# If $G$ is a nilpotent torsion group of nilpotence class $c$, then $\exp(G)$ divides $\exp(G^{\operatorname{ab}})^c$

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.

• 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 May 23, 2021 at 21:38
• @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 May 23, 2021 at 22:02

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 $$N^{(c-1)+1}$$.

• Where do you get the identity $[r,s]^N \equiv [r^N, s] \pmod{G_3}$ from? May 23, 2021 at 22:26
• @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). May 23, 2021 at 22:29