Let $C_9$ denote the the cyclic group of order $9$. What is the number of subgroups of the Cartesian product $C_9 \times C_9$ that are isomorphic to $C_9$. (NBHM 2023).
The answer is 12. But I can't get that.... I found 10 subgroups (which are generated by the elements of the group $C_9 \times C_9$) that are isomorphic to $C_9$. I can't get the exact approch.
Any hint or solution is highly appreciated. Thank you in advance.
There are $9\times9-1=80$ non-trivial elements, in which $9-1=8$ of them have order $3$, and $80-8=72$ of them have order $9$.
Each order $9$ element generates a copy of $C_9$, $C_9$ contains one proper subgroup $C_3$, therefore in each copy of $C_9$ there are $9-3=6$ generators
Counting the number of generators you get $72/6=12$ such subgroups, they are generated by (1,0),(1,3),(1,6) (0,1),(3,1),(6,1) (1,1),(1,4),(1,7) (1,2),(1,5),(1,8)
To find the number of cyclic subgroup of order $9$, first find the number of elements of order $9$.
Now $|(a,b)| = lcm(|a|,|b|)$ ,for $(a,b) \in C_9×C_9$.
So order $9$ elements are those where
$1.\; |a|=9 ,|b|=1$
$2.\; |a|=1 ,|b|=9$
$3.\; |a|=3 , |b|=9 $
$4.\; |a|= 9, |b|=3 $
$5.\; |a|=9 , |b|=9 $
Now find number of these elements using the fact that there are $\;\phi(k)$ ($\phi$ is Euler phi function) elements of order $k$ in a cyclic group,where $k$ divides order of the group.
You will get $72$ elements of order $9$.So number of cyclic subgroup of order $9 = \frac{72}{\phi(9)}$ =$12$
