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Given $ G'' $ is a cyclic group, how to show that $ G'' \subseteq Z(G') $ ? I can prove the result that $ G'\subseteq C_G(N) $ for a normal, cyclic subgroup $N$ using $N/C$ theorem but can't relate with this problem.

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What is $G''$? $[G',G]$ or $[G',G']$? –  Boris Novikov Mar 14 '13 at 21:53
    
It usually denotes the second derived group, which is the second option you list. –  Geoff Robinson Mar 14 '13 at 22:12

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up vote 3 down vote accepted

Hint: $G^{"} \lhd G.$ Therefore the conjugation action of $G$ induces automorphisms of $G^{"}.$ Therefore, $G^{"}$ can play the role of the cyclic normal subgroup $N$ in the situation which you seem to understand,

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But how to relate $ C_{G'}(G'') $ with $ Z(G') $ ? –  smiley06 Mar 14 '13 at 21:27
    
It is certainly the case that $G^{"} \subseteq G^{'},$ and you have proved that $G^{'} \subseteq C_{G}(G^{"}),$ so you have all the information you need. –  Geoff Robinson Mar 14 '13 at 22:11
    
It maybe very simple but still I am not getting it, how do you show that $ G'' $ is normal in $G$ ? –  smiley06 Mar 14 '13 at 23:01
    
If you know about characteristic subgroups, notice that you have $G^{"} {\rm char} G^{'} \lhd G.$ –  Geoff Robinson Mar 15 '13 at 1:32
    
Thanks ! I got it :) –  smiley06 Mar 15 '13 at 10:05

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