Welcome to MathOverflow

Question

6

1

I've been contemplating various conjectures on various fields that a priori don't have anything to do with group theory; and yet these heuristics (that it would take too long to go over, and are besides the point here) seem to imply the following very weird proposition:

Fix a prime $p\geq 3$. Let $G$ be a finite, prime-to-$p$, group generated by $r$ elements: $a_1,...,a_r$. Assume $r\leq p$. Then there exists an automorphism $\phi \in Aut(G)$ such that $G/$the group normally generated by $a_i \phi(a_i)^{-1}$ (for every $i$) is generated by elements whose order divides $p-1$.

This is true for $r=1$ (pick $\phi(a_1)=a_1^{-1}$).

I can prove it for some easy examples. Can you think of a reason for this to be true/of a counterexample?

flag	asked Mar 25 2011 at 22:44 Makhalan Duff 3,837●5●36

1

Not true, take p=7, G = Frobenius group of order 20. Then every automorphism is inner, so your normal subgroup is always the commutator (or trivial), so there is always an element of order 4 (not dividing p-1=6). – Steve D Mar 25 2011 at 22:56

1

Could the normal subgroup be all of G? Because then we're fine. – Makhalan Duff Mar 25 2011 at 23:02

1

Of course, the idea behind it is fairly general: Take primes $q$ and $s$ such that $s$ divides $q-1$, and form the semidirect product $G=C_q\rtimes C_s$. Then all automorphisms of $G$ come from conjugation inside $Hol(C_q)$, so your normal subgroup is always in the $C_q$ factor, so the only possible quotients are $C_s$ or $G$. Now just choose $p$ so that $q$ and $s$ don't divide $p-1$. – Steve D Mar 25 2011 at 23:16

1

It is a fairly standard exercise to show all automorphisms come from conjugation in the holomorph. Basically you have two generators (one for $C_q$ and one for $C_s$), and an automorphism $\phi$, and you ask yourself "where can they go"? Well the generator for $C_q$ (let's call it $x$) needs to stay in $C_q$, because that's where all the elements of order $q$ live. The generator of $C_s$ (let's call it $y$) can go anywhere else, but we must have $\phi(y)$ acting on $\phi(x)$ the same way $y$ acts on $x$. so the only choice is the coset $yC_q$. Thus there at most $q(q-1)$ automorphisms. – Steve D Mar 26 2011 at 0:26

1

(continued) But the holomorph has size exactly $q(q-1)$ and $C_q\rtimes C_s$, as a subgroup, has trivial centralizer. So all automorphisms are brought about by conjugation in $Hol(C_q)$. – Steve D Mar 26 2011 at 0:26

show 4 more comments

Existence of weird automorphisms in finite groups

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.