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.


1 Answer 1


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? .......


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.