Suppose we are given a finite group $G$, and we know the order of $G$ as well as the orders of each of its elements. Does this information alone uniquely determine the group up to isomorphism? What if we add extra hypotheses about $G$? If not, can we at least extract some interesting information regarding its structure?

For example, if $|G| = 8$ and $G$ contains $3$ elements of order $2$ and $4$ elements of order $4$, then I know $G \cong \mathbb{Z}_4\times\mathbb{Z}_2$.

  • $C_2\times C_2$ and the Klein 4-group both have 3 elements of order 2, but are non-isomorphic. So in general this is insufficient to completely determine a group. – ChocolateAndCheese Dec 21 '16 at 1:18
  • 6
    @ChocolateAndCheese: the Klein 4-group is just another name for $C_2\times C_2$. – carmichael561 Jan 15 '17 at 5:40
up vote 44 down vote accepted
+100

In the following, denote by $a_n(G)$ the number elements satisfying $x^n = 1$ in $G$ ($n$ any integer). Now two groups $G$ and $H$ have the same order sequence (ie. same order, and the same number of elements of each order) if and only if $a_n(G) = a_n(H)$ for all integers $n$. To see this, you can use the fact that $a_n(G) = \sum_{d \mid n} o_d(G)$, where $o_d(G)$ is the number of elements of order $d$ in $G$.


Suppose that $G$ and $H$ have the same order sequence. As Alexander mentions in his answer, this does not imply that $G$ and $H$ are isomorphic.

But it is true if we assume that $G$ is simple! This follows from a stronger result that was proven in 2009. See the slides here for more information.

If $G$ is simple, then $G$ and $H$ are isomorphic.

The proof of the above statement uses the classification of finite simple groups (not surprising).

There are a few properties that are determined by order sequences, and many that are not.

If $G$ is cyclic, then $H$ is cyclic.

Proof: This is a common lemma. If $a_n(G) \leq n$ for all integers $n$, then $G$ is a cyclic group. See answers in this question.

If $G$ is nilpotent, then $H$ is nilpotent.

Proof: A finite group $G$ is nilpotent if and only if it every Sylow subgroup of $G$ is normal. Let $p$ be a prime divisor of $|G|$ and $p^\alpha$ the order of a Sylow $p$-subgroup in $G$. Then $G$ has a normal Sylow $p$-subgroup if and only if $a_{p^{\alpha}}(G) = p^\alpha$.

Counterexamples for some properties:

  • $G$ abelian, $H$ nonabelian: $G = C_p \times C_p \times C_p$, $H$ Heisenberg group of order $p^3$
  • $G$ supersolvable, $H$ not supersolvable: $G = \operatorname{SmallGroup}(72,35)$ $H = \operatorname{SmallGroup}(72, 40)$.
  • In the previous example $G$ also satisfies the converse to Lagrange's theorem, and $H$ does not.
  • $G$ and $H$ have different numbers of Sylow $p$-subgroups: $G = \operatorname{SmallGroup}(216,34)$, $H = \operatorname{SmallGroup}(216,100)$. Here $G$ has $27$ Sylow $2$-subgroups, while $H$ has $9$ Sylow $2$-subgroups.
  • (Thompson) $G$ and $H$ have different composition factors. The Mathieu group $M_{23}$ (wikipedia) has two maximal subgroups $G$ and $H$ of order $40320$ such that $G$ and $H$ have the same order sequence, but $G = \operatorname{PSL(3,4)}:2$ and $H = 2^4 : A_7$.
  • etc..

There is a theorem of Frobenius (1895) which states that if $n$ divides the order of $G$, then $a_n(G)$ is a multiple of $n$. Frobenius conjecture states that if $n$ divides the order of $G$ and $a_n(G) = n$, then $G$ has a unique subgroup of order $n$. This conjecture was proven in 1991, and the proof uses the classification of finite simple groups. Hence

If $G$ has a unique subgroup of order $n$, then $H$ has a unique subgroup of order $n$.

Using Frobenius conjecture, we can also deduce the following.

If $G$ is supersolvable, then $H$ is solvable.

Proof: (from this article) Suppose $G$ has order $|G| = p_1^{a_1} \ldots p_t^{a_t}$, where $p_1 < p_2 < \ldots < p_t$ are primes. We know that $|G| = |H|$. Furthermore, since $G$ supersolvable, we see that $a_n(G) = n$ for every $n = p_i^{a_i} p_{i+1}^{a_{i+1}} \ldots p_t^{a_t}$. Hence $a_n(H) = n$ for every such $n$ since $G$ and $H$ have the same order sequence. By Frobenius conjecture, $H$ has a unique subgroup of order $p_i^{a_i} p_{i+1}^{a_{i+1}} \ldots p_t^{a_t}$ for every $i$ and thus $H$ must be solvable.

So what about when $G$ is solvable? Can we deduce that $H$ must be solvable? Nobody knows. This is an open problem, due to J. G. Thompson (Kourovka Notebook, 12.37).

Finally, let me mention that there are some conditions on the order sequence which force solvability. For example, if $a_n(G) \leq 7n$ for all integers $n$ dividing the order of $G$, then $G$ is solvable. Note that when $G = A_5$, we have $a_n(G) \leq 8n$ for all $n$ dividing the order of $G$.

  • 2
    Ooohh, thanks for that conjecture of Thompson's. I hadn't heard about that. Great answer! – Alexander Gruber Mar 28 '14 at 22:09
  • That Kourovka Notebook is super interesting, by the way. I recall seeing the first new question (18.1) come up on MO. – Alexander Gruber Mar 28 '14 at 22:16
  • Yeah, Kourovka Notebook is pretty amazing. If you are interested, there are some references and information about Thompson's conjecture in these slides. Pages 11-12 contain a letter of Thompson, where he mentions this problem and gives the example above ($G$ and $H$ with same order sequence, different composition factors). It seems that most of the material about Thompson's conjecture is unpublished or in obscure Chinese journals.. – Mikko Korhonen Mar 28 '14 at 22:25
  • This deserves more votes! – Pedro Tamaroff Mar 29 '14 at 20:36

No.

The list of orders of each element a group $G$ sorted in ascending order is called the order sequence of $G$. Here is a list of order sequences for groups of order up to $512$. In particular, the smallest counterexamples have order $16$, corresponding to the following order sequences:

$$\begin{array}{cccccccccccccccl}(a)& 1, &2, &2, &2, &4, &4, &4, &4, &4, &4, &4, &4, &4, &4, &4, &4\\\\ (b)& 1, &2, &2, &2, &2, &2, &2, &2, &4, &4, &4, &4, &4, &4, &4, &4\\\\ (c)& 1, &2, &2, &2, &4, &4, &4, &4, &8, &8, &8, &8, &8, &8, &8, &8 \end{array}$$

There are three isomorphism classes of groups with order sequence $(a)$, three with order sequence $(b)$, and two with order sequence $(c)$.

You may also want to read this answer.

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