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Let be $m$ an integer, $m>0$, $n=p_1^{a_1}...p_t^{a_t}$. Define $f(n)=a_1+...+a_t$. Let $G$ be a finite group and define $f(G)=\max\left\{\,f(\lvert g\rvert):g\in G\,\right\}$.

Is there a finite solvable group $G$ with $f(\exp (G))>3\,f(G)?$

And, in general, is there a finite solvable group $G$ with $f(\exp(G))>n\,f(G)$, for each $n \in \mathbb{N}?$

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  • $\begingroup$ Why $3f(G)$? Do you know examples with $f({\rm exp}(G)) = 3f(G)$? $\endgroup$
    – Derek Holt
    Commented Jul 11, 2014 at 18:53
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    $\begingroup$ With Frobenuis group, we can construct a solvable group with $ f(exp(G))/f(G) \rightarrow 3$. But I don't know examples with $f(exp(G))=3f(G)$. $\endgroup$
    – user55910
    Commented Jul 11, 2014 at 19:46
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    $\begingroup$ Which Frobenius groups bring the quotient close to $3$? $\endgroup$
    – Peter Mueller
    Commented Jul 12, 2014 at 17:46
  • $\begingroup$ $H$ the semidirect product of a cyclic group $A=C_{p^{m}}$ by a cyclic group $B=C_{q^{n+1}}$ which induces an automorphism of order $q^{n}$ on $A$ ($p$ and $q$ primes appropriate). $N$ the nilpotent group $C_{l^{d}}$ with $l \neq p,q$. Let $G=NH$ of Frobenius. If $n=m=d-1$, then $f(exp(G))=3d-1$ and $f(G)=d$ $\endgroup$
    – user55910
    Commented Jul 13, 2014 at 10:18
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    $\begingroup$ I think $N$ should be a direct product of copies of $C_{l^d}$ rather than just $C_{l^d}$. I checked that the construction works for $d=2$ with $A=C_7$, $B=C_9$ and $N = C_4^6$, so it seems plausible that it works for higher $d$. Assuming that to be the case, I would say that the question is interesting! $\endgroup$
    – Derek Holt
    Commented Jul 15, 2014 at 20:18

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