Find a subgroup of $\mathbb{Z}_{12} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}_{15}$ that is of order 9.

By a certain theorem, $\mathbb{Z}_{4}\oplus\mathbb{Z}_{15} \cong \mathbb{Z}_{60}$ from the fact that 4 and 15 are relatively prime.

This gives

$\mathbb{Z}_{12} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}_{15}\cong \mathbb{Z}_{12}\oplus \mathbb{Z}_{60}$

It now suffices to find all elements of order 9 in the group $\mathbb{Z}_{12} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}_{15}$; knowing that the order of any element in a group is the order of the cyclic subgroup generated by that element.

But, aside from listing the elements manually-which is tedious- what is a more efficient way to go about?

  • $\begingroup$ Note that this isn't asking for a cyclic subgroup of order $9$: so there could be a subgroup of order $9$, but no element of order $9$. You should also keep an eye out for subgroups of the form $\Bbb Z_3\oplus\Bbb Z_3$. $\endgroup$ – Stahl Apr 9 '17 at 4:52
  • $\begingroup$ yes, you are correct. Apparently, the browser 'saves' the entire draft of the thread upon reopening and I conveniently forgotten this fact, assuming that I had already input a fitting title. $\endgroup$ – Mathematicing Apr 9 '17 at 4:54

The structure theorem for abelian groups is the right way to go but you are using it in an unhelpful direction. It implies that you can split up that sum further: $$\mathbb{Z}_{12} \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_{15} \cong \mathbb{Z}_4 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5.$$ You can read off the subgroup of order $9$. To be explicit, the subgroup $\mathbb{Z}_3 \subseteq \mathbb{Z}_{12}$ is cyclic generated by $4$ and the subgroup $\mathbb{Z}_3 \subseteq \mathbb{Z}_{15}$ is cyclic generated by $5$.


While the decomposition into minimal cyclic subgroups as in the answer by user399601 is probably the easiest way to go, here is a somewhat difference approach. Whatever the structure of your subgroup, its elements have orders that are powers of$~3$, so you can restrict attention to the subgroup of such elements. The property of having order a power of$~3$ is equivalent to the same property for the components in each of the given factors in the direct sum. In $\Bbb Z_4$ only the identity qualifies, in the other two factors the subgroup with the property is of order$~3$. That gives $3\times1\times3=9$ elements in all, and you have to search no longer.


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