A condition of abelian groups related to an automorphism

Let $G$ be a group. If for any $a, b\in G\backslash\{1\}$ there exists an automorphism $\sigma$ of $G$ such that $\sigma (a)=b$ then $G$ is abelian.

I am not the best of the algebra and I have a problem with the proof of the fact above although I guess it must be very simple.. I will be grateful for hint.

Solution 1:

Every element of G has the same order (either finite or infinite). G has no proper non-identity characteristic subgroups (it is characteristically simple), so considering the derived subgroup, G is either abelian or perfect. If G is finite, then it must be a p-group. Since a finite non-identity p-group is not perfect (maximal subgroups are normal), G is an elementary abelian p-group.

In general, if G is abelian, it must either be an elementary abelian p-group (abelian and all elements of the same finite order), or it must be a torsion-free divisible group (a vector space over $\mathbb{Q}$; choose f so that $f(nx) = x$, but then $n f(x) = f(nx) = x$ so x is divisible).

However, there are characteristically simple infinite p-groups, and I don't know that they do not have transitive automorphism groups. Three important examples are McLain groups (locally finite, lots of subnormals, upper triangular matrices over GF(p)), Hall groups (locally finite, few subnormals, wreath products), and Tarski monsters (2-generated simple groups).

As Steve mentions, in the torsion-free case, there are many non-abelian examples due to Higman (and Higman–Neumann–Neumann), since every torsion-free group embeds into a torsion-free group in which the inner automorphism group acts transitively on the non-identity elements.

Solution 2:

After one proves as KCd suggested that $Z(G)$ shouldn't be trivial then the thesis follows from this: let $G$ be a group such for each $a,b \in G \setminus\{1\}$ exists $\sigma \in \text{Aut}(G)$ such that $\sigma(a)=b$. We have that $Z(G)$, the center of the group $G$, is characteristic subgroup of $G$, so for each $a \in Z(G)$ we must have $\sigma(a) \in Z(G)$, but for hypothesis given $a \in Z(G)\setminus \{1\}$ and for each $b \in G\setminus\{1\}$ there must be a $\sigma \in \text{Aut}(G)$ such that

$$b = \sigma(a) \in Z(G)$$

because this holds for each $b \in G$ this means $G \subseteq Z(G)$. Thus $Z(G)=G$ and so $G$ must be abelian.

Edit:

Let's see that $Z(G)$ is not trivial for $G$ finite: because $G$ is finite all the elements in $G \setminus \{1\}$ have the same order $p$, $p$ must be a prime, otherwise given $a \in G \setminus \{1\}$ the subgroup $\langle a \rangle$ should contain element, different from the identity, of order lesser then $p$; thus $G$ must be a $p$-group, and so it has non trivial center.

Solution 3:

This was a problem which i found in the book: "Berkeley problems in Mathematics". The solution is as follows:

Since $G$ is finite, every element of $G$ has finite order. Since any two elements of $G \setminus \{e\}$ are related by an automorphism of $G$, all such elements have the same order, say $q$. Since all powers of an element of $G \setminus \{e\}$ have order $q$ or $1,q$ is prime. By sylow's theorem the order of $G$ is a power of $q$. Therefore $Z(G)$ contains an element other than $e$. Since the center of $G$ is invariant under all automorphisms, $Z(G) = G$ which implies $G$ is abelian.

Solution 4:

Let $a,b$ be any two elements of $G$ and put $x:= aba^{-1}b^{-1}$. By assumption there exists an automorphism $\sigma \colon G \to G$ such that $\sigma(x) = e$, whence $\sigma(ab) = \sigma(ba)$ whence $ab = ba$ (since $\sigma$ is 1-1).