# Finding all subgroups of direct product of cyclic groups

Find all subgroups of $$\mathbb{Z}_{9} \oplus \mathbb{Z}_{3}$$ of order $$3$$.

I have been having some confusion with these types of problems.

I know all of the subgroups for this problem would be isomorphic to $$\mathbb{Z}_{3}$$, hence cyclic, and would be of the form $$\langle (a,b) \rangle$$ where $$|a|=3,|b|=3$$, $$|a|=1,|b|=3$$,$$|a|=3,|b|=1$$.

Now I would start out by listing the elements of the form above, and the cyclic subgroup generated by each element.

$$\langle (3,1) \rangle$$,$$\langle (0,1) \rangle$$,$$\langle (3,0) \rangle$$

now my main confusion comes with the cyclic group generated by the two elements:

$$\langle (3,2) \rangle=\langle (6,1) \rangle$$

How can we know that this group is generated by these two elements?

This stirs up some confusion. I know a cyclic group that is not a direct product of groups like $$\mathbb{Z}_{n}$$ can be represented by any of its generators. Now with direct products it seems like combinations of different generators that generate the same cyclic group in non-direct products yield different groups. For instance in the last subgroup with the two generators, I guess I am confused because normally in $$\mathbb{Z}_3$$, $$\langle 1 \rangle =\langle 2 \rangle$$ however in the direct product the subgroups $$\langle (3,1) \rangle \neq \langle (3,2) \rangle$$. I cannot pinpoint precisely what I want to say here, I guess I just need someone to explain why different combinations of ordered pairs, with elements that would generate the same cyclic subgroup in a group that is not the direct product can generate a different subgroup while in a group that is in a direct product.

• The subgroup generated by $\langle 1,0\rangle$ is not isomorphic to $\Bbb Z_3$. Nov 9, 2020 at 3:02
• Alright my bad.
– user835291
Nov 9, 2020 at 3:40

Note that your list of elements of order $$3$$ is incomplete. The elements of order $$3$$ in $$\mathbb{Z}_3$$ are $$1$$ and $$2$$, not just $$1$$. The elements of order $$3$$ in $$\mathbb{Z}_9$$ are $$3$$ and $$6$$, not just $$3$$.

Now, the key here is that the subgroup generated by $$x$$ in a group is always the same as the subgroup generated by $$x^{-1}$$ (or in additive notation, $$-x$$). In the context here, the subgroups generated by $$(a,b)$$ and by $$(a^{-1},b^{-1}$$) (or in additive notation, $$(-a,-b)$$) are the same.

So the subgroup generated by $$(3,1)$$ is the same as the subgroup generated by $$(6,2)$$ because $$6=-3$$ in $$\mathbb{Z}_9$$ and $$2=-1$$ in $$\mathbb{Z}_2$$.

So the full list should be something like this:

1. Elements of order $$3$$ in $$\mathbb{Z}_9$$: $$3$$ and $$6=-3$$.
2. Elements of order $$3$$ in $$\mathbb{Z}_3$$: $$1$$ and $$2=-1$$.
3. Elements of order $$1$$ in $$\mathbb{Z}_9$$: $$0$$.
4. Elements of order $$1$$ in $$\mathbb{Z}_3$$: $$0$$.

So:

1. Pairs in which both entries have order $$3$$: $$(3,1)$$, $$(6,1)$$, $$(3,2)$$, and $$(6,2)$$. Of these, $$(6,2)=(-3,-1)$$, and $$(6,1)= (-3,-2)$$. So these provide two subgroups of order $$3$$.

2. Pairs in which the first entry is of order $$3$$ and the second of order $$1$$: $$(3,0)$$ and $$(6,0)$$; but $$(6,0)=(-3,-0)$$, so they give the same subgroup.

3. Pairs in which the first entry is of order $$1$$ and the second of order $$3$$: $$(0,1)$$ and $$(0,2)$$; but $$(0,2)=(-0,-1)$$, so they give the same subgroup.

• Sorry for the confusion, thanks for this it makes much more sense now.
– user835291
Nov 9, 2020 at 3:57
• @PegasusMitchell: More generally, note that if $\langle x\rangle$ has order $n$, then $\langle x\rangle = \langle x^k\rangle$ if and only if $\gcd(k,n)=1$. For $n=3$, that means you only need to worry about $k=1$ and $k=2$ (and of course $2\equiv -1\pmod{3}$). But for other sizes, you may have more work to do. Nov 9, 2020 at 4:31