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Given a finite group $G$ and $N$ a normal subgroup such that $N\cong K$ where $K$ is Klein group and $G/N\cong \mathbb{Z}/7\mathbb{Z}$, prove $N\leq Z(G)$

I managed to deduce from both of the informations given that $|G|=28$

Edit:
$|G| = 28$, From Sylow 2nd theorm we get $n_7\in \{1,2,4\}$ but $n_7=1 \pmod{7}$ so $n_7 = 1$ => The sylow group of order 7 is normal in G. But also the sylow group of order 4 is also normal in G from the assumption so we get $G\cong K\times P_7$ Since $P_7$ is cyclic and K is abelian we get G to be abelian (which is stronger then the requested claim)

Arturo Magidin
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    What size can the orbit of a non-identity element of $N$ have when conjugating by $g \in G$ with order $7$? – Robert Shore Jun 27 '22 at 16:08
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    The requested claim implies $G$ is abelian. It is well-known and oft-asked that if $N\leq Z(G)$ and $G/N$ is cyclic, then $G$ is abelian. – Arturo Magidin Jun 27 '22 at 16:59
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    While your argument is good, you do not need Sylow's Theorems, just Cauchy. By Cauchy's Theorem, $G$ has an element of order $7$, call it $g$, and let $H=\langle g\rangle$. Since $N$ is normal, conjugation by $g$ induces an automorphism of $N$, $n\mapsto gng^{-1}$ for each $n\in N$, which gives a morphism $H\to \mathrm{Aut}(N)$. But the automorphism group of the Klein $4$-group is isomorphic to $S_3$, so it has no elements of order $7$. Thus, the map is trivial, so conjugation by $g$ must act like the identity on $N$. Thus, $G=N\times H$. – Arturo Magidin Jun 27 '22 at 17:16

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