Show that these two definitions of a nilpotent group are equivalent. A group $G$ is nilpotent if there exists of a normal series $1 = G_0 \subseteq G_1 \subseteq G_2 \subseteq \dots \subseteq G_n = G$ such that $G_{i + 1}/G_i \subseteq Z(G/G_i)$.
But now I'm seeing another definition in this lecture video by Richard E. Borcherds. A group $G$ is nilpotent if you can "kill" it by repeatedly "killing" the center. That is, $G_0 = G, G_1 = G_0/Z(G_0), G_2 = G_1/Z(G_1), \dots, G_n = 1$.
Is it clear that these two definitions are equivalent?
Does it help to look at the quotient maps $(G_0 = G) \to (G_1 = G_0/Z(G_0)) \to (G_2 = G_1/Z(G_1)) \to \dots \to (G_n = 1)$ to form a normal series by taking kernels $K_i$ of the maps $G \to G_i$? Maybe $K_{i + 1}/K_i = Z(G/K_i)$?
 A: These are essentially the same definitions. The series described by Borcherds is called the "upper central series": it can also be defined as the series
$$\{1\} = Z_0(G)\leq Z_1(G)\leq\cdots\leq Z_n(G)$$
where $Z_k(G)$ is called the $k$-th center of $G$, and is the subgroup of $G$ such that $Z_k(G)/Z_{k-1}(G) = Z(G/Z_{k-1}(G))$. A group $G$ is nilpotent if the upper central series has $Z_k(G)=G$ for some $k$. Note that because $Z(G/Z_{k-1}(G))$ is normal in $G/Z_{k-1}(G)$, it follows that $Z_k(G)$ is normal in $G$, so this is a normal series.
The normal series you describe is said to be a "central series" for $G$:
$$1 = N_0\leq N_1\leq\cdots \leq N_k$$
with $N_i\triangleleft G$ and such that $N_i/N_{i-1}\subseteq Z(G/N_{i-1})$. A group is nilpotent if there is a central series that terminates at $G$, i.e., $N_m=G$ for some $m$.
To verify the two definitions are the same, note that the upper central series is a central series that terminates at $G$: it's just that instead of having $N_i/N_{i-1}\subseteq Z(G/N_{i-1})$, you have equality. So a group that satisfies the upper central series definition also satisfies the definition you knew.
To show that any group satisfying the definition you knew also satisfies the one given by Borcherds, assume that $G$ has a central series
$$ 1 = N_0 \leq N_1\leq\cdots \leq N_m=G$$
with $N_i/N_{i-1}\subseteq Z(G/N_{i-1})$. I claim that $N_k\leq Z_k(G)$ for all $k$. We proceed by induction on $k$.
By construction, $N_1 = (N_1/N_0)\leq Z(G/N_0) = Z(G)= Z_1(G)$, so the result holds for $k=1$.
Assume that $N_k\leq Z_k(G)$; then $G/N_k$ maps onto $G/Z_k(G)$, and therefore the image of $Z(G/N_k)$ maps into $Z_{k+1}(G)/Z_k(G)$ (since the image of the center is contained in the center of the image). Thus, every element of $N_{k+1}$ maps into $Z_{k+1}(G)$, yielding the desired result. Thus, a group having a central series must satisfy the definition given by Borcherds.

Another definition of nilpotent uses the "lower central series", where $G_0=G$, $G_{n+1}=[G_n,G]$, and the group is nilpotent if and only if the lower central series terminates in the trivial group: there exists $k$ such that $G_k=\{e\}$. The reason for the names is that any central series lies, in a sense, between the lower central series and the upper central series. Proving the upper and lower central series definitions are equivalent is a little trickier than the one you are asking about.
cf. 5.1.9 in Derek Robinson's A Course in the Theory of Groups 2nd edition:

Let $1=N_0\leq N_1\leq\cdots\leq N_n = G$ be a central series in a (necessarily nilpotent) group. Then:

*

*$G_i\leq N_{n-i+1}$ for each $i$, so $G_{n+1}=1$.

*$N_i\leq Z_i(G)$, so that $Z_n(G)=G$.

*The upper and lower central series of $G$ have the same length.


