Something obvious I am missing about the Fundamental theorem of finitely generated abelian groups

I must be missing something very obvious, but I am not sure what. According to the Fundamental theorem of finitely generated abelian groups, a finitely generated abelian group can be written as $G\cong \mathbb Z^r\oplus \mathbb Z_{n_1}\oplus ...\oplus\mathbb Z_{n_s}$, where $r\geq 0,\,\,\,\,\, n_{i+1}|n_i$ for $1\leq i\leq s-1$.

What I don't get is I think $\mathbb Z_6\cong \mathbb Z_3\oplus\mathbb Z_2$ but $(2,3)=1$ so it doesn't satisfy the theorem??

• Yes it does. If $G=\mathbb{Z}/6\mathbb{Z}$, then $G\cong \mathbb{Z}^r \oplus \mathbb{Z}/n_1\mathbb{Z}$, with $r=0$ and $n_1=6$. – Álvaro Lozano-Robledo Mar 15 '12 at 19:50
• I think you'll find that the expression $\mathbb Z_6$ suffices for the theorem. The theorem as expressed does not say that this expression is the only one possible, just that the expression in the theorem is always possible. – Mark Bennet Mar 15 '12 at 19:51
• @MarkBennet: Thank you, Mark, but according to Wikipedia doesn't the expression have to be unique? – Ringo Mar 15 '12 at 19:59
• @Ringo: Decompositions aren't unique in general, but the invariant factors form is unique. Clearly $Z_6$ is already of that form all by itself. Do you see any other decompositions that count as a second, distinct invariant factor form? Because $Z_2\oplus Z_3$ is invalid. – anon Mar 15 '12 at 20:06
• @Ringo: The expression is unique in the sense that if you can write $G$ as $$\mathbb{Z}^r\oplus \mathbb{Z}_{n_1}\oplus\cdots\oplus \mathbb{Z}_{n_t}\text{ with }n_r|n_{r-1}|\cdots|n_1$$ and you can also write it as $$\mathbb{Z}^s\oplus\mathbb{Z}_{m_1}\oplus\cdots\oplus\mathbb{Z}_{m_u}\text{ with }m_u|m_{u-1}|\cdots|m_1$$then $s=r$, $t=u$, and $m_i=n_i$ for each $i$. That is, any two expressions that satisfy the conditions are identical. – Arturo Magidin Mar 15 '12 at 20:06

The Fundamental Theorem of Finitely Generated Abelian groups is like the Fundamental Theorem of Arithmetic: it describes a "canonical way" of expressing a finitely generated abelian group as a direct sum (in fact, two different ways), in a way that is "essentially unique", and where two groups are isomorphic if and only if they have the same "canonical way of being described."

The analogy with the Fundamental Theorem of Arithmetic is that the latter tells you that there is a unique way (up to order) of expressing a positive integer as a product of powers of distinct primes; it does not tell you that there is only one way of expressing a positive integer as a product. So, the fact that we can write $36$ as $6\times 6$, with neither factor a prime power, does not contradict the Fundamental Theorem of Arithmetic. The Fundamental Theorem of Arithmetic is reflected in the fact that we can write $36$ as a product of powers of distinct primes (namely $2^2\times 3^2$) and that this is the only way to express $36$ as a product of powers of distinct primes (up to order). But it says nothing about other ways of expressing $36$ as a product. You can also use the Fundamental Theorem of Arithmetic to say that every positive integer can be written as $n=q_1q_2\cdots q_m$, where $q_1\leq q_2\leq\cdots\leq q_m$ are all primes, and this expression is unique in that if $n=p_1p_2\cdots p_n$ with $p_1\leq p_2\leq\cdots\leq p_n$ primes, then $m=n$ and $p_i=q_i$ for each $i$. Even though you have two different expressions, each one is "unique within its domain".

The Fundamental Theorem for Finitely Generated Abelian groups says that you have two different "canonical decompositions": one into cyclic groups of prime power order, and one into numbers that divide each other:

1. Every finitely generated abelian group $G$ can be written as $$G\cong \mathbb{Z}^r \oplus \mathbb{Z}_{p_1^{a_1}}\oplus\cdots\oplus \mathbb{Z}_{p_k^{a_k}}$$ where $r,k$ are nonnegative integers, $p_1,\ldots,p_k$ are primes, and $a_1,\ldots,a_k$ are positive integers. Moreover, this expression is unique in the sense that if $$G\cong\mathbb{Z}^s\oplus\mathbb{Z}_{q_1^{b_1}}\oplus\cdots\oplus \mathbb{Z}_{q_{\ell}^{b_{\ell}}}$$ with $s,\ell$ nonnegative integers, $q_1,\ldots,q_{\ell}$ primes, and $b_1,\ldots,b_k$ positive integers, then $r=s$, $k=\ell$, and there is a permutation $\sigma$ of $\{1,\ldots,k\}$ such that $p_i=q_{\sigma(i)}$ and $a_i=b_{\sigma(i)}$ for all $i$.

2. Every finitely generated abelian group $G$ can be written as $$G\cong \mathbb{Z}^r\oplus\mathbb{Z}_{n_1}\oplus\cdots\oplus\mathbb{Z}_{n_t}$$ where $r,t$ are nonnegative integers, $n_1,\ldots,n_t$ are positive integers greater than $1$, and $n_t|n_{t-1}|\cdots|n_1$; moreover, the expression is unique in the sense that if $G$ can also be written as $$G\cong \mathbb{Z}^s\oplus\mathbb{Z}_{m_1}\oplus\cdots\oplus \mathbb{Z}_{m_u}$$ where $s,u$ are nonnegative integers, $m_1,\ldots,m_u$ are positive integers greater than $1$, and $m_u|m_{u-1}|\cdots|m_1$, then $r=s$, $t=u$, and $m_i=n_i$ for each $i$.

For $\mathbb{Z}_6$, the first format of the decomposition says that we can write it as $\mathbb{Z}_2\oplus\mathbb{Z}_3$, and that this is the only way to write it as a direct sum of cyclic groups of prime power order (except for the trivial $\mathbb{Z}_3\oplus\mathbb{Z}_2$, which is really "the same way"). The second part says that we can also write it as $\mathbb{Z}_6$, and that this is the only way to write it as a direct sum of cyclic groups in such a way that the order of each one divides the order of the previous one. That is, we have two different "unique factorizations", depending on which format you want to use.

Notice the condition that $n_{i+1}\vert n_i$. This condition does not hold for $\mathbb{Z}_3\oplus \mathbb{Z}_2$ since as you said 3 and 2 are coprime.

As you said, a finitely generated abelian group can be written in that form. That doesn't mean every way of expressing the group must be in that form.

In your example, $\mathbb{Z}_6$ is already in the form described in the theorem.

From Wikipedia:

We can also write any finitely generated abelian group G as a direct sum of the form $$\mathbb{Z}^n \oplus \mathbb{Z}_{k_1} \oplus \cdots \oplus \mathbb{Z}_{k_u}$$ where $k_1$ divides $k_2$, which divides $k_3$ and so on up to $k_u$. The rank $n$ and the invariant factors $k1,...,ku$ are uniquely determined by $G$ ( with a unique order).

• According to Wikipedia the expression is unique? – Ringo Mar 15 '12 at 19:57
• This form is unique, it's the invariant factor decomposition. The other one is the primary decomposition. – minimalrho Mar 15 '12 at 19:58
• @minimalrho: Ah, so the fundamental theorem applies to two distinct types of decomposition...? – Ringo Mar 15 '12 at 20:00
• Once the group is written in the invariant factor decomposition as you described, then the $n_i$ are unique. But there still may be other ways of writing the group as a direct sum. – Cardboard Box Mar 15 '12 at 20:03

Sorry, buzzkill: it's $\cong\mathbb{Z}^0\oplus Z_{\,6}$.

The novelty of the theorem is that when encountering a finite abelian group that you don't already know to be decomposable into cyclics a priori, the theorem tells you it is.

Indeed, the invariant factor decomposition is unique. However, $Z_2\oplus Z_3$ is not a valid invariant factor form, so there is no contradiction with the statement of uniqueness. Also, the other form of the theorem involves the primary decomposition, which is also unique as such but a distinct form.

The structure theorem provides an alternate form of decomposing our finitely generated abelian groups, where we instead express the torsion elements as $G\cong \mathbb Z^r\oplus \mathbb Z_{{p_1}^{n_1}}\oplus ...\oplus\mathbb Z_{{p_s}^{n_s}}$, where the $p_i$'s are prime and the $n_i$'s are natural numbers. Every such primary decomposition is unique up to rearrangements, but this isn't the only decomposition of any form. This is more conducive to your presentation of $\mathbb{Z}_6 \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3$, while $\mathbb{Z}_6$ is our only decomposition in the way that your version of the structure theorem describes. Both are valid.