2
$\begingroup$

I am trying to solve the following problem:

A partition of n is a sequence of positive integers $(k_1,\dots,k_m)$ such that $k_1≥k_2≥\dots$ and $\sum k_i=n$.

Prove that there is a bijection between the isomorphism classes of abelian groups of order $2^n$ and the partitions of $n$.

What I have tried so far:

I know I need to use the Fundamental Theorem of Finite Abelian Groups. Since the order of the group is a prime power ($2^n$), the theorem states that any such group is isomorphic to a direct sum of cyclic groups of prime-power orders. But once I get here I'm stuck. Any help would be appreciated.

$\endgroup$
2
  • $\begingroup$ here is a good tutorial on formatting for this site. As to the question, I suggest analyzing it in detail for small $n$ to get a sense of it. $\endgroup$ Commented yesterday
  • 1
    $\begingroup$ What are the options for the cyclic direct summands, for such an abelian group? $\endgroup$ Commented yesterday

1 Answer 1

Reset to default
2
$\begingroup$

It is easiest to just imagine a concrete example.

Suppose you want to classify all commutative groups of order $2^5$.

Here are some examples:

$(\mathbb{Z}/2^2\mathbb{Z})\oplus (\mathbb{Z}/2^3\mathbb{Z})$ and $(\mathbb{Z}/2\mathbb{Z})\oplus (\mathbb{Z}/2\mathbb{Z})\oplus (\mathbb{Z}/2\mathbb{Z})\oplus (\mathbb{Z}/2^2\mathbb{Z})$ and $(\mathbb{Z}/5\mathbb{Z})$.

Each of those examples correspond to partitions of $5$:

$5 = 3 + 2$

$5 = 2 + 1 + 1 + 1 + 1$

$5 = 5$

Now you just need to take that idea and formalize it into a proof.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.