I am trying to prove the following statement:

If $G$ is a torsion group and nilpotent of class $c$, then $\exp(G)$ divides $\exp(G^{\operatorname{ab}})^c$ (where $\exp$ is the exponent, i.e. the lcm of the order of the group elements, and $G^{\operatorname{ab}}$ is the abelianization of $G$).

I thought about using induction on $c$. The base case is easy: for $c=1$, $G = G^{\operatorname{ab}}$, so the statement follows. I have no idea how to do the induction step though.

  • 3
    $\begingroup$ I think this is in robinson's textbook on group theory. induction down the lower central series using a tensor product if i remember right $\endgroup$ May 23, 2021 at 21:38
  • $\begingroup$ @JackSchmidt I can find a similar result in Robinson's textbook (p. 138): if $\exp(Z(G)) = e$, then $\exp(G) \mid e^c$. Is that the one you mean? It looks like it shouldn't be that hard to go from one of them to the other, but I'm kind of lost here $\endgroup$
    – tolUene
    May 23, 2021 at 22:02

1 Answer 1


Consider the lower central series of $G$, $G_1=G$, $G_2=[G,G]$, $G_{n+1} = [G_n,G]$. Since $G$ is of class $c$, $G_{c+1}=\{e\}$. Let $N=\mathrm{exp}(G^{\rm ab})$.

We have that $G/G_2\cong G^{\rm ab}$ has exponent $N$. Thus, for every $x\in G$, $x^N\in G_2$.

Let us look next at $G/G_3$. If $x\in G$, then $x^N\in G_2$. Now, $G_2$ is generated by elements of the form $[r,s]$ with $r\in G_1$, $s\in G$; and in $G_2/G_3$ we have $$[r,s]^N \equiv [r^N,s]\pmod{G_3}$$ and $[r^N,s]\in [G_2,G]=G_3$ so $[r,s]^N\in G_3$. And since $G_2/G_3$ is abelian, generated by elements of exponent $N$, it is itself of exponent $N$.

Thus, if $x\in G$, then $x^N\in G_2$, and $x^{N^2}=(x^N)^N\in G_3$.

Now look at $G_3/G_4$. This is an abelian group; since $G_3$ is generated by elements of the form $[g,s]$ with $g\in G_2$, then $G_3/G_4$ is generated by their images modulo $G_4$. But we have that $$[g,s]^N \equiv [g^N,s]\pmod{G_4}$$ and $g^N\in G_3$, since $g\in G_2$. Therefore $[g,s]^N\in [G_3,G]=G_4$. Thus, for every $x\in G$, $x^{N^3}\in G_4$.

Continuing this way, we obtain that for every $x\in G$, $x^{N^c}\in G_{c+1}=\{e\}$, as desired.

You can probably frame this as a “descending induction”, by starting with elements in $G_c$ and proving they are of exponent $N$; then showing that if the elements of $G_{c-k}$ are of exponent $N^{k+1}$, with $k\lt c-1$, then the elements of $G_{c-(k+1)}$ are of exponent $N^{k+2}$; and thus concluding that the elements of $G= G_1 = G_{c-(c-1)}$ is of exponent $N^{(c-1)+1}$.

  • $\begingroup$ Where do you get the identity $[r,s]^N \equiv [r^N, s] \pmod{G_3}$ from? $\endgroup$
    – tolUene
    May 23, 2021 at 22:26
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    $\begingroup$ @tolUene: Basic property of nilpotent groups. You can deduce it from the product formulas for commutators: $[xy,z] = [x,z]^y[y,z] = [x,z][[x,z],y][y,z]$, so modulo the next term of the lower central series, you just get $[xy,z]=[x,z][y,z]$. (This is where the tensor that Jack mentioned comes in, because there is a multilinear map $G_{r}/G_{r+1}\times G/G_{r+1}\to G_{r+1}/G_{r+2}$ given by the commutator bracker). $\endgroup$ May 23, 2021 at 22:29

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