If $N$ is normal in $G$, show $Z_{i}(G)N/N \leq Z_{i}(G/N)$ where $Z_{i}(G)$ is the $i$th term in the upper central series for $G$.

Solution 1:

As you note, if $K$ is a group, then showing that $x\in Z_{i+1}(K)$ is equivalent to showing that $[x,y]\in Z_i(K)$ for all $y\in K$.

You are assuming that $Z_i(G)N/N\leq Z_i(G/N)$. You want to prove that $Z_{i+1}(G)N/N\leq Z_{i+1}(G/N)$. That means that you want to show that for all $x\in Z_{i+1}(G)$, $[xN,yN]\in Z_i(G/N)$ for all $yN\in G/N$.

So, let $x\in Z_{i+1}G$. We want to show that $xN\in Z_{i+1}(G/N)$.

Let $y\in G$ be arbitrary. Then $[xN,yN] = [x,y]N$ in $G/N$. If $x\in Z_{i+1}(G)$, then $[x,y]\in Z_i(G)$, so $[xN,yN] = [x,y]N\in Z_i(G)N$. But $Z_i(G)N$. But $Z_i(G)N/N\leq Z_i(G/N)$ by the induction hypothesis, so $[xN,yN]\in Z_i(G/N)$.

As this holds for all $yN\in G/N$, it follows that $xN\in Z_{i+1}(G/N)$, as required.