# Fundamental Theorem of Finite Abelian Groups

Fundamental Theorem of Finite Abelian Groups indicates that $\mathbb{Z}_{n}$ is isomorphic to $\mathbb{Z}_{p_1^{k_1}} \times \mathbb{Z}_{p_2^{k_2}}\times$ ... $\times\mathbb{Z}_{p_n^{k_n}}$ where $p_i$ are prime and not necessarily distinct.

But I know that $\mathbb{Z}_4$ is not isomorphic to $\mathbb{Z}_2 \times\mathbb{Z}_2$ because $\mathbb{Z}_{mn}$ is only isomorphic to $\mathbb{Z}_m\times\mathbb{Z}_n$ only if $m$ and $n$ are relatively prime.

Can anyone clarify and correct my misunderstanding on this subject.

Thanks

• It says the $p_i$ are not necessarily distinct. So your two examples both fit the general description. Commented Nov 12, 2013 at 1:51
• It is not entirely clear what your misunderstanding is. Is it that $Z_4$ is not isomorphic to $Z_2 \times Z_2$? Or is your misunderstanding related to not understanding why this is not a contradiction to the fundamental theorem? Commented Nov 12, 2013 at 1:55
• Some time ago I wrote something about it, elaborating in many details in my try to understand this. The only problem is that it is in spanish. I can share it with you if you can read it.
– leo
Commented Nov 12, 2013 at 2:17
• 4 is a prime power.
– lhf
Commented Nov 12, 2013 at 2:26
• Oooops...I think I got it now...thanks everyone for the enlightment and assisstance...Malo... Commented Nov 12, 2013 at 3:01

Given a finite Abelian group $G$, $G$ is isomorphic to a direct product of cyclic groups: $Z_{{p_1}^{e_1}}\times Z_{{p_2}^{e_2}}\times Z_{{p_3}^{e_3}}\times ...Z_{{p_n}^{e_n}}.$ But these primes may not be distinct.

It seems that your misunderstanding stems from thinking that any finite Abelian group must decompose entirely into a product of cyclic groups whose orders are powers of distinct primes.

• Note that not every finite group is isomorphic to $Z_n$ for some $n$. Commented Nov 12, 2013 at 2:03
• That is correct...that is where my misunderstanding lies...as stated on my comment above... Commented Nov 12, 2013 at 2:47
• Above, if say $p_1 = p_2$, then we cannot combine these terms. $Z_{mn} is only isomorphic to Z_m×Z_n$ only if m and n are relatively prime. For example, $Z_9 \times Z_25$ is not isomorphic to $Z_3 \times Z_3 \times Z_5 \times Z_5$. Since 3 and 3 are not relatively prime, we cannot combine them to create $Z_9$ for example. The same goes with the 5-groups. Commented Nov 12, 2013 at 3:16

There is nothing wrong with the statement. You have to note that $\mathbb{Z}_4\cong\mathbb{Z}_{2^2}$ which agrees with the statement.

Your misunderstanding may have arisen from thinking that the factorisation of the group follows the factorisation of the order of the group, $n$ in your case. So you would think that $\mathbb{Z}_4\cong\mathbb{Z}_2\times\mathbb{Z}_2$ which is wrong.

Hope this helps.

• The only issue with the statement is that it is not the Fundamental Theorem. It is a sort of corollary. Not every group of order n is isomorphic to $Z_n$. Commented Nov 12, 2013 at 2:17
• I agree with you. Commented Nov 12, 2013 at 2:35
• I was thinking that Z_2 X Z_2 is a product of cyclic group and Z_4 is a cyclic abelian group...on the same confusion, I thought that Z_12 is isomorphic to Z_4 X Z_3 which is isomorphic to Z_2 X Z_2 X Z_3 ... on the same principle, I am lost...please elaborate more on the matter... Commented Nov 12, 2013 at 2:52
• Z_12 is isomorphic to Z_4 X Z_3. But this is not isomorphic to Z_2 X Z_2 X Z_3. If terms are NOT relatively prime, you cannot combine them. Commented Nov 12, 2013 at 3:15
• I think @Vladhagen hit the nail on the head there. "If terms are NOT relatively prime, you cannot combine them". Commented Nov 12, 2013 at 4:33