# Showing a Lower Central Series is a Central Series

I have the following definition for a central series of a group:

A group $G$ has a central series if there exists a normal series,$$G=A_0\triangleright A_1 \triangleright A_2 \triangleright\dots \triangleright A_m=\{e\},$$ with each $A_i$ normal in $G$, such that $A_{i-1}/A_i$ is in the center of $G/A_i$.

A lower central series for $G$ is defined as:

the sequence of normal subgroups of $G$ defined by $G^0 = G$, $G^i=(G^{i-1},G)$, so that $$G^0\triangleright G^1 \triangleright \dots \triangleright G^i \dots .$$

Here if $H$ and $K$ are two normal subgroups of $G$ we define $(H,K)$ to be the subgroup $\langle hkh^{-1}k^{-1} | h\in H, k\in K \rangle$. Note that $(H,K)\triangleleft G$.

I would like to show that if $G$ has a lower central series such that for some $k$, $G_k=\{e\}$, then the lower central series is in fact a central series.

We do not need to check that the $G^i$ are all normal in G. What I am having a difficult time showing is that $G^{i-1}/G^i$ is contained in the center of $G/G^i$. So far all I have been able to show is that because $(G^{i-1},G^{i-1})\leq G^{i}$ we have that $G^{i-1}/G^i$ is abelian.

I don't know if this is the best approach to showing what I want. Any help would be greatly appreciated.

Consider two consecutive terms in lower central series: $$G^i \geq G^{i+1}=(G^i,G).$$ We show the factor $G^i/G^{i+1}$ is central; but where? Since we are looking this factor as some subgroup "modulo $G^{i+1}$, we should see "in $G/G^{i+1}$" whether above factor is central. Very simple:
Consider any $gG^{i+1}$ in $G/G^{i+1}$, and any $xG^{i+1}$ in $G^i/G^{i+1}$. What is means here? It is simply that $g\in G$ is any element and $x\in G^i$ is any element. Then we know that $[g,x]$ is an element of $G^{i+1}$. Then write this in slightly different way:
$$[g,x]\in G^{i+1} \Leftrightarrow (gxg^{-1}x^{-1}) G^{i+1}=G^{i+1} \Leftrightarrow (gG^{i+1})(xG^{i+1})=(xG^{i+1})(gG^{i+1})$$ Thus, every $xG^{i+1}$ commutes with every $gG^{i+1}$. Then? .......