Chapter 07

Medium

Direct Products & Direct Sums

00 · Symbol Glossary

$G \times H$External direct product

The set of ordered pairs $\{(g,h) : g\in G, h\in H\}$ , with operation $(g_1,h_1)(g_2,h_2) := (g_1g_2, h_1h_2)$ — combine in each coordinate independently. The most direct way to build a new, larger group from two known ones.

$(g,h)$An element of a direct product

An ordered pair, one coordinate from each factor. Two pairs are equal exactly when both coordinates match — no interaction between the coordinates is assumed.

$\mathrm{lcm}(a,b)$Least common multiple

The smallest positive integer divisible by both $a$ and $b$ . Governs exactly the order of an element in a direct product, as Theorem 7.1 shows.

$G \oplus H$Direct sum

Alternate notation for $G\times H$ , used almost exclusively when both groups are abelian (and written additively). For two factors, $\oplus$ and $\times$ denote the identical construction — the different symbol is a stylistic signal, not a different operation.

$HK$Product set of two subgroups (recalled)

$\{hk : h\in H, k\in K\}$ , from Chapter 06. Central to recognizing when a group already splits as a direct product internally, without building one from scratch.

01 · The External Direct Product

Definition — Direct Product

For groups $G,H$ , the direct product $G\times H$ is the set of ordered pairs $\{(g,h):g\in G,h\in H\}$ with operation:

(g_1,h_1)\ast(g_2,h_2) := (g_1g_2,\, h_1h_2)

(each coordinate combined using its own group's operation).

Theorem 7.1 — G × H Is a Group

$G\times H$ is a group, with identity $(e_G,e_H)$ , inverse $(g,h)^{-1}=(g^{-1},h^{-1})$ , and $|G\times H| = |G|\cdot|H|$ .

Proof

Closure: coordinatewise, $g_1g_2\in G$ and $h_1h_2\in H$ , so $(g_1g_2,h_1h_2)\in G\times H$ . Associativity: each coordinate's computation is associative independently, so the pair-wise operation is too. Identity: $(g,h)\ast(e_G,e_H) = (ge_G,he_H) = (g,h)$ , and likewise on the left. Inverses: $(g,h)\ast(g^{-1},h^{-1}) = (gg^{-1},hh^{-1}) = (e_G,e_H)$ , and likewise on the left. Order: distinct pairs correspond exactly to independent choices of a $G$ -coordinate ( $|G|$ options) and an $H$ -coordinate ( $|H|$ options), giving $|G|\cdot|H|$ total pairs.

Example — $\mathbb{Z}_2 \times \mathbb{Z}_3$

Elements: $(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)$ — six pairs, matching $|\mathbb{Z}_2\times\mathbb{Z}_3| = 2\times3=6$ . Operation is coordinatewise addition, each mod its own modulus: $(1,1)+(1,2) = (1+1 \bmod 2,\ 1+2\bmod 3) = (0,0)$ .

02 · Order of an Element in a Direct Product

Theorem 7.2 — Order Formula for Direct Products

For $(g,h) \in G\times H$ with $|g|,|h|$ both finite: $|(g,h)| = \mathrm{lcm}(|g|,|h|)$ .

Proof

$(g,h)^k = (g^k,h^k)$ by definition of the coordinatewise operation, repeated $k$ times. So $(g,h)^k = (e_G,e_H)$ exactly when $g^k=e_G$ and $h^k=e_H$ simultaneously.

$g^k=e_G$ holds precisely when $|g|$ divides $k$ (a standard fact: powers of $g$ that return to the identity are exactly the multiples of $|g|$ , else the division algorithm applied to $k$ against $|g|$ would produce a smaller positive power equal to $e$ , contradicting minimality of $|g|$ ). Similarly $h^k=e_H$ iff $|h|$ divides $k$ .

So $(g,h)^k=(e_G,e_H)$ iff $k$ is a common multiple of $|g|$ and $|h|$ . The smallest positive such $k$ is, by definition, $\mathrm{lcm}(|g|,|h|)$ .

Step-by-step — Order of $(2,1)$ in $\mathbb{Z}_4 \times \mathbb{Z}_6$

Find $|2|$ in $\mathbb{Z}_4$ :

1\cdot2=2

2\cdot2=4\equiv0

. So

|2|=2

Find $|1|$ in $\mathbb{Z}_6$ :

1

generates all of

\mathbb{Z}_6

(Chapter 03), so

|1|=6

Apply Theorem 7.2:

|(2,1)| = \mathrm{lcm}(2,6)

Compute the lcm: multiples of 2:

2,4,6,8,\ldots

; multiples of 6:

6,12,\ldots

; smallest common one is

6

Conclusion:

|(2,1)| = 6

\mathbb{Z}_4\times\mathbb{Z}_6

Common mistake — Guessing the order is the product $|g|\cdot|h|$

For $(2,1)$ above, $|g|\cdot|h| = 2\times6=12 \neq 6=\mathrm{lcm}(2,6)$ . The product overcounts whenever $|g|$ and $|h|$ share a common factor. Only when $\gcd(|g|,|h|)=1$ does $\mathrm{lcm}(|g|,|h|)=|g|\cdot|h|$ — exactly the coprimality condition that governs the next theorem.

03 · When Is $\mathbb{Z}_m \times \mathbb{Z}_n$ Cyclic?

Theorem 7.3 — $\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn}$ iff $\gcd(m,n)=1$

$\mathbb{Z}_m \times \mathbb{Z}_n$ is cyclic (hence isomorphic to $\mathbb{Z}_{mn}$ ) if and only if $\gcd(m,n)=1$ .

Proof

( $\Leftarrow$ ) Suppose $\gcd(m,n)=1$ . Consider $(1,1)\in\mathbb{Z}_m\times\mathbb{Z}_n$ . By Theorem 7.2, $|(1,1)| = \mathrm{lcm}(|1|,|1|) = \mathrm{lcm}(m,n)$ . A standard identity relates lcm and gcd: $\mathrm{lcm}(m,n)\cdot\gcd(m,n) = mn$ , so $\mathrm{lcm}(m,n) = mn/\gcd(m,n) = mn/1 = mn$ . So $|(1,1)| = mn = |\mathbb{Z}_m\times\mathbb{Z}_n|$ (by Theorem 7.1's order formula). By Theorem 3.4, $|\langle(1,1)\rangle| = mn$ , matching the size of the whole group — so $(1,1)$ generates everything, and $\mathbb{Z}_m\times\mathbb{Z}_n$ is cyclic. Any two cyclic groups of the same finite order are isomorphic (both are, in structure, "the integers mod that order," matching generators to generators), so $\mathbb{Z}_m\times\mathbb{Z}_n \cong \mathbb{Z}_{mn}$ .

( $\Rightarrow$ ) Suppose $d=\gcd(m,n)>1$ . For any $(a,b)\in\mathbb{Z}_m\times\mathbb{Z}_n$ , the order of $a$ divides $m$ and the order of $b$ divides $n$ , so by Theorem 7.2, $|(a,b)| = \mathrm{lcm}(|a|,|b|)$ divides $\mathrm{lcm}(m,n) = mn/d < mn$ (strict, since $d>1$ ). So no element of $\mathbb{Z}_m\times\mathbb{Z}_n$ has order $mn$ , meaning no element generates the whole group (order $mn$ ) — the group is not cyclic. Since $\mathbb{Z}_{mn}$ is cyclic, and isomorphic groups share the property "is cyclic," $\mathbb{Z}_m\times\mathbb{Z}_n \not\cong \mathbb{Z}_{mn}$ .

Example — Distinguishing $\mathbb{Z}_4$ from $\mathbb{Z}_2 \times \mathbb{Z}_2$

Both have order 4, resolving Exercise 5.2's open question. Since $\gcd(2,2)=2\neq1$ , Theorem 7.3 says $\mathbb{Z}_2\times\mathbb{Z}_2$ is not cyclic — confirmed directly: every nonidentity element $(1,0),(0,1),(1,1)$ has order $2$ (e.g. $(1,1)+(1,1)=(0,0)$ ), never order $4$ . Meanwhile $\mathbb{Z}_4$ has the element $1$ of order $4$ . Two genuinely different groups of order 4 — these are, in fact, the only two groups of order 4 up to isomorphism, a small case of the classification machinery developed further in Chapter 09.

04 · Recognizing an Internal Direct Product

Building $G\times H$ from scratch is one direction. The reverse question — does a given group $G$ secretly split apart as a direct product of two of its own subgroups? — is often more useful.

Theorem 7.4 — Internal Direct Product Criterion

Let $H,K\trianglelefteq G$ with $H\cap K=\{e\}$ and $HK=G$ . Then:

G \;\cong\; H\times K

Proof

Step 1 — elements of $H$ and $K$ commute. For $h\in H,k\in K$ , consider the element $x=hkh^{-1}k^{-1}$ . Since $K\trianglelefteq G$ , $hkh^{-1}\in K$ , so $x=(hkh^{-1})k^{-1} \in K$ . Since $H\trianglelefteq G$ , $kh^{-1}k^{-1}\in H$ , so $x = h(kh^{-1}k^{-1}) \in H$ . So $x\in H\cap K=\{e\}$ , giving $hkh^{-1}k^{-1}=e$ , i.e. $hk=kh$ .

Step 2 — define $\varphi: H\times K \to G$ by $\varphi(h,k)=hk$ .

Homomorphism: $\varphi(h_1,k_1)\varphi(h_2,k_2) = h_1k_1h_2k_2 = h_1h_2k_1k_2$ (swapping $k_1,h_2$ using Step 1) $=\varphi(h_1h_2,k_1k_2) = \varphi\big((h_1,k_1)(h_2,k_2)\big)$ .

Injective: if $\varphi(h,k)=e$ , then $hk=e$ , so $h=k^{-1}$ . But $h\in H$ and $k^{-1}\in K$ , so $h\in H\cap K=\{e\}$ , giving $h=e$ and then $k=h^{-1}=e$ . So $\ker\varphi=\{(e,e)\}$ ; by Theorem 6.4, injective.

Surjective: $HK=G$ means every $g\in G$ is $hk$ for some $h\in H,k\in K$ — exactly $\varphi(h,k)$ .

$\varphi$ is a bijective homomorphism: an isomorphism. $G\cong H\times K$ .

Example — $\mathbb{Z}_6$ as an internal direct product

In $\mathbb{Z}_6$ , let $H=\{0,3\}$ (order 2) and $K=\{0,2,4\}$ (order 3), both normal (abelian group, Chapter 05). $H\cap K = \{0\}$ (their only shared element). $HK$ : sums $h+k$ for $h\in H,k\in K$ give $0,2,4,3,5,7\equiv1$ — all six elements of $\mathbb{Z}_6$ , so $HK=\mathbb{Z}_6$ . By Theorem 7.4:

\mathbb{Z}_6 \;\cong\; \mathbb{Z}_2\times\mathbb{Z}_3

consistent with Theorem 7.3, since $\gcd(2,3)=1$ .

Common mistake — Forgetting to check the trivial-intersection condition

Take $G=\mathbb{Z}_4$ , $H=\{0,2\}$ , $K=\{0,2\}$ (the same subgroup twice). $HK = \{0,2\}=H\neq G$ already fails the second condition, but even setting that aside, $H\cap K = \{0,2\} \neq \{0\}$ — the trivial-intersection condition fails badly (they're not even distinct subgroups). Theorem 7.4 requires both conditions; a group failing either does not split as the corresponding direct product, and indeed $\mathbb{Z}_4 \not\cong \mathbb{Z}_2\times\mathbb{Z}_2$ by Theorem 7.3's proof (Section 03).

05 · Direct Sums: A Note on Notation

Same construction, different symbol

For exactly two factors, $G\oplus H$ and $G\times H$ describe the identical group — the same set of pairs, the same coordinatewise operation. The $\oplus$ symbol is conventionally reserved for abelian groups (almost always written additively), signaling to the reader "these pieces combine independently, coordinate by coordinate, with no cross-interaction," which is exactly the intuition behind ordinary addition. You will see $\mathbb{Z}_2\oplus\mathbb{Z}_2$ and $\mathbb{Z}_2\times\mathbb{Z}_2$ used interchangeably across different textbooks describing the exact same group from Section 03. The distinction matters more once infinitely many factors are combined — direct sums restrict to elements with only finitely many nonzero coordinates, while direct products allow all coordinates simultaneously — but for the finite constructions in this course, the two notations name the same object.

06 · Exercises

EXERCISE 7.1

Find the order of each coordinate separately first, then apply Theorem 7.2.

In $\mathbb{Z}_3$ : $|2|=3$ (since $2,4\equiv1,6\equiv0$ — smallest positive multiple of 2 hitting 0 mod 3 is at $k=3$ ). In $\mathbb{Z}_4$ : $|3|$ : $3,6\equiv2,9\equiv1,12\equiv0$ , so $|3|=4$ .

By Theorem 7.2, $|(2,3)| = \mathrm{lcm}(3,4) = 12$ .

Find $|(2,3)|$ in $\mathbb{Z}_3 \times \mathbb{Z}_4$ .

EXERCISE 7.2

Apply Theorem 7.3 directly — you only need to check $\gcd(6,10)$ .

$\gcd(6,10)=2\neq1$ . By Theorem 7.3, since the gcd is not 1, $\mathbb{Z}_6\times\mathbb{Z}_{10}$ is not cyclic, hence not isomorphic to $\mathbb{Z}_{60}$ .

Is $\mathbb{Z}_6\times\mathbb{Z}_{10} \cong \mathbb{Z}_{60}$ ? Justify using Theorem 7.3.

EXERCISE 7.3

Check the two conditions of Theorem 7.4 directly: is $H\cap K=\{0\}$ , and does $HK$ give every element of $\mathbb{Z}_{12}$ ?

$H=\langle4\rangle=\{0,4,8\}$ (order 3), $K=\langle6\rangle=\{0,6\}$ (order 2). Both are subgroups of the abelian group $\mathbb{Z}_{12}$ , hence normal.

$H\cap K$ : comparing $\{0,4,8\}$ and $\{0,6\}$ , the only shared element is $0$ . So $H\cap K=\{0\}$ . ✓

$HK$ : all sums $h+k$ : $0,4,8,6,10,14\equiv2$ — giving $\{0,2,4,6,8,10\}$ , only six elements, not all twelve. $HK\neq\mathbb{Z}_{12}$ . ✗

Since $HK\neq G$ , Theorem 7.4 does not apply — $\mathbb{Z}_{12}$ does not split as $H\times K$ for this particular choice. (This makes sense: $|H|\cdot|K|=3\times2=6\neq12=|G|$ , so $H\times K$ couldn't have matched $G$ 's order in the first place.)

In $\mathbb{Z}_{12}$ , let $H=\langle4\rangle$ and $K=\langle6\rangle$ . Check whether the hypotheses of Theorem 7.4 hold for $H,K$ .

EXERCISE 7.4

Try $H=\langle3\rangle$ (order 4) and $K=\langle4\rangle$ (order 3) instead — check $\gcd(4,3)$ first to predict the outcome via Theorem 7.3.

$H=\langle3\rangle = \{0,3,6,9\}$ (order 4, since $|3|=4$ by Theorem 3.4 and Section 04 of Chapter 03's divisor logic). $K=\langle4\rangle=\{0,4,8\}$ (order 3).

$H\cap K$ : comparing the two lists, only $0$ is shared. $H\cap K=\{0\}$ . ✓

$HK$ : since $|H|\cdot|K|=4\times3=12=|\mathbb{Z}_{12}|$ and $H\cap K$ is trivial, $HK$ must have exactly $12$ elements (no overlap-driven repeats), so $HK=\mathbb{Z}_{12}=G$ . ✓

Both conditions hold: $\mathbb{Z}_{12} \cong \mathbb{Z}_4\times\mathbb{Z}_3$ — consistent with $\gcd(4,3)=1$ and Theorem 7.3.

Find a pair of subgroups $H,K\le\mathbb{Z}_{12}$ that do satisfy Theorem 7.4's hypotheses, and state the resulting isomorphism.

07 · Chapter Summary

Concept	Statement
Direct product $G\times H$	Pairs $(g,h)$ ; coordinatewise operation; $
Order of $(g,h)$	$\mathrm{lcm}(\lvert g\rvert,\lvert h\rvert)$ (Thm 7.2)
$\mathbb{Z}_m\times\mathbb{Z}_n$ cyclic	Iff $\gcd(m,n)=1$ ; then $\cong\mathbb{Z}_{mn}$ (Thm 7.3)
Internal direct product	$H,K\trianglelefteq G$ , $H\cap K=\{e\}$ , $HK=G$ $\implies G\cong H\times K$ (Thm 7.4)
Direct sum $\oplus$	Same as $\times$ for two factors; conventional notation for abelian groups
$\mathbb{Z}_4$ vs $\mathbb{Z}_2\times\mathbb{Z}_2$	Same order, genuinely different groups (one cyclic, one not)

Next: Chapter 08 — Group Actions studies how a group can move the elements of an entirely different set, unifying symmetry, counting, and conjugation under a single framework that powers the Sylow theorems in Chapter 09.

00 · Symbol Glossary

01 · The External Direct Product

02 · Order of an Element in a Direct Product

03 · When Is Zm×Zn\mathbb{Z}_m \times \mathbb{Z}_nZm​×Zn​ Cyclic?

04 · Recognizing an Internal Direct Product

05 · Direct Sums: A Note on Notation

06 · Exercises

07 · Chapter Summary

03 · When Is $\mathbb{Z}_m \times \mathbb{Z}_n$ Cyclic?