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I am reading "A First Course in Algebra", and there, I am trying to solve the exercises, but there is something i don't understand. How do we understand whether two groups are isomorphic or not? For example, there is an example that asks whether the given groups are isomorphic:

$\Bbb Z_8 \times \Bbb Z_{10} \times \Bbb Z_{24}$ and $\Bbb Z_4 \times \Bbb Z_{12} \times \Bbb Z_{40}$

$\Bbb Z_4 \times \Bbb Z_{18} \times \Bbb Z_{15}$ and $Z_3 \times \Bbb Z_{36} \times \Bbb Z_{10}$

Are the given groups isomorphic? How can I decide this? All I can think now is that $8\cdot 10\cdot 24=4\cdot 12\cdot 40$ and $4\cdot 18\cdot 15=3\cdot 36\cdot 10$. What can I do more?

Thanks

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  • $\begingroup$ I tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. $\endgroup$ – A.P. Apr 7 '13 at 20:21
  • $\begingroup$ @A.P. thank you, it is fine $\endgroup$ – Yasin Razlık Apr 7 '13 at 20:22
  • $\begingroup$ So, for the general case, do you understand that two groups are isomorphic if and only if there is an isomorphism between them, @bigO? $\endgroup$ – Alexander Gruber Apr 7 '13 at 20:23
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You can use the fundamental theorem of finitely generated abelian groups, which states that each of your groups can be expressed uniquely (up to isomorphism) as the product cycles of order the power of primes and compare factors and the prime powers of the cycles of any two such decompositions to determine whether they are isomorphic, up to the order of their factors.

In doing so, recall that $\;\mathbb Z_{m} \times \mathbb Z_n\;$ is cyclic and $$\;\mathbb Z_{mn} \cong \mathbb Z_m\times \mathbb Z_n \;\;\text{if and only if}\;\; \gcd(m, n) = 1$$

For example, let's take $\mathbb Z_8 \times \mathbb Z_{10} \times \mathbb Z_{24}$, and decomposing this gives us:

$$\mathbb Z_8 \times \color{blue}{\bf \mathbb Z_{10}} \times \color{red}{\bf \mathbb Z_{24}} \quad \cong \quad \mathbb Z_8 \times \color{blue}{\bf (\mathbb Z_{2} \times \mathbb Z_5)} \times \color{red}{\bf (\mathbb Z_{3}\times \mathbb Z_{8})} \cong {\bf \mathbb Z_{2} \times \mathbb Z_{3}\times \mathbb Z_5 \times \mathbb Z_{8}^2}\tag{1}$$

Now compare this to the decomposition you get for the expression you are to compare with $(1)$: $$\Bbb Z_4 \times \Bbb Z_{12} \times \Bbb Z_{40} \quad \cong \quad \mathbb Z_4 \times (\mathbb Z_3 \times \mathbb Z_4)\times (\mathbb Z_8\times \mathbb Z_5) \cong {\bf \mathbb Z_3 \times \mathbb Z_4^2 \times \mathbb Z_5 \times \mathbb Z_8}\tag{2}$$

Note that there the cyclic factors of $(1), (2)$ are not equivalent, up to the order of their appearance, hence not isomorphic.

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    $\begingroup$ Shouldn't the last one be $Z_8^2$? $\endgroup$ – N. S. Apr 7 '13 at 20:16
  • $\begingroup$ @N.S. Yes, indeed. Corrected! $\endgroup$ – amWhy Apr 7 '13 at 20:20
  • $\begingroup$ @amWhy: Very nice answer +1 $\endgroup$ – Amzoti May 11 '13 at 0:23
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There is the Chinese remainder theorem, which, in one form, says that

$$\mathbb Z/n\cong \mathbb Z/k\times \mathbb Z/l$$

if and only if $n=kl$ and $k$ and $l$ are coprime. You can understand the same phenomenon as a version of the structure theorem for finitely generated abelian groups. Using the isomorphism above you will be able to answer the question above. Maybe you will want to try and prove it.

Hint: For one direction you explicitly write down a map which can be shown to be an isomorphism. For the other direction you could argue with the existence of elements of a certain order.

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  • $\begingroup$ +1: As a further hint (works, if the groups ARE isomoprhic): you can break the factors into prime power orders, like $\mathbb{Z}/36\simeq\mathbb{Z}/9\times\mathbb{Z}/4$, $\mathbb{Z}/18\simeq\mathbb{Z}/9\times\mathbb{Z}/2$,$\ldots$, and see how it plays out. $\endgroup$ – Jyrki Lahtonen Apr 7 '13 at 20:11

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