# Existence of weird automorphisms in finite groups

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?

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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 '11 at 22:56
Could the normal subgroup be all of G? Because then we're fine. –  Makhalan Duff Mar 25 '11 at 23:02
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 '11 at 23:16
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 '11 at 0:26
(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 '11 at 0:26