PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (194)
Orphanage
Unclass'd
Unproven (498)
Corrections (25)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: High Entry average rating: No information on entry rating
[parent] abelian groups of order $120$ (Example)

Here we present an application of the fundamental theorem of finitely generated abelian groups.

Example (Abelian groups of order $ 120$):

Let $ G$ be an abelian group of order $ n=120$. Since the group is finite it is obviously finitely generated, so we can apply the theorem. There exist $ n_1,n_2,\ldots,n_s$ with

$\displaystyle G\cong \mathbb{Z}/n_1\mathbb{Z}\oplus\mathbb{Z}/n_2\mathbb{Z}\oplus\ldots\oplus\mathbb{Z}/n_s\mathbb{Z}$
$\displaystyle \forall i, n_i\geq 2;\quad n_{i+1}\mid n_i\ $   for $\displaystyle 1\leq i\leq s-1$
Notice that in the case of a finite group, $ r$, as in the statement of the theorem, must be equal to 0. We have
$\displaystyle n=120=2^3\cdot3\cdot5=\prod_{i=1}^s n_i=n_1\cdot n_2\cdot \ldots \cdot n_s$
and by the divisibility properties of $ n_i$ we must have that every prime divisor of $ n$ must divide $ n_1$. Thus the possibilities for $ n_1$ are the following
$\displaystyle 2\cdot 3\cdot 5,\quad 2^2\cdot 3 \cdot 5,\quad 2^3\cdot 3\cdot 5$
If $ n_1=2^3\cdot 3\cdot 5=120$ then $ s=1$. In the case that $ n_1=2^2\cdot 3 \cdot 5$ then $ n_2=2$ and $ s=2$. It remains to analyze the case $ n_1=2\cdot 3 \cdot 5$. Now the only possibility for $ n_2$ is $ 2$ and $ n_3=2$ as well.

Hence if $ G$ is an abelian group of order $ 120$ it must be (up to isomorphism) one of the following:

$\displaystyle \mathbb{Z}/120\mathbb{Z},\quad \mathbb{Z}/60\mathbb{Z}\oplus \mat... ...\mathbb{Z}/30\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$
Also notice that they are all non-isomorphic. This is because
$\displaystyle \mathbb{Z}/(n\cdot m)\mathbb{Z}\cong \mathbb{Z}/n\mathbb{Z}\oplus \mathbb{Z}/m\mathbb{Z} \Leftrightarrow \operatorname{gcd}(n,m)=1$
which is due to the Chinese Remainder theorem.



"abelian groups of order $120$" is owned by alozano.
(view preamble)

View style:

See Also: fundamental theorem of finitely generated abelian groups, abelian group


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: Chinese remainder theorem, isomorphism, divide, prime divisor, properties, divisibility, finite group, finitely generated, finite, group, order, abelian groups, fundamental theorem of finitely generated abelian groups, application

This is version 2 of abelian groups of order $120$, born on 2003-08-26, modified 2004-03-26.
Object id is 4654, canonical name is AbelianGroupsOfOrder120.
Accessed 3110 times total.

Classification:
AMS MSC: 20E34 (Group theory and generalizations :: Structure and classification of infinite or finite groups :: General structure theorems)
Pending Errata and Addenda
None.
[ View all 1 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add example | add (any)