1. ## cyclic group help

Hello, I want to make sure I understand the fundamental theorem of cyclic groups correctly.

For example, consider the group $(\mathbb{Z}_{45},+_{45})$. I need to construct a subgroup lattice for $(\mathbb{Z}_{45},+_{45})$. Since the order of $(\mathbb{Z}_{45},+_{45})$ is 45, I need to find all the natural numbers $k$ that divide 45. Of course, these numbers are 1, 3, 5, 9, 15, and 45, which have orders 45, 15, 9, 5, 3, and 1, respectively. Since $<1>=(\mathbb{Z}_{45},+_{45})$ we have $<1^{5}>=<5>$, $<1^{9}>=<9>$, $<1^{15}>=<15>$, and so on are the subgroups of $(\mathbb{Z}_{45},+_{45})$. From here I just decide which set is a subset of the other and I am done.

Am I using the theorem correctly? Does this mean that any natural number $k$ that does not divide the order of $(\mathbb{Z}_{45},+_{45})$ will create set that does not abide by group axioms? I know I can check this for the cyclic group in this example, but, in general, is the previous sentence correct?

2. Originally Posted by Danneedshelp
Hello, I want to make sure I understand the fundamental theorem of cyclic groups correctly.

For example, consider the group $(\mathbb{Z}_{45},+_{45})$. I need to construct a subgroup lattice for $(\mathbb{Z}_{45},+_{45})$. Since the order of $(\mathbb{Z}_{45},+_{45})$ is 45, I need to find all the natural numbers $k$ that divide 45. Of course, these numbers are 1, 3, 5, 9, 15, and 45, which have orders 45, 15, 9, 5, 3, and 1, respectively. Since $<1>=(\mathbb{Z}_{45},+_{45})$ we have $<1^{5}>=<5>$, $<1^{9}>=<9>$, $<1^{15}>=<15>$, and so on are the subgroups of $(\mathbb{Z}_{45},+_{45})$. From here I just decide which set is a subset of the other and I am done.

Am I using the theorem correctly? Does this mean that any natural number $k$ that does not divide the order of $(\mathbb{Z}_{45},+_{45})$ will create set that does not abide by group axioms? I know I can check this for the cyclic group in this example, but, in general, is the previous sentence correct?

All is fine, but if you write $\mathbb{Z}_{45}$ as an ADDITIVE group, then the subgroups are $5\mathbb{Z}_{45}=\{0,5,10,15,20,25,30,35,40\}=$ the subgroup of order 9,

$9\mathbb{Z}_{45}=\{0,9,18,27,36\}=$ the subgroup of order 5, etc.

After all, here $1=1^5=1^{15}\!\!\pmod {45}$

Tonio