We talking about $\mathbb{Z}_{66}\times \mathbb{Z}_{35}$.
$\gcd(66,35)=1 \Rightarrow\;\mathbb{Z}_{66}\times \mathbb{Z}_{35}\;$ is cyclic.

A. I need to find a subgroup with order 210, and tell how many subgroups with this order there is.
So: $ord(11,2) \in \mathbb{Z}_{66}\times \mathbb{Z}_{35}$ is $lcm(6,35)=210$.
And there is only one subgroup with this order.

B. I need to find an element at $\mathbb{Z}_{66}\times \mathbb{Z}_{35}$ with order 110, and tell how many elements at $\mathbb{Z}_{66}\times \mathbb{Z}_{35}$ have this order.
So: at $\mathbb{Z}_{66},ord(2)=33$ and at $\mathbb{Z}_{35},ord(7)=5$
So: $ord(3,7)=lcm(22,5)=110$. And there is $\varphi(110)=40$ elements with this order (110) at $\mathbb{Z}_{66}\times \mathbb{Z}_{35}$.

I'm right? Or I miss something?
Thank you!

  • $\begingroup$ Your approach is right for finding the subgroups but your lcm calculations are wrong. $\endgroup$ Dec 10 '13 at 7:44
  • $\begingroup$ @Callus - can you tell me exactly where it wrong? Thank you! $\endgroup$
    – CS1
    Dec 10 '13 at 7:46
  • $\begingroup$ @Callus - I try to correct it. Hope is fine now... $\endgroup$
    – CS1
    Dec 10 '13 at 7:50
  • $\begingroup$ yep, looks better $\endgroup$ Dec 10 '13 at 8:40

Consider the element $a=\left(\frac{66}{\gcd(n,66)},\frac{35}{\gcd(n,35)}\right)$, where $n$ is a factor of $66\cdot35$. Observe that


Put $n=210$. It follows that the element $\left(\frac{66}{6},\frac{35}{35}\right)=(11,1)$ generates a subgroup of order $210$. It is not difficult to see that this is the only subgroup of this order (this is true for any subgroup of a cyclic group).

Now, put $n=110$. It follows that the element $\left(\frac{66}{22},\frac{35}{5}\right)=(3,7)$ has order $110$, and there are $\varphi(110)=40$ such elements in the group.


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