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Given a finite nonabelian group $G$, let $p$ be the smallest divisor of $|G|$. Then I want to show that

  1. $p^2 |Z(G)| \le |G|$
  2. If $N \le G$ is normal and $|N|=p$ then $N \le Z(G)$.

For the first one I think that since $G$ is non-abelian, $|G/Z(G)|>1$ and since $|G/Z(G)| $ divides $|G|$ then $|G/Z(G)|$ is at least $p$. However, in this case $G$ would be abelian. Then we must have $|G/Z(G)| \ge p^2$ thus $|G|\ge p^2 |Z(G)|$.

How can I show the second one?

Ninja
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    What do you mean by ``in this case $G$ would be abelian''? In what case? Why not $p < |Z(G)| < p^2$? Part 2 follows from that fact that ${\rm Aut}(N)$ has order $p-1$, which is divisible only by primes less than $p$. – Derek Holt Dec 20 '20 at 13:32
  • $Z(G)$ is of order $pk$ for some integer $k$ so that $pk$ divides $|G|$ but the smallest possible $k$ except $1$ is $p$ that is why I wrote in this way – Ninja Dec 20 '20 at 13:42
  • Why $Aut(N)$ has order $p-1$? – Ninja Dec 20 '20 at 13:43
  • Why is $Z(G)$ of order $pk$ for some integer $k$? – Derek Holt Dec 20 '20 at 16:39

1 Answers1

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$G/Z(G)$ is not cyclic, and so $|G/Z(G)|$ is not a prime. Thus, $|G/Z(G)|\ge p^2$.

For the second question, it suffices to show that $C_G(N)=G$. Note that $N_G(N)/C_G(N)\lesssim\operatorname{Aut}(N)\cong \mathbb{Z}_{p-1}$ (see this question). As $p$ is the smallest prime divisor, it implies that $N_G(N)/C_G(N)=1$.

Groups
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