot3\cdot5. Collecting all the prime powers gives the elementary divisors 4,2,3,54, 2, 3, 5 — note the TWO factors of 22, one from each invariant factor. Check: 4⋅2⋅3⋅5=1204\cdot2\cdot3\cdot5=120, matching 2⋅602\cdot60.

Example
A column with nothing to pair against

A group has elementary divisors 8,3,38, 3, 3 (order 8⋅3⋅3=728\cdot3\cdot3=72). Prime 22 contributes only one entry, 88; prime 33 contributes two, 33 and 33. Since prime 33 has more entries, prime 22's short list must be padded with a 11 to match: 8,18, 1. Multiplying across columns: 8⋅3=248\cdot3=24, then 1⋅3=31\cdot3=3. The invariant factors are 33 and 2424 (3∣243\mid24), so G≅Z3×Z24G\cong\mathbb Z_3\times\mathbb Z_{24}, order 3⋅24=723\cdot24=72 — matching, because the padding 11 was included.

Example
Using invariant factors to settle an isomorphism question

Are Z2×Z4×Z9\mathbb Z_2\times\mathbb Z_4\times\mathbb Z_9 and Z6×Z12\mathbb Z_6\times\mathbb Z_{12} isomorphic? Both have order 7272. For the first, elementary divisors 4,2,94,2,9 give invariant factors 22 and 3636 (padding prime 33's missing second entry with 11: columns 4⋅9=364\cdot9=36 and 2⋅1=22\cdot1=2). For the second, first split 6≅2×36\cong2\times3 and 12≅4×312\cong4\times3, giving elementary divisors 4,2,3,34,2,3,3 and invariant factors 66 and 1212 (columns 4⋅3=124\cdot3=12, 2⋅3=62\cdot3=6). The invariant factor lists {2,36}\{2,36\} and {6,12}\{6,12\} do not match, so the two groups are not isomorphic, despite having the same order.

Side note
Where it goes wrong

Forgetting to pad a prime's shorter list with 11s silently loses a whole invariant factor. For elementary divisors 8,2,38,2,3 (order 8⋅2⋅3=488\cdot2\cdot3=48), prime 22 has two entries (8,28,2) but prime 33 has only one (33); padding gives prime 33's list 3,13,1, and the columns multiply to 8⋅3=248\cdot3=24 and 2⋅1=22\cdot1=2 — invariant factors 2,242,24, product 4848, matching the group's order. Skipping the pad and writing just "Z24\mathbb Z_{24}" alone loses the second factor entirely: 24≠4824\ne48, an immediate contradiction with the group's actual order.

Try it
Try it: invariant factors from elementary divisors

Find the invariant factor decomposition of the group with elementary divisors 9,3,5,29, 3, 5, 2.

Show the answer

Group by prime: 33 gives 9,39,3; 55 gives 55 (padded to 5,15,1); 22 gives 22 (padded to 2,12,1). Columns: 9⋅5⋅2=909\cdot5\cdot2=90, then 3⋅1⋅1=33\cdot1\cdot1=3. The invariant factors are 33 and 9090 (3∣903\mid90), so G≅Z3×Z90G\cong\mathbb Z_3\times\mathbf{\mathbb Z_{90}}. Check: 9⋅3⋅5⋅2=270=3⋅909\cdot3\cdot5\cdot2=270=3\cdot90.

Why It Works

Let (g,h)∈G×H(g,h) \in G\times H with ord⁡(g)=m\operatorname{ord}(g)=m, ord⁡(h)=n\operatorname{ord}(h)=n, and let N=lcm⁡(m,n)N=\operatorname{lcm}(m,n). For any kk, (g,h)k=(gk,hk)=(eG,eH)(g,h)^k = (g^k,h^k) = (e_G,e_H) exactly when gk=eGg^k=e_G AND hk=eHh^k=e_H, i.e. exactly when m∣km \mid k and n∣kn \mid k, i.e. exactly when N∣kN \mid k (a common multiple of mm and nn is a multiple of m,nm,n together precisely when it is a multiple of their least common multiple). So the positive integers kk with (g,h)k=(eG,eH)(g,h)^k=(e_G,e_H) are exactly the multiples of NN, and the smallest one is NN itself: ord⁡(g,h)=N=lcm⁡(m,n)\operatorname{ord}(g,h) = N = \operatorname{lcm}(m,n).

Let G,HG,H be finite groups. (⇐\Leftarrow) Suppose G=⟨g⟩G=\langle g\rangle, H=⟨h⟩H=\langle h\rangle are cyclic with gcd⁡(∣G∣,∣H∣)=1\gcd(|G|,|H|)=1. By the order formula, ord⁡(g,h)=lcm⁡(∣G∣,∣H∣)=∣G∣∣H∣\operatorname{ord}(g,h) = \operatorname{lcm}(|G|,|H|) = |G||H| (a least common multiple with gcd 11 equals the plain product), which is exactly ∣G×H∣|G\times H|. An element whose order equals the order of the whole group generates it, so G×H=⟨(g,h)⟩G\times H = \langle(g,h)\rangle is cyclic. (⇒\Rightarrow) Suppose G×HG\times H is cyclic, generated by some (g,h)(g,h); then ord⁡(g,h)=∣G∣∣H∣\operatorname{ord}(g,h)=|G||H|. But ord⁡(g)∣∣G∣\operatorname{ord}(g) \mid |G| and ord⁡(h)∣∣H∣\operatorname{ord}(h)\mid|H| by Lagrange (Topic 5), so ord⁡(g,h)=lcm⁡(ord⁡(g),ord⁡(h))≤lcm⁡(∣G∣,∣H∣)≤∣G∣∣H∣\operatorname{ord}(g,h)=\operatorname{lcm}(\operatorname{ord}(g),\operatorname{ord}(h)) \le \operatorname{lcm}(|G|,|H|) \le |G||H|, with equality throughout forcing ord⁡(g)=∣G∣\operatorname{ord}(g)=|G|, ord⁡(h)=∣H∣\operatorname{ord}(h)=|H| (so G,HG,H are cyclic, generated by g,hg,h) and lcm⁡(∣G∣,∣H∣)=∣G∣∣H∣\operatorname{lcm}(|G|,|H|)=|G||H|, which happens only when gcd⁡(∣G∣,∣H∣)=1\gcd(|G|,|H|)=1.

If a prime qq divides ∣G∣|G| for a finite abelian group GG, then GG has an element of order qq. Proof by strong induction on ∣G∣|G|. If ∣G∣=q|G|=q, any g≠eg\ne e has order dividing qq (Lagrange) and order ≠1\ne1, so order qq. Otherwise pick any g≠eg\ne e. If q∣ord⁡(g)q \mid \operatorname{ord}(g), then by the order formula for a cyclic group (Topic 3), gord⁡(g)/qg^{\operatorname{ord}(g)/q} has order ord⁡(g)/(ord⁡(g)/q)=q\operatorname{ord}(g)/(\operatorname{ord}(g)/q) = q, done. If q∤ord⁡(g)q \nmid \operatorname{ord}(g), let N=⟨g⟩N=\langle g\rangle, automatically normal since GG is abelian, and form the quotient G/NG/N (Topic 7), of order ∣G∣/ord⁡(g)<∣G∣|G|/\operatorname{ord}(g) < |G|. Since ∣G∣=ord⁡(g)⋅∣G/N∣|G|=\operatorname{ord}(g)\cdot|G/N| and q∣∣G∣q \mid |G| but q∤ord⁡(g)q \nmid \operatorname{ord}(g), Euclid's lemma forces q∣∣G/N∣q \mid |G/N|. By induction G/NG/N has an element Ng′Ng' of order qq. Let d=ord⁡(g′)d=\operatorname{ord}(g') in GG; then (Ng′)d=N(Ng')^d = N, so dd is a multiple of ord⁡(Ng′)=q\operatorname{ord}(Ng')=q (the order of a coset divides any exponent that collapses it to the identity coset). Writing d=qd1d=qd_1, the element (g′)d1(g')^{d_1} has order d/gcd⁡(d,d1)=d/d1=qd/\gcd(d,d_1) = d/d_1 = q, again by the cyclic order formula. Either way, GG has an element of order qq.

Let GG be a finite abelian group of order nn, and single out one prime p∣np \mid n, writing n=pamn=p^am with p∤mp\nmid m. Let Gp={g∈G:ord⁡(g) is a power of p}G_p=\{g\in G:\operatorname{ord}(g)\text{ is a power of }p\} and Gm={g∈G:ord⁡(g)∣m}G_m=\{g\in G:\operatorname{ord}(g)\mid m\}. Each is a subgroup: if g,h∈Gpg,h\in G_p then, GG being abelian, ord⁡(gh)\operatorname{ord}(gh) divides lcm⁡(ord⁡(g),ord⁡(h))\operatorname{lcm}(\operatorname{ord}(g),\operatorname{ord}(h)) (a short direct check: (gh)lcm⁡=glcm⁡hlcm⁡=e(gh)^{\operatorname{lcm}}=g^{\operatorname{lcm}}h^{\operatorname{lcm}}=e), itself a power of pp, and ord⁡(g−1)=ord⁡(g)\operatorname{ord}(g^{-1})=\operatorname{ord}(g); the same argument with mm in place of pap^a handles GmG_m. Since gcd⁡(pa,m)=1\gcd(p^a,m)=1, Bezout gives integers x,yx,y with xpa+ym=1xp^a+ym=1. For any g∈Gg\in G, g=gxpa+ym=gxpagymg=g^{xp^a+ym}=g^{xp^a}g^{ym}. Now (gym)pa=gympa=(gn)y=e(g^{ym})^{p^a} = g^{ymp^a} = (g^{n})^{y} = e since ord⁡(g)∣n\operatorname{ord}(g)\mid n (Lagrange), so gym∈Gpg^{ym}\in G_p; likewise (gxpa)m=gxpam=(gn)x=e(g^{xp^a})^m = g^{xp^am} = (g^n)^x = e, so gxpa∈Gmg^{xp^a}\in G_m. So G=GpGmG=G_pG_m: every element is a product of one piece from each. If g∈Gp∩Gmg\in G_p\cap G_m, then ord⁡(g)\operatorname{ord}(g) divides both pap^a and mm, hence divides gcd⁡(pa,m)=1\gcd(p^a,m)=1, forcing g=eg=e: the intersection is trivial. A group that is the product of two subgroups meeting only at ee, with every element of one commuting with every element of the other (automatic here, since GG is abelian), is their internal direct product, so G≅Gp×GmG\cong G_p\times G_m. Finally, if some prime q≠pq\ne p divided ∣Gp∣|G_p|, the lemma above (applied to the abelian group GpG_p) would produce an element of GpG_p of order qq — impossible, since every element of GpG_p has pp-power order. So ∣Gp∣|G_p| has no prime factor besides pp, i.e. ∣Gp∣=pb|G_p|=p^b for some b≤ab\le a; likewise ∣Gm∣|G_m| has no factor of pp. Since ∣Gp∣∣Gm∣=∣G∣=pam|G_p||G_m|=|G|=p^am and ∣Gm∣|G_m| contributes no factor of pp, b=ab=a: ∣Gp∣=pa|G_p|=p^a exactly. Repeating this argument on GmG_m for its next prime (induction on the number of distinct primes dividing nn) splits GG completely into one primary component per prime, G≅Gp1×⋯×GpkG\cong G_{p_1}\times\cdots\times G_{p_k} with ∣Gpi∣=piai|G_{p_i}|=p_i^{a_i}.

Let GG be a finite abelian pp-group and let a∈Ga\in G have maximal order pcp^c among all elements of GG (every other order is a smaller power of pp, hence divides pcp^c). Then there is a subgroup K≤GK\le G with G=⟨a⟩×KG=\langle a\rangle\times K. Proof by strong induction on ∣G∣|G|. If G=⟨a⟩G=\langle a\rangle, take K={e}K=\{e\}. Otherwise pick b∉⟨a⟩b\notin\langle a\rangle of minimal order pdp^d among elements outside ⟨a⟩\langle a\rangle (such bb exists since G≠⟨a⟩G\ne\langle a\rangle). Then bpb^p has order pd−1<pdp^{d-1}<p^d, so minimality forces bp∈⟨a⟩b^p\in\langle a\rangle, say bp=akb^p=a^k. Since d≤cd\le c, ord⁡(b)=pd\operatorname{ord}(b)=p^d divides pcp^c, so bpc=eb^{p^c}=e; raising bp=akb^p=a^k to the power pc−1p^{c-1} gives e=bpc=akpc−1e=b^{p^c}=a^{kp^{c-1}}, so pc∣kpc−1p^c \mid kp^{c-1}, forcing p∣kp\mid k. Write k=pk′k=pk' and set c0=b a−k′c_0=b\,a^{-k'}. Then c0p=bpa−k=aka−k=ec_0^p = b^pa^{-k} = a^ka^{-k}=e, so ord⁡(c0)∈{1,p}\operatorname{ord}(c_0)\in\{1,p\}; it cannot be 11 (else b=ak′∈⟨a⟩b=a^{k'}\in\langle a\rangle, contradiction), so ord⁡(c0)=p\operatorname{ord}(c_0)=p, and c0∉⟨a⟩c_0\notin\langle a\rangle for the same reason. Since ord⁡(c0)=p\operatorname{ord}(c_0)=p is prime, ⟨a⟩∩⟨c0⟩\langle a\rangle\cap\langle c_0\rangle is a subgroup of ⟨c0⟩\langle c_0\rangle that cannot be all of it (that would put c0∈⟨a⟩c_0\in\langle a\rangle), so it is trivial. Now pass to the quotient Gˉ=G/⟨c0⟩\bar G=G/\langle c_0\rangle (normal since GG is abelian), of order ∣G∣/p<∣G∣|G|/p<|G|. Because ⟨a⟩∩⟨c0⟩={e}\langle a\rangle\cap\langle c_0\rangle=\{e\}, the quotient map is injective on ⟨a⟩\langle a\rangle, so aˉ\bar a has order pcp^c in Gˉ\bar G; and for any g∈Gg\in G, gpc=eg^{p^c}=e gives gˉ pc=eˉ\bar g^{\,p^c}=\bar e, so ord⁡(gˉ)\operatorname{ord}(\bar g) divides pcp^c too — hence aˉ\bar a has maximal order pcp^c in Gˉ\bar G as well. By induction, Gˉ=⟨aˉ⟩×Kˉ\bar G=\langle\bar a\rangle\times\bar K for some Kˉ≤Gˉ\bar K\le\bar G. Let KK be the preimage of Kˉ\bar K in GG. If ai∈Ka^i\in K then aˉ i∈Kˉ∩⟨aˉ⟩={eˉ}\bar a^{\,i}\in\bar K\cap\langle\bar a\rangle=\{\bar e\}, so ai∈⟨c0⟩a^i\in\langle c_0\rangle; combined with ai∈⟨a⟩a^i\in\langle a\rangle and ⟨a⟩∩⟨c0⟩={e}\langle a\rangle\cap\langle c_0\rangle=\{e\}, this gives ai=ea^i=e — so ⟨a⟩∩K={e}\langle a\rangle\cap K=\{e\}. And for any g∈Gg\in G, gˉ=aˉ ikˉ\bar g=\bar a^{\,i}\bar k for some ii and some k∈Kk\in K, so g(aik)−1∈⟨c0⟩⊆Kg(a^ik)^{-1}\in\langle c_0\rangle\subseteq K, giving g∈⟨a⟩Kg\in\langle a\rangle K — so G=⟨a⟩KG=\langle a\rangle K. With trivial intersection, elementwise commuting (abelian), and ⟨a⟩K=G\langle a\rangle K=G, G≅⟨a⟩×KG\cong\langle a\rangle\times K.

Let GG be a finite abelian group of order pap^a. By strong induction on aa: if a=0a=0, G={e}G=\{e\}, the empty product. Otherwise let x∈Gx\in G have maximal order pcp^c (c≥1c\ge1, since GG has an element of order pp by Cauchy's lemma above). By the splitting lemma, G=⟨x⟩×KG=\langle x\rangle\times K for some K≤GK\le G with ∣K∣=pa−c<pa|K|=p^{a-c}<p^a. By induction, KK is itself a direct product of cyclic groups of pp-power order, so G≅Zpc×KG\cong\mathbb Z_{p^c}\times K exhibits GG as such a product too. Combined with the primary decomposition above, every finite abelian group is a direct product of cyclic groups of prime-power order — existence in the Fundamental Theorem.

Suppose G≅Zpλ1×⋯×ZpλrG\cong\mathbb Z_{p^{\lambda_1}}\times\cdots\times\mathbb Z_{p^{\lambda_r}} with λ1≥⋯≥λr≥1\lambda_1\ge\cdots\ge\lambda_r\ge1, a pp-group. For each j≥0j\ge0, let f(j)=∣{g∈G:gpj=e}∣f(j)=|\{g\in G : g^{p^j}=e\}| — a count depending only on GG itself, not on any chosen decomposition. Inside a cyclic factor Zpλi\mathbb Z_{p^{\lambda_i}}, the solutions of pjx≡0p^jx\equiv0 form the subgroup of order pmin⁡(j,λi)p^{\min(j,\lambda_i)} (Topic 3's subgroup count, with d=min⁡(j,λi)d=\min(j,\lambda_i)). Across the whole product, f(j)=∏ipmin⁡(j,λi)=p g(j)f(j)=\prod_i p^{\min(j,\lambda_i)} = p^{\,g(j)} where g(j)=∑imin⁡(j,λi)g(j)=\sum_i\min(j,\lambda_i). The increment d(j)=g(j)−g(j−1)=∑i(min⁡(j,λi)−min⁡(j−1,λi))d(j)=g(j)-g(j-1)=\sum_i\big(\min(j,\lambda_i)-\min(j-1,\lambda_i)\big) equals the number of λi\lambda_i with λi≥j\lambda_i\ge j, since each term is 11 if λi≥j\lambda_i\ge j and 00 otherwise. Since ff (hence gg, hence dd) is computable from GG alone, the sequence d(1),d(2),…d(1),d(2),\ldots — the number of parts ≥j\ge j, for every jj — is an invariant of GG. But that sequence determines the partition completely: the number of λi\lambda_i equal to exactly jj is d(j)−d(j+1)d(j)-d(j+1). So the multiset {λ1,…,λr}\{\lambda_1,\ldots,\lambda_r\}, and hence the elementary divisors pλ1,…,pλrp^{\lambda_1},\ldots,p^{\lambda_r}, are the same for every decomposition of GG of this form — they depend only on GG, not on how it happened to be written. The same argument applies independently at each prime dividing ∣G∣|G|, so the full list of elementary divisors across all primes is unique.