Section 11 Direct Products and Finitely Generated Abelian Groups

NOTE

| Operation | Notes | Example | | --- | --- | --- | | external direct product | ${\prod }_{i=1}^n{G}_i$ | ${\mathbb{Z}}_2\times {\mathbb{Z}}_3=(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)$ | | internal direct product | $G\cong H\times K$| ${\mathbb{Z}}_6\cong {\mathbb{Z}}_2\times {\mathbb{Z}}_3$|

Aspect	External Direct Product $H\times K$	Internal Direct Product in G
Starting data	Two separate groups HH, KK.	One big group G with subgroups $H,K\subseteq G$.
Underlying set	Ordered pairs $(h,k):h\in H,k\in K$.	Just the elements of G.
Group law	Component-wise: $(G,\ast ),(H,+)$ $(g_1,h_1)\times (g_2,h_2)=(g_1\ast g_2,h_1+h_2)$.	The original multiplication in G.
Construction	“External”: you build a brand-new group out of H and K.	“Internal”: you recognise that $G\cong H\times K$ because G splits as H K with $H\cap K=\{e\}$ and $hk=kh$.
Universality	Always exists for any two groups.	Only when within G those two subgroups satisfy the three internal-product axioms.
Isomorphism	$H\times K$ is the group.	You get an isomorphism $\phi :H\times K\stackrel{\sim }{\to }G,\phi (h,k)=hk$.

11.3 Example

Example

${\mathbb{Z}}_2\times {\mathbb{Z}}_3$ is cyclic with Generator $(1,1)$

Question

Is it true that ${\mathbb{Z}}_n\times {\mathbb{Z}}_m$ is cyclic if and only if n and m are coprime to each other?

It is true Case 1: $gcm(n,m)=1$ pick the Generator $(1,1)$ :$(1,1{)}^k=(k\mathrm{mod}n,k\mathrm{mod}m)=(0,0)$which means: $k=lcm(n,m)$ $k=lcm(n,m)=nm=|{\mathbb{Z}}_n\times {\mathbb{Z}}_m|$ meaning $(1,1)$ is a generator of $Z_{nm}$. Case 2: $gcm(n,m)>1$ For any element $a,b\in {\mathbb{Z}}_n\times {\mathbb{Z}}_m$ $ord(a,b)=lcm(ord(a),ord(b))$ according to the Theorem of Lagrange (The Theorem of Lagrange): $ord(a)\mid n,ord(b)\mid m$then: $ord(a,b)=lcm(ord(a),ord(b))\le lcm(n,m)<nm$This means no generator can generate the full ${\mathbb{Z}}_n\times {\mathbb{Z}}_m$ Thus it is not cyclic . $q.e.d.$

11.10 Example

Example

Find the order of (8, 4, 10) in the group ${\mathbb{Z}}_{12}\times {\mathbb{Z}}_{60}\times {\mathbb{Z}}_{24}$.

• 8 is of order 3 in ${\mathbb{Z}}_{12}$ : $3=12/gcd(8,12)$ (6.14 Theorem)
• 4 is of order 15 in ${\mathbb{Z}}_{60}$
• 10 is of order 12 in ${\mathbb{Z}}_{24}$ $lcm(3,12,15)=60$ (8,4,10) has an order of 60 in the Group ${\mathbb{Z}}_{12}\times {\mathbb{Z}}_{60}\times {\mathbb{Z}}_{24}$

11.12 Fundamental Theorem of Finitely Generated Abelian Groups

Fundamental Theorem of Finitely Generated Abelian Groups

Every finitely generated abelian group G is isomorphic to a direct product of cyclic groups in the form ${\mathbb{Z}}_{(p_1{)}^{r_1}}\times {\mathbb{Z}}_{(p_2{)}^{r_2}}\times \cdots \times {\mathbb{Z}}_{(p_n{)}^{r_n}}\times \mathbb{Z}\times \mathbb{Z}\times \cdots \times \mathbb{Z}$ where $p_i$ are primes, and $r_i$ are positive integers.

The direct product is unique except for possible rearrangement of the factors; that is, the number (Betti number of G) of factors $\mathbb{Z}$ is unique and the prime powers $(p_i{)}^{r_i}$ are unique.

Tip

Similar to factorizing an integer into prime powers, but $2^3\ne 2^2\times 2$

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=114&selection=42,0,43,1|p.109]]

The proof is omitted here.

LOL

Isomorphism of Cyclic Group

NOTE

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

NOTE

In general we have ${\mathbb{Z}}_n\times {\mathbb{Z}}_m\simeq {\mathbb{Z}}_{gcd(n,m)}\times {\mathbb{Z}}_{lcm(a,b)}$

11.13 Example

Example

Find all abelian groups, up to isomorphism, of order 360.

up to isomorphism: any abelian groups of order 360 should be isomorphic to one of the founded groups

make use of 11.12 Fundamental Theorem of Finitely Generated Abelian Groups $360=2^3{3}^{2}5$

1. ${\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_3\times {\mathbb{Z}}_3\times {\mathbb{Z}}_5$
2. ${\mathbb{Z}}_2\times {\mathbb{Z}}_4\times {\mathbb{Z}}_3\times {\mathbb{Z}}_3\times {\mathbb{Z}}_5$
3. ${\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_9\times {\mathbb{Z}}_5$
4. ${\mathbb{Z}}_2\times {\mathbb{Z}}_4\times {\mathbb{Z}}_9\times {\mathbb{Z}}_5$
5. ${\mathbb{Z}}_8\times {\mathbb{Z}}_3\times {\mathbb{Z}}_3\times {\mathbb{Z}}_5$
6. ${\mathbb{Z}}_8\times {\mathbb{Z}}_9\times {\mathbb{Z}}_5$

11.14 Definition

Note

A group G is decomposable if it is isomorphic to a direct product of two proper nontrivial subgroups. Otherwise G is indecomposable

11.15 Theorem

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=114&selection=238,0,239,17|p.109]]

The finite indecomposable abelian groups are exactly the cyclic groups with order a power of a prime.

Examples ${\mathbb{Z}}_2\times {\mathbb{Z}}_2\cong V_4$ (indecomposable)

Group G	cyclic	order	decomposable	conclusion
${\mathbb{Z}}_4$	✓	$2^2$	indecomposable	indecomposable
${\mathbb{Z}}_2\times {\mathbb{Z}}_2$	✗	$2^2$	Already decomposed	decomposable
${\mathbb{Z}}_6$	✓	$2\cdot 3$	$\cong {\mathbf{Z}}_2\times {\mathbf{Z}}_3$	decomposable
${\mathbb{Z}}_{15}$	✓	$3\cdot 5$	$\cong {\mathbf{Z}}_3\times {\mathbf{Z}}_5$	decomposable

Suppose ${\mathbb{Z}}_{p^r}$ could decompose as ${\mathbb{Z}}_{p^i}\times {\mathbb{Z}}_{p^j}$ with $i,j\ge 1$ and $i+j=r$. In that product, the order of any element is at most $lcm(p^i,p^j)=p^{\max (i,j)}<p^r$ (11.3 Example). But ${\mathbb{Z}}_{p^r}$ itself contains an element of order $p^r$. They cannot be isomorphic, so ${\mathbb{Z}}_{p^r}$ is indecomposable.

11.16 Theorem

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=114&selection=320,1,337,1&color=yellow|p.109]]

If m divides the order of a finite abelian group G, then G has a subgroup of order m.

Warning

Proof not understood

11.17 Theorem

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=115&selection=303,0,316,10&color=note|p.110]]

If m is a square free integer, that is, m is not divisible by the square of any prime, then every abelian group of order m is cyclic.

extension to 11.3 Example

$G={\mathbb{Z}}_{p_1}\times {\mathbb{Z}}_{p_2}\times \cdots \times {\mathbb{Z}}_{p_n}$ primes are coprime to each other. G is cyclic.