For a group $G$ and an element $a\in G,$ let $\langle a\rangle$ be the intersection of all subgroups of $G$ containing $a.$ If $G = \langle a\rangle$ for some $a \in G,$ then $G$ is cyclic. Determine with proof the number of cyclic subgroups of $\mathbb{Z}_9\times \mathbb{Z}_{15}$. Find, with justification, a non-cyclic proper subgroup of $\mathbb{Z}_9\times \mathbb{Z}_{15}.$

By Lagrange's Theorem, the order of each subgroup should divide $135,$ which has $4\cdot 2 = 8$ factors. I tried considering the number of cyclic subgroups of each possible order and deduced from considering the orders of elements of $\mathbb{Z}_9\times \mathbb{Z}_{15}$ that the orders for cyclic subgroups are $1, 3, 5, 9, 15, 45.$ The only subgroup with order $1$ is the trivial subgroup, and this is also a cyclic subgroup. The cyclic subgroups of order $3$ seem to be $\langle (0,5)\rangle = \{(0,0), (0,5), (0,10)\}, \langle (3,0)\rangle, \langle (3,5)\rangle, \langle (3,10)\rangle, \langle (6,5)\rangle, \langle (6,10)\rangle.$ However, I'm not sure how to find the number of cyclic subgroups of larger orders.

As for a non-cyclic subgroup, I think it suffices to find a subgroup of order $27$, since no cyclic subgroup has this order. Or there might be a simpler approach.

  • $\begingroup$ The subgroup consisting of $\{(0,0),(0,5),(0,10),(3,0),(3,5),(3,10),(6,0),(6,5),(6,10)\}$ has $9$ elements, each of order $3$, so it's not cyclic (it's like $\mathbb Z_3\times\mathbb Z_3$) $\endgroup$ – J. W. Tanner Jan 8 at 1:02
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    $\begingroup$ @J.W.Tanner Nit picky, but, all the nonidentity elements have order $3$. $\endgroup$ – Chris Custer Jan 8 at 5:14
  • $\begingroup$ Yes, @ChrisCuster $\endgroup$ – J. W. Tanner Jan 8 at 5:18

The only chances for cyclic subgroups are for those of order dividing the order of $G$. So, as one way of going at this, we could count the number of elements of order $1,3,5,9,15,27$ and $45$.

To do that, we can use the fact that the order of any element of $(a,b)\in G$ is $\rm{lcm}(|a|,|b|)$.

So, to get you started, of course we only have one element of order $1$. We have $\varphi(3)=2$ elements of order $3$ in $\Bbb Z_9$ and also, by the same reasoning, $2$ in the cyclic $\Bbb Z_{15}$. Forming ordered pairs, and pairing each element of order $3$ up with an element whose order divides $3$, we get a total of $8$ elements of $G$ of order $3$. Since, noting once again that each cyclic group of order $3$ has two elements of order $3$, we get a grand total of $4$ subgroups of order $3$.

We now have five more factors to do this for. Better you than me. Just kidding. Let's look at $5$. Since $\varphi(5)=4$, there are four elements of order five in $\Bbb Z_{15}$. Since we have to pair these with the identity in the other factor, we get a total of $4$ elements of order $5$, hence only one subgroup of $G$ of order $5$.

Now for $9$. We have $6$ elements of order nine in $\Bbb Z_9$. We cam pair each of them with anything of order dividing nine in the other factor, $\Bbb Z_{15}$. Those are the elements of order dividing three, of course, and there are three. So we get $18$ elements of order $9$. But each cyclic subgroup of order nine has $\varphi(9)=6$ elements of order nine. That makes $18/6=3$ cyclic subgroups of order nine.

For $15$, we have two elements of order $3$ in the first factor, and four elements of order $5$ in the second. Thus we have $8$ elements of order $15$, plus eight in the second factor times three elements of order dividing three in the first, to give $24$ more, for a total of $32$, just enough, $32/\varphi(15)=32/8=4$, to form four cyclic subgroups of order $15$.

I think I'll leave the last two, $27$ and $45$, for you.

Finally, for part two, it's fairly easy to see that you have a subgroup isomorphic to $\Bbb Z_3\times\Bbb Z_3$, or $\Bbb Z_9\times\Bbb Z_3$, say.


To be sure, Cauchy's theorem will guarantee that we have a cyclic subgroup with the order of any given prime divisor of 135. Some examples, if (j,9) = 1, then <(j,0)> has order 9 and if (m,15) = 1, then <(0,m)> has order 15 (if 15 divides km, then 15 divides k and the smallest such k is 15, same argument for 9). Also, the order of <(1,1)> is lcm(9,15) = 45 (if (k,k) = (0,0), k = 9R = 15S. Then 9 and 15 both divide k). So the order of <(a,b)> is completely contingent on the gcd's (9,a) and (15,b), so I would use that to get the rest of the cyclic subgroups of other orders dividing 135.


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