Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
Hint: If the group has a nontrivial center show the problem can be solved. Then think about which general kinds of groups you know have a nontrivial center. –  KCd Oct 20 '11 at 12:27
    
I also noticed this before but I do not know how to prove it, I know that for example p-groups have a nontrivial center but in this case the group may be of the infinite order. Ok, at least now I know it's a good clue –  dawid Oct 20 '11 at 12:42

3 Answers 3

up vote 2 down vote accepted

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.

share|improve this answer
    
this proves Z(G) = 1 or Z(G) = G. In some of Steve's examples, Z(G) = 1. –  Jack Schmidt Oct 20 '11 at 18:32
    
@JackSchmidt My bad. I'll try to complete the answer, for the moment I added some specifications, thanks for the correction. –  Giorgio Mossa Oct 20 '11 at 19:01
    
All good. KCd's suggestion is correct for finite groups, and since the thesis is false for infinite groups, I assume the original asker forgot a hypothesis. –  Jack Schmidt Oct 20 '11 at 19:17

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.

share|improve this answer
    
In the torsion-free case there can be no hope of classification, since every torsion-free group will embed in one with exactly two conjugacy classes. –  user641 Oct 20 '11 at 15:26
    
@Steve: thanks, added. Note the same is true of torsion-free abelian groups, yet torsion-free abelian divisible groups are at least reasonably classified by dimension. However, at the very least it shows there are many non-abelian examples. –  Jack Schmidt Oct 20 '11 at 15:48
    
+1${}{}{}{}{}{}$ –  t.b. Oct 20 '11 at 23:57

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.

share|improve this answer
    
thank you all very much for your excellent response! –  dawid Oct 20 '11 at 22:11
    
Sylow's theorem is a slight overkill. Cauchy's is enough. –  j.p. Oct 28 '11 at 15:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.