Is there a finite abelian group $G$ such that the product of the orders of all its elements is $2^{2013}$? Clearly, we can assume the Structure Theorem for finite abelian groups.

Edited Later: All I know at the moment is, all elements must have the order in the form of $2^n$, and the size of the group must be in that form as well, and $|G|\leq{2^{10}}=1024\leq{2014}$

  • $\begingroup$ What else have you considered? What kind of primes could possibly be involved? What kind of powers on them? $\endgroup$ Nov 10, 2014 at 1:35
  • 2
    $\begingroup$ $\mathbb{Z}_2^{11}=\mathbb{Z}_2\times\cdots\mathbb{Z}_2$ gives you $2^{2047}=2^{2^{11}-1}$. Can you see why this is? (Hint: powersets.) You then want to see what happens when you remove some of the $\mathbb{Z}_2$s and add in a $\mathbb{Z}_4$, or $\mathbb{Z}_8$, and so on. $\endgroup$
    – user1729
    Nov 10, 2014 at 12:52

2 Answers 2


The answer is no, but the solution is somewhat involved.

Given a finite abelian group $G$, let $T_n(G)=\{g\in G | ng=0 \}$ denote the $n$-torsion of $G$. Let $T(G)$ be the infinite sequence whose $n$th entry is $\log_2 |T_{2^n}(G)|$. We will call an infinite sequence of non-negative integers admissible if it is of the form $T(G)$ for some finite abelian group $G$ with order a power of $2$. Our main result is the following.

Theorem: A sequence is admissible if and only if it is bounded, non-decreasing, and the difference between succesive elements non-increasing. In symbols, $(b_1,b_2,\ldots)$ with $b_i\in \mathbb Z$ is admissible if, setting $b_0=0$, we have $b_i\geq 0, 0\leq b_{i+1}-b_i\leq b_i-b_{i-1}$ for all $i\in \mathbb N$.

Let us use the theorem to solve the problem. Suppose that $G$ is a group of order a power of $2$ such that $T(G)=(b_1, b_2, \ldots, b_{n-1}, b_n, b_n, b_n, \ldots)$ where $b_n \neq b_{n-1}$. Set $b_0=0$. Then the number of elements of $G$ of order $2^k$ is $2^{b_k}-2^{b_{k-1}}$, and so

$$\prod_{g\in G} o(g)=\prod_{i=1}^n (2^i)^{2^{b_k}-2^{b_{k-1}}}=2^{n2^{b_n}-\sum_{i=0}^{n-1}2^{b_i}}. $$

Our question becomes whether there is an admissible sequence as above such that

$$2013 = n2^{b_n}-\sum_{i=0}^{n-1}2^{b_i}.$$

We will assume there is and derive a contradiction. The conditions on our sequence imply that it is strictly increasing up until $b_n$, and so the term $\sum 2^{b_i}<2^{b_n}$. Therefore,


and so if we had the value $b_n$, we must have

$$n=\lceil \frac{2013}{2^{b_n}} \rceil. $$

Additionally, since the $b_i$ are all distinct (up through $b_n$), if we write $n2^{b_n}-2013$ in binary, the result must have exactly $n$ $1$'s in it.

Since $b_n \geq n$, we cannot have $b_n \leq 7$. If $b_n>10$, then $n=1$. However, since we have $b_0=0$, we would have $2^{b_n}=2013+1$, which has no integer solutions. Therefore, any solution would have $b_n=8, 9$, or $10$, in which case we have $n=8, 4$, or $2$ respectively. In all three cases, we have $$n2^{b_n}-2013=35=1+2+32,$$

and so we would have to have $n=3$. However, this does not occur in any of these cases.

The proof of the theorem can be done through a series of mostly easy lemmas and calculations. The proofs of most of them will be intentionally omitted. In what follows, all groups are assumed to be finite and abelian, and after the first lemma, they are also assumed to have order a power of $2$. $G$ and $H$ will denote groups, all sequences will be infinite sequences of non-negative integers. Sums of sequences will denote component-wise addition. $C_n$ will denote the cyclic group of order $n$.

Lemma: $|T_n(G\times H)|=|T_n(G)| \cdot |T_n(H)|$

Lemma: $T(G\times H)=T(G)+T(H)$

Definition/calculation $A_n:=T(C_{2^n})=(1,2,3, \ldots, n, n, n, \ldots)$, i.e., $A_n(i)=\min(n,i)$.

Definition: A partition is non-increasing sequence that is eventually zero. We will often omit the zero terms and write a partition as a finite, non-increasing sequence. Associated to the partition $\lambda = (\lambda_1, \lambda_2, \ldots)$, we have a Young Diagram, which is a collection of left justified square boxes (or dots arranged in a square lattice type configuration) where we have a box at position $(i,j)$ if and only if $\lambda_i\geq j$. The conjugate of $\lambda$ is the partition $\lambda^*=(\lambda_1^*, \lambda_2^*, \ldots)$ such that the $i$th row of the Young diagram associated to $\lambda^*$ has the length of the $i$th column of the diagram associated to $\lambda$. For example, the conjugate of the partition $(4,2,2)$ is $(3,3,1,1)$. See the link for pictures and a more thorough description.

By the structure theorem for finitely generated abelian group, if $|G|$ is a power of $2$, we can write $G=\prod C_{2^{\lambda_i}}$ for a unique partition $\lambda$. By our lemmas, $T(G)=\sum A_{\lambda_i}$.

Lemma: Using the notation of the previous two paragraphs, $$\sum A_{\lambda_i}=(\lambda_1^*, \lambda_1^* + \lambda_2^*, \ldots, \sum_{i=1}^n \lambda_i^*, \ldots).$$

For a proof, draw a Young diagram.

Sketch of proof of the Theorem: By the lemmas, admissible sequences are exactly sums of $A_i$'s. Furthermore, we the map sending a partition to its conjugate and the map sending a partition to a non-decreasing sequence with non-increasing differences $(\lambda_i)\mapsto (\sum_{i=1}^n \lambda i)$ are both invertable.


The following answer is based on the fundamental theorem for finitely generated abelian groups. If you have never heard of it, then you won't understand why this general idea works.

$\mathbb{Z}_2^{11}=\mathbb{Z}_2\times\cdots\mathbb{Z}_2$ gives you $2^{2047}=2^{2^{11}-1}$. Can you see why this is? (Hint: powersets.) You then want to see what happens when you remove some of the $\mathbb{Z}_2$s and add in a $\mathbb{Z}_4$, or $\mathbb{Z}_8$, and so on.

For example, $\mathbb{Z}_2^n\times\mathbb{Z}_8$ gives you $2^{2^{n}\times 18-1}=2^{4\times3\times 2^n+2\times2\times 2^n+1\times 2^n+(2^n-1)}$ (pair off each element of order 1 or two obtained from the $\mathbb{Z}_2^n$, there are $2^{n}$ of them by above, with each power from the $\mathbb{Z}_8$), which can never equal $2^{2013}$! So, what about $\mathbb{Z}_2^n\times\mathbb{Z}_4$? You try :-)

  • $\begingroup$ Why is $2^{13}$ already too big? $\endgroup$ Nov 10, 2014 at 13:02
  • $\begingroup$ @Tobias Err. Because I was confused? I'll edit it. (I was going $2^{13}>2^{11}$.) $\endgroup$
    – user1729
    Nov 10, 2014 at 13:05
  • $\begingroup$ I am not sure what powersets you are referring to here. $\endgroup$ Nov 10, 2014 at 13:18
  • $\begingroup$ @Tobias The powerset of all the components of $\mathbb{Z}^n$ (for the first bit)...for example, $\mathbb{Z}_2\times\mathbb{Z}_2=\langle a, b\rangle$. Powerset is $\{\{a\}, \{b\}, \{a,b\}, \{1\}\}$. Then remove the identity as it does not have order two. $\endgroup$
    – user1729
    Nov 10, 2014 at 13:23
  • $\begingroup$ The number you get is way too small (if you set $n=1$ then it is still smaller than what it should be for $n=0$ where we get the product of the orders of the elements of $\mathbb{Z}_8$ which is $(2^3)^4\times (2^2)^2\times 2 = 2^{3\times 4 + 2\times 2 + 1} = 2^{17}$. $\endgroup$ Nov 10, 2014 at 13:31

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