# List all subgroups of $\mathbb Z_6$ and $\mathbb Z_8$

List all the subgroups of $\mathbb Z_6$ and $\mathbb Z_8$.

I think this implies that the operation is addition because that makes the sets above groups.

I was thinking that for $\mathbb Z_6$, the groups $\mathbb Z_1$, $\mathbb Z_2$, $\mathbb Z_3$, $\mathbb Z_4$, and $\mathbb Z_5$ are all subgroups because they are all still groups and are subsets of $\mathbb Z_6$. In addition though, $G={0,2,4}$ is also a subset of $\mathbb Z_6$ and is a group under addition.

1. Is the above correct and exhaustive for $\mathbb Z_6$?

2. Is there any easier way to do this or just to enumerate all possible subsets of the group and see if they satisfy group rules?

(I saw something online for finding the subgroups of $\mathbb Z_n$ as a k such that $gcd(n,k)=1$ and then $\langle k \rangle$ is a subgroup but that didn't make much sense. Is that relevant?)

• Lagrange's Theorem says that the order of a subgroup divides the order of the group itself. For $\mathbb{Z}_6$ you need to find subgroups with order equal to each of the divisors of $6$. – Edward Evans Mar 29 '16 at 1:04
• Okay, so then a subgroup of $\mathbb Z_6$ must have either 1,2, or 3 elements because 1, 2, and 3 divide 6. In the case of 1, the subgroup is just the identity, 0. In the case of 2, the subgroup must contain the identity and half of the value, so it is the set of {0,3} and isomorphic to $\mathbb Z_2$. Finally, in the case of 3, the subgroup must contain the identity, a third and two/thirds of the value in our case giving {0,2,4} and is isomorphic to $\mathbb Z_3$. These are the three subgroups of $\mathbb Z_6$. Am I correct/on the right track? – User123454321 Mar 29 '16 at 1:32
• You are correct. Also, don't forget that any group is considered a subgroup itself and since 6 is a divisor of 6, $\mathbb{Z}_6$ is also a subgroup. – Edward Evans Mar 29 '16 at 10:50

Well $\mathbb Z_5$ is not a subgroup of $\mathbb Z_6$, since no member of $\mathbb Z_6$ has order $5$. Since you know $\mathbb Z_6$ is cyclic, every subgroup is cyclic. So, look at the subgroups of $\mathbb Z_6$ which are generated by single elements.
Generally, for the cyclic group 𝑍𝑛, the unique subgroups are in a bijective correspondence with the divisors of 𝑛. Specifically, for every 𝑘∣𝑛, ∃!𝐻⩽𝐺 such that |𝐻|=𝑘. Furthermore, we know that the order of a cyclic (sub)group is equal to the order of its generator. Therefore, find the subgroups generated by $$x^1,x^2,x^4,x^8=1 \in Z_8$$ and $$x^1,x^2,x^3,x^6=1 \in Z_8$$. Hint: these subgroups should be of isomorphism type $$Z_8,Z_4,Z_2,Z_1$$ and $$Z_6,Z_3,Z_2,Z_1$$, respectively.
First of all, since $$\mathbb{Z}_{6}$$ and $$\mathbb{Z}_{8}$$ are two cyclic subgroups, so, by Converse of Lagrange's theorem, for every divisors of order of group, there should be unique cyclic subgroup !
For $$\mathbb{Z}_{6}$$, subgroups are : $$<2>$$ is only cyclic subgroup generated by $$2$$ of order $$3$$ , $$<3>$$ is the only cyclic subgroup generated by $$3$$ of order $$2$$, and $$<0>$$ is trivial subgroup of order $$1$$ , $$\mathbb{Z}_{6}$$ is improper cyclic subgroup of order $$6$$.
For $$\mathbb{Z}_{8}$$ , subgroups are : $$<2>$$ is only cyclic subgroup generated by $$2$$ of order $$4$$,subgroups are : $$<4>$$ is only cyclic subgroup generated by $$4$$ of order $$2$$,and $$<0>$$ is trivial subgroup of order $$1$$ , $$\mathbb{Z}_{8}$$ is improper cyclic subgroup of order $$8$$.