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**NonCommAlg** let $\displaystyle N=<x>$ and $\displaystyle y \in G.$ if $\displaystyle o(y)=p^2,$ then $\displaystyle G$ would be cyclic, hence abelian, and we're done. also if $\displaystyle y=1_G,$ then $\displaystyle xy=yx.$ so we may assume that $\displaystyle o(y)=p.$ now we have $\displaystyle yxy^{-1}=x^j,$ for

some $\displaystyle j,$ because $\displaystyle N \lhd G.$ then $\displaystyle x=y^pxy^{-p}=x^{j^p}$ and therefore $\displaystyle x^{j^p - 1} = 1.$ hence $\displaystyle p \mid j^p -1,$ because $\displaystyle o(x)=p.$ but, by Fermat's little theorem, we also have that $\displaystyle p \mid j^p - j.$ thus $\displaystyle p \mid j - 1$ and

so $\displaystyle x^j = x.$ therefore $\displaystyle yxy^{-1}=x^j=x.$ hence $\displaystyle yx=xy. \ \Box$

__Remark__: a similar argument can be used to prove a more general result: if $\displaystyle G$ is any group of order $\displaystyle p^n,$ whre $\displaystyle p$ is prime, and $\displaystyle N \lhd G$ with $\displaystyle |N|=p,$ then $\displaystyle N \subseteq Z(G).$