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If a normal subgroup $N$ of order $p$ ($p$ prime) is contained in a group $G$ of order $p^n$, then $N$ is in the center of $G$.

My attempt: Since $|N|$ is prime $p$, we have $N=\langle a \rangle$. Since $N\lhd G$, we have $gNg^{-1}=N$, $\forall g\in G$. In particular, $gag^{-1}=a^i$ for some $1\leq i\lt p$. So $|a^i|=|gag^{-1}|=|a|$ (second equality is easy to check). Since $\langle a^i\rangle \subseteq \langle a\rangle$ and $|\langle a^i\rangle|= |a^i|=|a|=| \langle a\rangle|$, we have $\langle a^i\rangle = \langle a\rangle$. If $i\gt 1$, then $a\notin \langle a^i\rangle =\langle a\rangle$. We reach contradiction. So $i=1$. Thus $ga=ag$, for all $g\in G$. That is $a\in C(G)$. Hence $N=\langle a\rangle\subseteq C(G)$. Is my proof correct?

In above proof we didn’t use $|G|=p^n$ condition, did we? Here is an alternative proof of above exercise. In my opinion, it is bit complicated proof. There must be some relatively easy solution (using every hypothesis) for this exercise that author had in mind.

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  • $\begingroup$ Why would $a\notin\langle a^i\rangle$ if $i>1$? $\endgroup$ Commented Sep 16, 2023 at 13:40
  • $\begingroup$ @user10354138 opsss I made a mistake. I think I can fix my proof. Give me some time. $\endgroup$
    – user264745
    Commented Sep 16, 2023 at 13:48
  • $\begingroup$ @user10354138 By theorem 6 section 1.3, $(p,i)=1$. For $n=1$, statement holds trivially. For $n\geq 2$, $p^n\gt pi$. By this post, $\exists r,s\geq 0$ such that $p^n=rp+si$. Note $|a^{p^n}|=\frac{p}{(p,p^n)}=\frac{p}{p}=1$. So $a^{p^n}=e$. Then $$e=a^{p^n}=a^{rp+si}=(a^p)^r(a^i)^s=(a^i)^s.$$ Can’t progress from here. $\endgroup$
    – user264745
    Commented Sep 16, 2023 at 14:31
  • $\begingroup$ Do you know the normalizer centralizer theorem? $\endgroup$
    – Steve D
    Commented Sep 16, 2023 at 16:11
  • $\begingroup$ Also note this isn't true in general, think of the smallest non-abelian group you know. $\endgroup$
    – Steve D
    Commented Sep 16, 2023 at 16:12

1 Answer 1

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Following proof is not mine. Original author wish to delete their answer. I am adding it (not word to word) for future reference.

Proof: Since $|N|$ is prime $p$, we have $N=\langle a \rangle$. Since $N\lhd G$, we have $gNg^{-1}=N$, $\forall g\in G$. In particular, $gag^{-1}=a^i$ for some $1\leq i\lt p$. By induction on $k$, it is easy to see that $g^k ag^{-k}=a^{i^k}$. By Lagrange theorem, $|g|$ divides $|G|=p^n$. So $|g|=p^r$ for some $r\geq 0$. Put $k=p^r$, we have $g^{p^r} ag^{-p^r}=eae=a=a^{i^{p^r}}$. So $i^{p^r}\equiv 1\pmod p$.

We claim $i\equiv i^{p^k}\pmod p$ for all $k\geq 0$. Base case: $k=0$. Clearly $i\equiv i\pmod p$. Inductive step: Suppose $i\equiv i^{p^k}\pmod p$ , where $k\geq 0$. Fermat little theorem applied to $a=i^{p^k}$, we have $i^{p^k}\equiv i^{p^{k+1}}\pmod p$. Since $\equiv \pmod p$ is equivalence relation, $i\equiv i^{p^k}\pmod p$ and $i^{p^k}\equiv i^{p^{k+1}}\pmod p$ implies $i\equiv i^{p^{k+1}}\pmod p$. By principle of mathematical induction, $i\equiv i^{p^k}\pmod p$ for all $k\geq 0$.

In particular $i\equiv i^{p^r}\pmod p$. Then $i\equiv i^{p^r}\pmod p$ and $i^{p^r}\equiv 1\pmod p$ implies $i\equiv 1\pmod p$. Thus $gag^{-1}=a^i=a$ for all $g\in G$. That is $a\in C(G)$. Hence $N=\langle a\rangle\subseteq C(G)$.

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