A Theorem in our textbook says:

If R is a PID, then every finitely generated torision R-module M is a direct sum of cyclic modules

$$M= R/(c_1) \bigoplus R/(c_2) \bigoplus \cdots \bigoplus R/(c_t)$$

where $t \geq 1$ and $c_1 | c_2 | \cdots | c_t$.

If we want to classify groups of order 32 and put them in invariant factor form then the abelian groups of order 32 are:


$Z_{16} \bigoplus Z_2$

$Z_8 \bigoplus Z_4$

$Z_8 \bigoplus Z_2 \bigoplus Z_2$

$Z_4 \bigoplus Z_2 \bigoplus Z_2 \bigoplus Z_2$

$Z_2 \bigoplus Z_2 \bigoplus Z_2 \bigoplus Z_2 \bigoplus Z_2$

My question is: The theorem only tells us that "M is a direct sum of cyclic modules". But how do we know what those cyclic modules are?

  • 1
    $\begingroup$ You accidentally omitted $Z_4\oplus Z_4 \oplus Z_2$. $\endgroup$ – rschwieb Jan 16 '13 at 18:09

The invariant factors are part of the structure of the module itself. You wouldn't be able to say much without looking at the structure of the module directly.

Let's take your 32 elements abelian group as an example.

If one had an abelian group of order 32 prepared, they might go about figuring out which of the listed groups it is by checking orders of elements. Finding an element of order 16, for example, would narrow it down to the top two possibilities.

Why did they generate that list using the factorization of 32? They were simply trying to write down every possible set of invariant factors. Because $(32)$ annihilates $M$, we know that $c_n$ is a divisor of 32 (because $(32)\subseteq ann(M)=(c_n)$.) But $c_{n-1}$ divides $c_n$, so it's going to be in that list of divisors of 32. If $c_{n-1}$ happened to be 4, well then $c_{n-2}$ would need to be a divisor of 4, etc. If you go on this way, remembering keeping the size of the group in mind, you will be able to list all possible lists of invariant numbers, as they did:

$$ 32\\ 2|16\\ 4|8\\ 2|4|4\\ 2|2|8\\ 2|2|2|4\\ 2|2|2|2|2\\ $$

In a nutshell, if you can find anything that annihilates $M$, say $(x)$, the invariant factors have to be divisors of $x$.

It's probably going to be easier, though, if you first try to find the primary decomposition of $M$, and then compute the invariant factors from that. (I think that is possible... how was it done again...? Hmm...)

  • $\begingroup$ Thanks a lot, but I'm not sure if I understand the last paragraph. Is it possible for you to elaborate? Also, as far as I know, the ones that I listed actually are in invariant factor form (except that we would need to write them from left to right, instead of from right to left...so that the divisors come first), right? I'm just asking about how they came up with this answer. How did they know that they were supposed to write down the prime factorization of 32 and then use that to find the invariant factors? How is that connected to the theorem? $\endgroup$ – user58289 Jan 16 '13 at 17:46
  • $\begingroup$ @Artus OK, I tried to make the connection between the factorization and the method they used to list all invariant factors clearer. $\endgroup$ – rschwieb Jan 16 '13 at 18:10
  • $\begingroup$ Oh, so you're saying that I can use primary decomposition? There is another theorem in our textbook that says "Every finite abelian group G is the direct sum of its p-primary components", and it defines a p-primary component as $\{a \in G : p^na=0$ for $n \geq 1\}$. We know that for $ap^n=0$, p needs to be a divisor of 32, right? Is that why we find the prime factorization first? And then after finding this, we just write it in a way so that $c_k$ divides $c_{k+1}$, right? So invaraint factors is just about reording/rewriting the primary decomposition in a spacial way. Is that correct? $\endgroup$ – user58289 Jan 16 '13 at 18:19
  • $\begingroup$ @Artus Yes, I think you're right. I keep feeling like there was an easy way to jump from primary factors to invariant factors, but it's been a while. In the case of an abelian group, this is the same as finding elements of prime power order (which is all we need, in a group of order $2^5$...) $\endgroup$ – rschwieb Jan 16 '13 at 18:26
  • $\begingroup$ @Artus If it were a group of order $2^43^4$, then you could first split it into components with prime power orders. Then you could work to get them into invariant form, as we did with a group of order 32. $\endgroup$ – rschwieb Jan 16 '13 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy