If N is normal in G, show Zi(G)NN≤Zi(GN) where Zi(G) is the i-th term in the upper central series for G.

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bubble53zjr · Accepted Answer

As you note, if K is a group, then showing that x∈Zi+1(K) is equivalent to showing that [x,y]∈Zi(K) for all y∈K. You are assuming that Zi(G)NN≤Zi(GN). You want to prove that Zi+1(G)NN≤Zi+1(GN). That means that you want to show that for all x∈Zi+1(G),[xN,yN]∈Zi(GN) for all yN∈GN. So, let x∈Zi+1G. We want to show that xN∈Zi+1(GN). Let y∈G be arbitrary. Then [xN,yN]=[x,y]N∈GN. If x∈Zi+1(G),then [x,y]∈Zi(G), so [xN,yN]=[x,y]N∈Zi(G)N.But Zi(G)N. But Zi(G)NN≤Zi(GN) by the induction hypothesis, so [xN,yN]∈Zi(GN). As this holds for all yN∈GN, it follows that xN∈Zi+1(GN), as required.

If N is normal in G, show Z_{i}(G) \frac{N}{N} \leq

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