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I see here and in somewhat different terms here that if the index of $Z(G) = 4$, $G/Z(G)$ cannot be cyclic.

Forgive me if the question should be reworded, as I may be asking more along the lines of how to prove $G/Z(G)$ cannot be cyclic rather than the main question, but given this idea seems implied in previous answers I've seen, I cannot understand the proofs I've found.

Thank you.

zq_Aux
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    If $G/Z(G)$ is cyclic then $G$ is abelian. See, e.g., this. – lulu Nov 15 '20 at 20:18
  • @lulu thanks for commenting. Does that then imply that if G is abelian, Z(G) cannot have index 4? Why would that be? Perhaps my main lack of understanding lies in group centers. – zq_Aux Nov 15 '20 at 20:19
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    If $G$ is abelian then $Z(G)=G$ (well, that's really the definition of "abelian"). It therefore follows from the claim I cited that, if $G/Z(G)$ is cyclic it must in fact be trivial. – lulu Nov 15 '20 at 20:19
  • It might help to know that $$G/Z(G)\cong {\rm Inn}(G),$$ where ${\rm Inn}(G)$ is the group of inner automorphisms of $G$ under composition. – Shaun Nov 15 '20 at 20:21
  • Ah, I see now that my understanding of the index of a subgroup is what was holding me back. If G is abelian, then G/Z(G) = 1, therefore the index of Z(G) being 4 is a contradiction, yes? Thanks everyone for the help. – zq_Aux Nov 15 '20 at 20:26
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    Yes, exactly - this is the definition of the index! Indeed, $4=(G:Z(G))=\frac{|G|}{|Z(G)|}$. – Dietrich Burde Nov 15 '20 at 20:30
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    It is perfectly possible that $Z(G)$ is an index $4$ subgroup (happens for example for both non-abelian groups of order $8$). But in those cases the quotient $G/Z(G)$ cannot be a cyclic group (in fact, it must be isomorphic to the Klein four). That's all you can conclude. – Jyrki Lahtonen Nov 15 '20 at 20:47
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    The index $4$ is interesting here because $4$ is the smallest non-prime. It is correct to conclude that $[G:Z(G)]$ cannot be any prime number $p$. For otherwise the argument Lulu explained would kick in. $G/Z(G)$ would necessarily be cyclic (because a group of prime order is always cyclic), and hence $G$ would be abelian, and $[G:Z(G)]=1\neq p$. A contradiction. – Jyrki Lahtonen Nov 15 '20 at 20:50

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