# Are $\Bbb Z_{8} \times \Bbb Z_{10} \times \Bbb Z_{24}$ and $\Bbb Z_{4} \times \Bbb Z_{12} \times \Bbb Z_{40}$ isomorphic? [closed]

Are the groups $\Bbb Z_{8} \times \Bbb Z_{10} \times \Bbb Z_{24}$ and $\Bbb Z_{4} \times \Bbb Z_{12} \times \Bbb Z_{40}$ isomorphic? Why or why not?

(Here $\times$ means the direct product or direct sum.)

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## closed as off-topic by Daniel Rust, Zev Chonoles, Davide Giraudo, Shuchang, Grigory MDec 20 '13 at 14:10

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Hint: use the classification of finite abelian groups.

The former is isomorphic to

$$\Bbb Z_{2} \times \Bbb Z_{3}\times \Bbb Z_{5}\times \Bbb Z_{8} \times \Bbb Z_{8}$$

While the latter is isomorphic to

$$\Bbb Z_{3} \times \Bbb Z_{4}\times \Bbb Z_{4}\times \Bbb Z_{5} \times \Bbb Z_{8}$$

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How to determine , what are the theorems playing in its Background...Can you explain? –  EuReka Dec 20 '13 at 13:38
@EuReka You should take a moment and study the Chinese Remainder Theorem (and this is not about a doggy bag at your favorite restaurant). With the aid of this theorem you can prove the isomorphism that amWhy mentioned. –  Nicky Hekster Dec 20 '13 at 15:53
As far I knew CRT is concerned with solving a system of linear congruences. I have just read the equivalent algebraic form of this statement and its nice to see the connection with it(But How?). –  EuReka Dec 20 '13 at 16:09

No.

To see why not:

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

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ok. Then ?? i am not getting, Please explain –  EuReka Dec 20 '13 at 13:40
See the link I added, EuReka. You can decompose each group into its invariant prime decomposition. The second bullet should show you that $\mathbb Z_8 \not\cong \mathbb Z_2\times \mathbb Z_4$, since $\gcd(2, 4) = 2\neq 1.$ To see more clearly, note that the first group is isomorphic to $\large \Bbb Z_{2} \times \Bbb Z_{3}\times \Bbb Z_{5}\times \Bbb Z_{2^3} \times \Bbb Z_{2^3}$. The second group is isomorphic to $\large \Bbb Z_{3} \times \Bbb Z_{2^2}\times \Bbb Z_{2^2}\times \Bbb Z_{5} \times \Bbb Z_{2^3}$ –  amWhy Dec 20 '13 at 13:44
EuReka Read the Wikipedia link and make sure you understand it and the examples given there. See if you can apply what you read there, and the second bullet of my answer, to understand why your two groups are NOT isomorphic. If you're still struggling after doing that, post a comment, and I'll get back to you. –  amWhy Dec 20 '13 at 13:51
Short but nice!+1 –  Sami Ben Romdhane Dec 20 '13 at 14:24
Nicely and sufficiently hinted. +1 –  DonAntonio Dec 20 '13 at 14:33

The first lemma you should prove is the following:

If $gcd(a,b) = 1$, then $\mathbb{Z}_{ab} \cong \mathbb{Z}_a \times \mathbb{Z}_b$.

Then consider the number of elements of order $4$.

You can also use the theorem of structure of finite abelian groups.

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