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Let $n$ be an integer, $n>1$, $n=p_1^{a_1}...p_t^{a_t}$. I define $f(n)=a_1+...+a_t$. Let be $G$ a finite group and I define $f(G)=max(f(|g|):g\in G)$.

I have to prove that if $G$ is a metabelian group with $Z(G)\neq 1$ then $f(G')+f(G/G')<2f(G)$ (clearly, $f(G')+f(G/G')\leq 2f(G)$ holds)

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    $\begingroup$ Why do you "have to prove" this -- is this homework? $\endgroup$
    – Stefan Kohl
    Jun 16, 2014 at 15:55
  • $\begingroup$ In fact what makes you believe that it is true? $\endgroup$
    – Derek Holt
    Jun 16, 2014 at 16:58
  • $\begingroup$ I'm sorry, I wonder if it's true. $\endgroup$
    – user51044
    Jun 16, 2014 at 19:07

1 Answer 1

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The inequality does not hold in general.

Let $p$ be an odd prime, and let $G$ be the nonabelian group of order $p^3$ and exponent $p$ (or more generally, any nilpotent $p$-group of class $2$, $3$, or $4$, and exponent $p$). This is a $p$-group, so $Z(G)\neq 1$; being nilpotent of class $2$, it is metabelian.

Now, because $G$ is of exponent $p$, $f(G)=1$; since $G'$ is nontrivial and of exponent $p$, we also have $f(G') = 1$; and as $G/G'$ is abelian of exponent $p$, we again have $f(G/G') = 1$. Thus, $f(G')+f(G/G') = 1+1 = 2$, and $2f(G)=2$, giving equality. We don't get strict inequality here (or in any $p$-group of class$\leq 4$ and exponent $p$).

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