Finite groups with elements of order n

Question

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.

Answer 1

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 $m$).

Answer 2

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.

Answer 3

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.)

Answer 4

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.

Answer 5

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.

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