Let's call a group $G$ homogeneous if for every two distinct, non-identity elements $a$ and $b$ there is an automorphism $\phi$ of $G$ such that $\phi(a)=b$.
Examining this definition, we can see that the underlying additive group of any field (indeed, any division ring) is homogeneous by examining the automorphism $\phi(x)=ba^{-1}x$.
Thus for any $n$ and any prime $p$, we have that $Z_p^n$ is homogeneous. Conversely, if we also know that $G$ is finite, we can show that $G$ is of the form $Z_p^n$ for some $n$ and $p$.
We can clearly see that all elements of a homogeneous group have the same order. And by Cauchy's theorem, we get that $G$ must have prime-power order. Say $|G|=p^n$. The Conjugacy Class Equation then implies that $G$ has non-trivial center. And since $G$ is homogeneous, all elements are in the center and thus $G$ is abelian. Since $G$ is finite, abelian, and all elements have the same order, we can conclude that $G$ is isomorphic to $Z_p^n$. If it weren't the fundamental theorem of abelian groups would imply there were an element of order $p^i$ for $i>1$ contradicting the fact that all elements have the same order (Cauchy's theorem still implies there's an element of order $p$).
Thus finite, homogeneous groups are kind of boring. Infinite examples are more interesting as we can garner a couple examples just from fields, and they are more difficult to investigate. We can still surmise a couple things when $G$ is infinite. For example, if $G$ has a single non-identity element of finite order then all (non-identity) elements of $G$ have prime order.
Let $x\neq e$ have finite order $n$. Let $n=p^em$ for some prime $p$ with $e\geq 1$ and $p\nmid m$. Then the element $x^{p^{e-1}m}$ has order $p$, and thus all elements have order $p$.
Also, in general,
- If $G$ has a non-trivial center then it is abelian.
- Any quotient by a characteristic subgroup is also homogeneous.
I am having difficulty figuring out other, even basic, things though. My question is kind of general:
What else can we determine about homogeneous groups?
I am specifically interested in these questions:
Does $G$ have to be abelian?If not, can $G$ be perfect?Can $G$ be free? It clearly can't have rank 1, but what about other ranks?- If all elements have prime order, is $G$ a direct sum of multiple copies of $Z_p$?
But any addition is welcome and appreciated.
-UPDATE
Qiaochu Yuan has deftly answered three of my original questions, but one still remains though it may be difficult. However his answer has left me intrigued for a proof of the following claim:
- Every abelian, homogenous group is the underlying additive group of some field.
I don't see how you could construct the multiplication operation, but it seems the choice for which element would become unity is a free choice.
Best Answer
$G$ does not have to be abelian. There exist examples of infinite (necessarily highly nonabelian) groups with exactly two conjugacy classes; that is, even the action of inner automorphisms on non-identity elements is transitive.
As observed in the comments, since $[G, G]$ is a characteristic subgroup, if $G$ is nonabelian then it is necessarily perfect.
As observed in the comments, nonabelian free groups are not perfect.
I don't know the answer to the last question. It seems potentially difficult due to the existence of Tarski monsters.
Groupprops claims that for abelian groups this property is equivalent to being the underlying additive group of a field, although I don't immediately see a proof. As observed in the comments, as groups these are just vector spaces over $\mathbb{F}_p$ or $\mathbb{Q}$.
Edit: Here's a proof that if $G$ is abelian then it's a vector space over $\mathbb{F}_p$ or $\mathbb{Q}$. As you observed, if $G$ has an element of finite order then every element has order $p$ for some prime $p$, hence $G$ is an $\mathbb{F}_p$-vector space. Otherwise, every element has infinite order. If $g \in G$ and $n \in \mathbb{N}$, then by assumption there is an automorphism $\varphi : G \to G$ such that $\varphi(ng) = n \varphi(g) = g$. Hence $G$ is divisible, and an abelian torsion-free divisible group is a $\mathbb{Q}$-vector space.