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How can I determine if the group $G=(\Bbb Z_{16}\times\Bbb Z_{18}, +)$ is isomorphic to $(\Bbb Z_4\times\Bbb Z_{72}, +)$?

I came up with that $(1,1)$ in $G$ has order $144$ because $\operatorname{lcm}(16,18)=144$.

Highest possible order for an element in $(\Bbb Z_4\times\Bbb Z_{72}, +)$ is $72$ because $\operatorname{lcm}(4,72)=72$.

Therefore they cant be isomorphic because there is no element with order $144$ in $(\Bbb Z_4\times\Bbb Z_{72}, +)$.

Is there any other way I can prove it?

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1 Answer 1

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Of course, you provided an excellent and elementary way to see this. In general, you can use the following, but it requires a classification result, which is non-trivial to prove:

You can invoke the fundamental theorem of finite abelian groups. Decompose $$\mathbb{Z}_{16} \times \mathbb{Z}_{18}\cong \mathbb{Z}_{2^4} \times \mathbb{Z}_2 \times \mathbb{Z}_{3^2}$$ $$\mathbb{Z}_{4} \times \mathbb{Z}_{72}\cong \mathbb{Z}_{2^2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{2^3}$$ and such a decomposition is unique (up to permutation of the factors). Since the decompositions do not agree, the groups are not isomorphic.

Here, I used the fact that $\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{nm}$ when $m$ and $n$ are coprime.

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