Group Theory: Finite Abelian Groups - An element of order

[Lelouch]

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.

 

[fresh_42]

You surely need a subgroup $U$ with $45\,|\,|U|\,.$ So what about subgroups which contain $\mathbb{Z}_{45}\,?$

 

[Lelouch]

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 $|U| = 45, 90, 135, ...$

 

[fresh_42]

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.