# subgroup of $\mathbb{Z}_9 \times \mathbb{Z}_{15}$

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.

• 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 $\mathbb Z_3\times\mathbb Z_3$) – J. W. Tanner Jan 8 at 1:02
• @J.W.Tanner Nit picky, but, all the nonidentity elements have order $3$. – Chris Custer Jan 8 at 5:14
• Yes, @ChrisCuster – J. W. Tanner Jan 8 at 5:18

## 2 Answers

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.