Group Theory: Finite Abelian Groups - An element of order

  • #1
18
0

Homework Statement


Decide all abelian groups of order 675. Find an element of order 45 in each one of the groups, if it exists.

Homework Equations

/propositions/definitions[/B]
Fundamental Theorem of Finite Abelian Groups
Lagrange's Theorem and its corollaries (not sure if helpful for this problem)

The Attempt at a Solution


I used the Fundamental Theorem of Finite Abelian Groups to find the abelian groups. The prime factorization of 675 is
$$
\begin{split}
675 &= 3 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \\
& = 3^{2} \cdot 3 \cdot 5 \cdot 5 = 9 \cdot 3 \cdot 5 \cdot 5 \\
& = 3^{3} \cdot 5 \cdot 5 = 27 \cdot 5 \cdot 5 \\
&= 3 \cdot 3 \cdot 3 \cdot 5^{2} = 3 \cdot 3 \cdot 3 \cdot 25 \\
&= 3^{2} \cdot 3 \cdot 5^{2} = 9 \cdot 3 \cdot 25 \\
&= 3^{3} \cdot 5^{2} = 27 \cdot 25 .\\
\end{split}
$$

and the groups are

$$
\begin{split}
\mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5} \quad & \land \quad \mathbb{Z}_{9} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5} \\
\mathbb{Z}_{27} \times \mathbb{Z}_{5} \times \mathbb{Z}_{5} \quad & \land \quad \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{25} \\
\mathbb{Z}_{9} \times \mathbb{Z}_{3} \times \mathbb{Z}_{25} \quad & \land \quad \mathbb{Z}_{27} \times \mathbb{Z}_{25} .
\end{split}
$$

I am stuck on the second question "Find an element of order 45 in each one of the groups, if it exists.". I know that I have to find an ##a \in G## (where G is each of the above abelian groups) such that ##order(a) := \#(<a>) = \#(\{k \cdot a : k \in \mathbb{Z}\}) = 45##, where ##\#(\bullet)## is the cardinality of a set; or show that such an element a does not exist.

Hints are very much appreciated.
 
Last edited:

Answers and Replies

  • #2
14,414
11,726
You surely need a subgroup ##U## with ##45\,|\,|U|\,.## So what about subgroups which contain ##\mathbb{Z}_{45}\,?##
 
  • #3
18
0
You surely need a subgroup ##U## with ##45\,|\,|U|\,.## So what about subgroups which contain ##\mathbb{Z}_{45}\,?##
I don't quite understand. I am really lost on this one. If I am supposed to find a subgroup U with ##45 | |U|##, then this subgroup must have ##|U| = 45, 90, 135, ... ##
 
  • #4
14,414
11,726
Given an element ##g## of order ##45## in ##G##, it must be part of a subgroup ##U## with at least ##\mathbb{Z}_{45} \subseteq U##. We have in addition ##|U|\,|\,|G|=675## so ##|U|=90## is not possible.
 

Related Threads on Group Theory: Finite Abelian Groups - An element of order

Replies
3
Views
5K
Replies
5
Views
3K
  • Last Post
Replies
0
Views
1K
Replies
2
Views
3K
Replies
8
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
736
Top