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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.

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

1 Answer 1

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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.

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  • $\begingroup$ Sorry for the confusion, thanks for this it makes much more sense now. $\endgroup$
    – user835291
    Nov 9, 2020 at 3:57
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    $\begingroup$ @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. $\endgroup$ Nov 9, 2020 at 4:31

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