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Let $A$ be an abelian group of order $n = p_1^{\alpha_1} \cdot \ldots \cdot p_k^{\alpha_k}$ (i.e., $n$'s unique prime factorization). The Primary Decomposition Theorem states that $A \cong \mathbb{Z}_{p_1^{\alpha_1}} \times \ldots \times \mathbb{Z}_{p_k^{\alpha_k}}$. On the other hand, the Fundamental Theorem of Finitely Generated Albelian Groups states that $A \cong \mathbb{Z}_{n_1} \times \ldots \times \mathbb{Z}_{n_j}$ for some $\{n_j\}$ s.t. $n = n_1 \cdot \ldots \cdot n_j$ and $n_{i+1}\,|\,n_i$ for all $1 \le i < j-1$. Now I'm confused because it initially seems to me that both of these statements cannot be true at once.

For example, suppose that the order of $A$ gives rise to at least two unique isomorphism types given by the Fundamental Theorem of Finitely Generated Abelian Groups. That is, suppose that $|A_1| = |A_2| = n = p_1^{\alpha_1} \cdot \ldots \cdot p_k^{\alpha_k}$ whereby $A_1 \not\cong A_2$ so that by the Fundamental Theorem of Finitely Generated Groups we have

$$ A_1 \cong \mathbb{Z}_{n_1} \times \ldots \times \mathbb{Z}_{n_j} $$

and

$$ A_2 \cong \mathbb{Z}_{n_1} \times \ldots \times \mathbb{Z}_{m_k} $$

with $\{n_i\} \ne \{m_k\}$. But we know no matter what that by the Primary Decomposition Theorem we have that $A_1 \cong \mathbb{Z}_{p_1^{\alpha_1}} \times \ldots \times \mathbb{Z}_{p_k^{\alpha_k}} \cong A_2$, a contradiction.

What am I missing?

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"The primary decomposition theorem" in the form you stated is false. You must have misquoted it. –  Makoto Kato Dec 11 '12 at 15:04

3 Answers 3

The invariant factors (the $\mathbb{Z}_{n_i}$ in your FTFGAG decomposition) are also uniquely determined up to isomorphism. The Chinese remainder theorem gives the equivalence of these statements. I think the problem you're having is that in the primary decomposition statement the $p_k$'s don't necessarily have to be distinct; it's not the prime factorization of $n$, though they do multiply to $n$.

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It is not true that a finite Abelian group is determined up to isomorphism by it order $n$ only, and that is what your formulation of the Primary Decomposition Theorem would imply. So that is what is wrong; please look up the correct formulation of this theorem which apparently you know about (I don't recall any such theorem, other than the Fund. Th. of Finitely Gen. Abelian Groups). For instance a group of order $9$ could be either $\Bbb Z_9$ or $\Bbb Z_3\times\Bbb Z_3$, and these are not isommorphic.

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up vote 0 down vote accepted

Indeed, all of the comments indicating I was misquoting the Primary Decomposition Theorem were correct.

Pg. 161 of Dummit & Foote states the Primary Decomposition Theorem for finite abelian groups:

Let $G$ be an abelian group of order $n > 1$ and let the unique factorization of $n$ into distinct prime powers be $n = p_1^{\alpha_1} \cdot \ldots \cdot p_k^{\alpha_k}$. Then $G \cong A_1 \times A_2 \times \ldots \times A_k$, where $|A_i| = p_i^{\alpha_i}$.

In other words, a finite abelian group can be decomposed into a direct product of its (unique) Sylow p-subgroups. From this we can see that the error I was making was in assuming that each of the $A_i$ were already cyclic (instead each of the $A_i$ can be further decomposed into cyclic groups via the FTFGAG so that there is no discord).

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each A_i is cyclic, but not necessarily having prime power order. You can decompose them into prime powers using Sun Zi's theorem. –  user51427 Dec 11 '12 at 18:17
    
@user51427 No, quite to the opposite, the $A_i$ of this answer do have (the maximal possible) prime power order, but they are not necessarily cyclic. –  Marc van Leeuwen Apr 8 at 8:51

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