Assume $T$ is finite and non-abelian then why is $T/Z(T)$ non-cyclic? Where $Z(T)$ is the centre of the group $T$.
I've shown $Z(T)$ is a normal subgroup of T, but not sure what to do next or if that helps
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Lemma :$T/Z(T)$ is cyclic implies $T$ is abelian.
Proof:Proving that if $G/Z(G)$ is cyclic, then $G$ is abelian. Can you conclude now?
Arpit Kansal
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I'm not sure I understand how to think of $T/Z(T)$ do we think of it as a subgroup? – The Problem Mar 01 '15 at 18:16
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What do you mean by as a "subgroup"?Don't you about know quotient groups?Don't you know that if $H$ is a normal subgroup of $G$ then $G/H$ is a group? – Arpit Kansal Mar 01 '15 at 18:20
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1$T/Z(T)$ is the set of left (=right, since $Z(T)$ is normal in $T$) cosets of $Z(T)$ in $T$: ${Z(T), t_1Z(T), t_2Z(T),\dots,t_nZ(T)}$ – David Wheeler Mar 01 '15 at 18:23
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1You may read here about quotient groups:http://en.wikipedia.org/wiki/Quotient_group – Arpit Kansal Mar 01 '15 at 18:26