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Consider a finite group where all elements have the same order $n$. What could be said about such groups?

For $n=2$ it could be proved that such group is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^k$. Could it be somehow generalized on case $n>2$?

EDIT: Surely the identity has order 1, so we have to exclude it.

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$n$ must be a prime. – Steve D Jul 14 '10 at 5:33
Can I also ask if you were looking at this question from curiosity, or if it was suggested to you as something to think about? – Yemon Choi Jul 14 '10 at 5:35
If all elements have the same order, then since the identity has order $1$, the group must be trivial. So surely you mean to exclude the identity. Also, it is easy to see that if all non-identity elements have the same order, that order must be a prime number $p$. If $p>2$, then it is well-known that there are finite groups of exponent $p$ which are not commutative: e.g. the group of upper triangular $3 \times 3$ matrices over $\mathbb{F}_p$ with $1$'s on the main diagonal ("finite Heisenberg group"). – Pete L. Clark Jul 14 '10 at 5:35
@falagar: Could you please address what remains of your question in light of my comment above? "Could it be somehow generalized" is not a good question: virtually anything can be somehow generalized. What exactly are you looking for? – Pete L. Clark Jul 14 '10 at 5:47
The example of the Heisenberg group doesn't necessarily satisfy all elements having the same order. For example, the Heisenberg group over $C_2$ is $ D_8$. – Doug Oct 28 '14 at 1:25
up vote 18 down vote accepted

This question is closely related to the restricted Burnside problem: given numbers $m$ and $p$, is the restricted Burnside group $B_0(m,p)$ finite? Every group with $m$ generators of exponent $p$ is the quotient of the Burnside group $B(m,p)=F_m/\langle w^p\rangle,$ where $F_m$ is a free group with $m$ generators, and $B_0(m,p)$ is the quotient of $B(m,p)$ by the intersection of all subgroups of finite index (which is a normal subgroup). For the case of prime exponent, A.I. Kostrikin proved that the restricted Burnside problem has affirmative solution (and Efim Zelmanov proved it in general). Thus the answer to the original question is:

A finite group $G$ has the property that all non-unit elements have the same order $p$ if and only if $p$ is prime and $G\ne 1$ is a quotient of $B_0(m,p)$ for some $m.$

For small values of $p$, even the Burnside group $B(m,p),$ which is somewhat easier to study, is known to be finite ($p=2,3$) and one may hope to get a more precise answer (for $p=2$ the group is elementary abelian 2-group of rank $m$).

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I was going to mention the restricted Burnside problem earlier, but I wasn't sure if it was what the original questioner was looking for, or indeed if he or she was being led up to rediscovering parts of it by a third party. – Yemon Choi Jul 14 '10 at 6:48
It seems to me that the question is about group all of whose non trivial elements have order equal to $n$, not at most $n$. – Benoît Kloeckner Jul 14 '10 at 6:56
Benoit: as remarked above, if every non-identity element has order $n$, then $n$ has to be prime; so I think the distinction you allude to disappears, no? – Yemon Choi Jul 14 '10 at 6:58
Yemon: Regardless of OP's intentions, his/her question admits a precise mathematical answer. I am very cautious not to give away answers to problems or projects that someone should think over themselves, but I firmly believe in indicating to that person the right theory or tools. – Victor Protsak Jul 14 '10 at 8:19
@Benoit: de rien. – Yemon Choi Jul 14 '10 at 9:36

Adding to Victor's answer, the answer is "sort of yes" for $p=3$. The group $B(n,3)$ is nonabelian for $n>1$ but admits a normal form see "Group Theory" by M. Hall. If $p>3$ you are out of luck: $B_0(2,5)$ is known to have $5^{34}$ elements but $B_0(3,5)$ and $B_0(2,7)$ are too hard to handle with approximately $5^{2280}$ and $7^{10000}$ elements. See "Around Burnside" by Kostrikin for detailed discussion.

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This is not really on point, but it may be of some interest. The finite groups of prime exponent are exactly those finite groups that are the set-theoretic union of a collection of pairwise trivially intersecting proper subgroups, all of which have the same order. (See my paper in the Pacific J, of Math. 49, 1973.)

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The answer is $no$, even for $n=3$. The group $G$ whose presentation is $G= \langle x, y, z | x^3=1, y^3=1, z^3=1, [x,z]=1, [y,z]=1, [x,y]=z^{-1} \rangle$ is non-abelian of order $27$, and all its non-trivial elements have order $3$. This is the group whose label is $[27,3]$ in GAP or MAGMA list of small groups.

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This information is already contained in my comment above. – Pete L. Clark Jul 14 '10 at 15:31
Sorry, you are right. There are many comments on this post, and I missed yours... – Francesco Polizzi Jul 14 '10 at 16:37

There is a related paper for the general case, where all elements (nontrivial) has prime power order. Although, this paper does not answer your question in generally, but it has some good techniques for attacking to such problem. The paper name is:

"Classification of Finite Groups with all Elements of Prime Order" by "Marian Deaconescu".

In this paper, he studied the variant of above question and obtained some results as follows:

Let $\mathcal{P}$ be the class of the finite groups having all (nontrivial) elements of prime order. Let $G$ be a $\mathcal{P}$-group. Then one of the following cases occurs:

$\text{I}.$ $G$ is a $p$-group of exponent $p$.

$\text{II}.$ $(a)$ $|G|=p^aq$, $3\leq p<q$, $a\geq 3$, $|F(G)|=p^{a-1}$, $|G:G'|=p$.

$(b)$ $|G|=p^aq$, $3\leq q<p$, $a\geq 1$, $|F(G)|=|G'|=p^a$.

$(c)$ $|G|=2^ap$, $p\geq 3$, $a\geq 2$, $|F(G)|=|G'|=2^a$.

$(d)$ $|G|=2p^a$, $p\geq 3$, $a\geq 1$, $|F(G)|=|G'|=p^a$ and $F(G)$ is elementary abelian.

$\text{III}.$ $G\cong A_5$.

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