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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
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Could the normal subgroup be all of G? Because then we're fine. – Makhalan Duff Mar 25 '11 at 23:02
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No, it can't, but that example doesn't work because I misread your question (in the previous example it can be generated by elements of order 2). Better to take $C_11\rtimes C_5$, whose normal subgroup is again always in the commutator, but this time there are no elements of order 2 or 3. – Steve D Mar 25 '11 at 23:05
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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
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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

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