Exercise 15, Section 2.4 of Hungerford’s Abstract Algebra

Question

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.

@user10354138 opsss I made a mistake. I think I can fix my proof. Give me some time. — user264745, Sep 16 '23 at 13:48
@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. — user264745, Sep 16 '23 at 14:31
Also note this isn't true in general, think of the smallest non-abelian group you know. — Steve D, Sep 16 '23 at 16:12
@SteveD Yes I have proved normalizer centralizer theorem. Yaa my attempt is wrong. — user264745, Sep 16 '23 at 16:26

score 0 · Accepted Answer · answered Sep 18 '23 at 16:37

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 $N=\langle a\rangle\subseteq C(G)$.

Exercise 15, Section 2.4 of Hungerford’s Abstract Algebra

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