abstract-algebragroup-theory
I know how to find all the subgroups of $n$ for a specific values of $n$, for example for $n=18$, the subgroups of $\Bbb{Z}_{18}$ are:
<1>={0,1, ... ,17}
<2>={0,2, ... ,16}
<3>={0,3, ... ,15}
<6>={0,6,12}
<9>={0,9}
<18>={0}
But, how do we draw the subgroup diagram of this? What are the steps? Here is the subgroup diagram that is given:
Can anyone explain how to draw this? Thanks
Personally, I am familiar to the following presentation of $D_4$ (or $D_8$). I added it here maybe you find it easier. In fact, You can feel the elements in this presentation easier.
Hint: Recall the theorem highlighted below, and note that it follows that $$\quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 6} \quad \cong \quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 2}\times \mathbb Z_{\large 3}$$
This might help to make your task a bit more clear, noting that each of $\mathbb Z_2, \; \mathbb Z_3,$ and $\,\mathbb Z_6 \cong \mathbb Z_2 \times \mathbb Z_3$ are cyclic, but $\;\mathbb Z_2 \times \mathbb Z_2,\;$ of order $\,4,\,$ is not cyclic. Indeed, there is one and only one group of order $4$, isomorphic to $\mathbb Z_2\times \mathbb Z_2$, i.e., the Klein $4$-group.
Theorem: $\;\mathbb Z_{\large mn}\;$ is cyclic and $$\mathbb Z_{\large mn} \cong \mathbb Z_{\large m} \times \mathbb Z_{\large n}$$
if and only if $\;\;\gcd(m, n) = 1.$
This is how we know that $\mathbb Z_6 = \mathbb Z_{2\times 3} \cong \mathbb Z_2\times \mathbb Z_3$ is cyclic, since $\gcd(2, 3) = 1.\;$
It's also why $\,\mathbb Z_2\times \mathbb Z_2 \not\cong \mathbb Z_4,\;$ and hence, is not cyclic, since $\gcd(2, 2) = 2 \neq 1$.
Good-to-know Corollary/Generalization:
The direct product $\;\displaystyle \prod_{i = 1}^n \mathbb Z_{\large m_i}\;$ is cyclic and $$\prod_{i = 1}^n \mathbb Z_{\large m_i}\quad \cong\quad \mathbb Z_{\large m_1m_2\ldots m_n}$$ if and only if the integers $m_i\,$ for $\,1 \leq i \leq n\,$ are pairwise relatively prime, that is, if and only if for any two $\,m_i, m_j,\;i\neq j,\;\gcd(m_i, m_j)=1$.
Best Answer
notice it's exactly the same as the divisibility lattice (upside down):
form a statement about the relation between the subgroup lattice of groups and the divisibility lattice of numbers and prove it.