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.
Best Answer
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:
So:
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 $3$.
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.
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.