I have to determine the subgroup lattice for $\mathbb{Z}_{20}$.
The order of all subgroups divides the order the group.
So, since $|\mathbb{Z}_{20}|=20$, the orders of the subgroups are: $20,10,5,4,2,1$.
But how can I find these subgroups??
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1If $\gcd(n,m)=1$ then $\Bbb Z_{nm}\simeq\Bbb Z_n\times \Bbb Z_m$. – Jun 22 '14 at 15:06
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Hint: Consider the canonical projection $\mathbb Z \to \mathbb Z_{20}$ and use the isomorphism theorem that relates the subgroups of the image with the subgroups that contain the kernel.
lhf
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Thank you for the hint. Which one of the three isomorphism theorems is that one? – evaristegd May 05 '19 at 23:12
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Use the fact that if $C_n = \{1,x,x^2,\ldots,x^{n-1} \}$ is a cyclic group generated by $x$, then every subgroup of $C_n$ has the form $\langle x^d \rangle$ for some $d$ dividing $n$. Thus, $C_{20}$ has exactly six subgroups: $\langle x^d \rangle$ for $d=1,2,4,5,10,20$. The lattice should be clear.
Ashwin Ganesan
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So, since the group $\mathbb{Z}_{20}$ has the operation $"+"$, the generator is $<[1]>$, right? So, are the subgroups the following?? $$<1 \cdot [1]>=<[1]>, \ \ <2 \cdot [1]>=<[2]>, \ \ <4 \cdot [1]>=<[4]>, \ \ <5 \cdot [1]>=<[5]>, \ <10 \cdot [1]>=<[10]>, \ \ <20 \cdot [1]>=<[20]>$$ – Mary Star Jun 22 '14 at 15:28
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And the order of each subgroup is as followed: $$ord(<[1]>)=20, \ \ ord(<[2]>)=10, \ \ ord(<[4]>)=5, \ \ ord(<[5]>)=4, \ \ ord(<[10]>)=2, \ \ ord(<[20]>=<[0]>)=1$$ Is this correct?? – Mary Star Jun 22 '14 at 15:36
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