Finding subgroups and their generators of cyclic group

[Ragnarok7]

List every generator of each subgroup of order 8 in $$\mathbb{Z}_{32}$$.

I was told to use the following theorem:

Let $$G$$ be a cyclic group of order $$n$$ and suppose that $$a\in G$$ is a generator of the group. If $$b=a^k$$, then the order of $$b$$ is $$n/d$$, where $$d=\text{gcd}(k,n)$$.

However, I am unsure how this helps. By inspection, I've found the only subgroup of order 8 in $$\mathbb{Z}_{32}$$ is $$\{0,4,8,12,16,20,24,28\}$$. I have also found its generators by inspection to be 4, 12, 20, and 28. But how is one supposed to find these without doing all the calculations? Thanks!

 

[Euge]

List every generator of each subgroup of order 8 in $$\mathbb{Z}_{32}$$.

I was told to use the following theorem:

Let $$G$$ be a cyclic group of order $$n$$ and suppose that $$a\in G$$ is a generator of the group. If $$b=a^k$$, then the order of $$b$$ is $$n/d$$, where $$d=\text{gcd}(k,n)$$.

However, I am unsure how this helps. By inspection, I've found the only subgroup of order 8 in $$\mathbb{Z}_{32}$$ is $$\{0,4,8,12,16,20,24,28\}$$. I have also found its generators by inspection to be 4, 12, 20, and 28. But how is one supposed to find these without doing all the calculations? Thanks!

Hi Ragnarok,

Recall that the order of an element $x$ in a group $G$ is the order of the cyclic subgroup generated by $x$. Also keep in mind that $\Bbb Z_{32}$ is a group under addition, not multiplication. So the result you mentioned should be viewed additively, not multiplicatively.

To solve the problem, first find all elements of order 8 in $\Bbb Z_{32}$. Since gcd(32,4) = 4, the order of 4 is 32/4 = 8. Now we can find the other elements of order 8 by adding multiples of 8 to 4: 12, 20, 28. We stopped at 28, because the next number is 36, which is 4 in $\Bbb Z_{32}$. So there are four elements of order 8: 4, 12, 20, 28. Note that each of these elements generate the same cyclic subgroup. So there is only one subgroup of order 8. In fact, in every finite cyclic group $\Bbb Z_n$, there is a unique subgroup of order $d$ for every positive divisor $d$ of $n$.

 

[Ragnarok7]

Thank you so much! That makes a lot of sense.

 

[Deveno]

I suspect that you are supposed to use the theorem like so:

$k \in \Bbb Z_{32}$ has order 8 if $\text{gcd}(k,32) = 4$.

In particular, this means $k$ is an odd multiple of 4, that is 4,12,20 or 28.