Hello,

I am given that G is a finite abelian group that contains

8 elements of order three

18 elements of order nine

and an identity

And am told to describe of possibilities for G, by giving explicit decomposition into cyclic groups up to an isomorphism

So it seems like there are three possibilities for G

$\displaystyle \mathbb{Z}_{27}$

$\displaystyle \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$

$\displaystyle \mathbb{Z}_9 \times \mathbb{Z}_3$ (is this isomorphic to $\displaystyle \mathbb{Z}_3 \times \mathbb{Z}_9$?)

Now G cannot be $\displaystyle \mathbb{Z}_{27}$, for G has no element of order 27.

I also think G cannot be $\displaystyle \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$, since no element of that group has order nine. I pretty sure every element of this group must have order less than or equal to 3, since if $\displaystyle (a,b,c) \in$ $\displaystyle \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$, then $\displaystyle 3*(a,b,c) = (0,0,0)$

So I am left with $\displaystyle \mathbb{Z}_9 \times \mathbb{Z}_3$. Is there a quicker way to test the elements of this group besides considering the order of all 27 elements?

i am pretty sure the element (1,a) \in $\displaystyle \mathbb{Z}_9 \times \mathbb{Z}_3$, would have order 9, but I dont think there's 18 of those.

Thank you for your time.