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Let $N$ be a normal subgroup of a group $G$ such that $N\cap G'=\{e\}$, where $G'$ is the derived/commutator subgroup of $G$. Then
i.) $N\subseteq Z(G)$, where $Z(G)$ is the center of $G$
ii.) $Z(G/N)=Z(G)/N$

Help me figure this out please... Thank you..

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1 Answer 1

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For (1), show that $[G,N]$ is a subset of both $N$ (using its normality) and $[G,G]$.

And for (2), use $[zN,gN]=N\iff [z,g]N=N\iff [z,g]\in N$ (put $\forall g\in G$ in front if it helps).

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  • $\begingroup$ Thank you Anon, may I ask how will the second statement be used to establish the equality for (ii)? $\endgroup$
    – Philip Benj
    Mar 19, 2013 at 6:26
  • $\begingroup$ @PhilipBenjMarcobyEragon We can write "$zN\in Z(G/N)$" as $\forall g\in G,[zN,gN]=N$, and write the statement "$z\in Z(G)$" as $\forall g\in G,[z,g]=1$. Note that $[z,g]\in[G,G]$ is automatic. In these sorts of elementary problems, to get an intuitive grip on the meanings behind these things, you'll want to practice and experiment with writing out statements in as many equivalent ways as possible. Note that $Z(G/N)=Z(G)/N$ may be written as $zN\in Z(G/N)\iff z\in Z(G)$. $\endgroup$
    – anon
    Mar 19, 2013 at 6:35
  • $\begingroup$ Thank you very much Anon, I get it now... $\endgroup$
    – Philip Benj
    Mar 19, 2013 at 6:51
  • $\begingroup$ It was just a play of equivalent statements. I learned a lot. $\endgroup$
    – Philip Benj
    Mar 19, 2013 at 6:52

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