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Let $G$ be a group and $[G:Z(G)]=n<\infty,$ where $Z(G)$ is the center of $G$. A theorem says that in this case

$$o([G,G])\leq(n^2)^{n^3}, $$

where $o(\cdot)$ denotes order and $[G,G]=G'$ is the commutator subgroup of $G$.

The bounds from the proof seem crude and I would like to ask if we can improve upon the result. Of course, the theorem is valuable with any bound because it implies that $G'$ is finite.

Proof. We have exactly $n$ cosets $\{x_iZ(G)\}_{i=1}^n$. We will be using the representatives $x_i.$ Let $$c_{ij}=[x_i,x_j]=x_i^{-1}x_j^{-1}x_ix_j.$$

First, we note the following

Lemma 1.

$$\{c_{ij}\,|\,i,j=1,...,n\}=\{[x,y]\,|\,x,y\in G\}.$$

Proof of lemma. For $x,y\in G$ we have some $i,j\in\{1,...,n\}$ and $z_1,z_2\in Z(G)$ such that

$$x=x_iz_1\text{ and }y=x_jz_2.$$


$$ \begin{eqnarray} [x,y]&=&x^{-1}y^{-1}xy=z_1^{-1}x_i^{-1}z_2^{-1}x_j^{-1}x_iz_1x_jz_2\\ &=& (x_i^{-1}x_j^{-1}x_ix_j)(z_1^{-1}z_1z_2^{-1}z_2)\\ &=& x_i^{-1}x_j^{-1}x_ix_j\\ &=& [x_i,x_j]. \end{eqnarray} $$

Lemma 2. For $x,y\in G$ we have


Proof of lemma. Since $G/Z(G)$ has order $n$, we have $[x,y]^n\in Z(G)$. Therefore

$$ \begin{eqnarray} [x,y]^{n+1}&=&x^{-1}y^{-1}xy[x,y]^n\\ &=& x^{-1}y^{-1}x[x,y]^ny\\ &=& x^{-1}y^{-1}x(x^{-1}y^{-1}xy)[x,y]^{n-1}y\\ &=& (x^{-1}y^{-2}xy^2)y^{-1}[x,y]^{n-1}y\\ &=& [x,y^2][y^{-1}xy,y^{-1}yy]^{n-1}\\ &=& [x,y^2][y^{-1}xy,y]^{n-1} \end{eqnarray} $$

Proof cont'd. Let $g\in G'.$ It is a product of a a finite number $m$ of elements $c_{ij}$ (with possible repetitions) because $[x,y]^{-1}=[y,x]$ and so we do not need to consider inverses. Suppose $m>n^3$. We have

$$\operatorname{card}(\{c_{ij}\,|\,i,j=1,...,n\})\leq n^2$$

so some element $c_{ij},$ say $c=[x,y],$ must appear at least $n+1$ times in the product. There is no reason to believe however that $c^{n+1}$ occurs in the product since the occurances of $c$ may be scattered. We fix it by noting that for any $x',y'\in G$ we have


This allows us to rewrite $g$ as a product of commutators which begins with $[x,y]^{n+1}:$


But this equals


which is a product of $m-1$ commutators. Repeating this procedure, we can decrease the number of factors to $n^3.$ Therefore we can write any element of $G'$ as a product of at most $n^3$ commutators. We recall that there can be at most $n^2$ distinct commutators in $G$ and therefore there are at most $(n^2)^{n^3}$ elements in $G'.$

EDIT I have laid my hands on a copy of Passman's book on group rings and I see that the theorem and the proof come from it. The author attributes the theorem to Schur.

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what is your source? – Babak Miraftab Feb 12 '12 at 19:54
@babgen Lecture notes from a course in group rings. – user23211 Feb 12 '12 at 20:30
It looks like you could improve the bound to $(n+1)^{n^2}$ by writing elements in the form $c_{11}^{p_{11}}\cdots c_{nn}^{p_{nn}}$ with $0\leq p_{ij}\leq n$. – Colin McQuillan Feb 12 '12 at 20:31
@ColinMcQuillan Do you mean distinct $c_{ij}?$ If so, how can I obtain that? A simple procedure based on (1) doesn't seem to work... – user23211 Feb 12 '12 at 21:37
ymar: $G/Z(G)$ is generated by $\log(n)$ elements, and $[G,G]$ has exponent $n$, so if it is abelian, it is of size at most $\exp(\log(n)^3)$. When $[G,G]$ is not abelian, it seems to increase $G/Z(G)$ faster than it does $o([G,G])$ (if $G$ is nilpotent of class $c\geq 2$, then I think one get something like $\exp(\log(n)^{2+1/(c-1)})$). I didn't check the non-nilpotent case carefully though, hence my request for an example. – Jack Schmidt Feb 13 '12 at 6:53
up vote 3 down vote accepted

Here's an argument for $(n+1)^{n^2}$. I'll try and be a bit formal, at the risk of making it sound mote complicated than it is. Let $c_1,\cdots,c_k$ be an enumeration of the set $\{c_{11},\cdots,c_{nn}\}$. We will write an arbitrary element $g$ of the commutator subgroup in the form $c_1^{p_1}\cdots c_k^{p_k}$ with $0\leq p_1,\cdots,p_k \leq n$.

There exist $p_1,\cdots,p_k\geq 0$ and a sequence $i_1,\cdots,i_m$ such that $g=c_1^{p_1}\cdots c_k^{p_k}c_{i_1}\cdots c_{i_m}$ (for example with $p_1=\cdots=p_k=0$). So we can pick such a representation of $g$ such that $p_1+\cdots+p_k+m$ is minimal, and then fixing that, such that $p_1$ is maximal, and then such that $p_2$ is maximal and so on. Suppose for contradiction that $m>0$. Then we can move the $c_{i_1}$ term back using rule 1 to get a representation of $g$ such that $p_{i_1}$ is larger, which is a contradiction. Suppose for contradiction that $p_i>n$. Then we can use rule 2 and get a contradiction to the minimality of $p_1+\cdots+p_k+m$.

Another small improvement is $(2n+1)^{n(n-1)/2}$ by letting $-n\leq p_i\leq n$ and using $[x,y]=[y,x]^{-1}$.

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You can do slightly better based on the fact that $[G,G]$ has exponent $n$. This follows from the transfer map $f:\ G\rightarrow Z(G)$, given by $f(g)=g^n$. – user641 Feb 12 '12 at 23:39
Thank you for your answer. Now I can express my doubt more clearly than I did in the comment. Why can we move $c_{i_1}$ back? My problem is that using rule (1) to move it back, we change the terms we move it through. Rule (1) only says they remain commutators, but do we have control over whether, say, the last term we jump over doesn't change to, say, $c_1$? Wouldn't that ruin the whole argument? – user23211 Feb 12 '12 at 23:40
@SteveD May I ask what happened to the comment you posted before? I've been struggling to prove that and I haven't managed. – user23211 Feb 12 '12 at 23:43
@ymar: the terms we jump over can become any junk of the same length - we still get a contradiction to the minimality of $p_{i_1}$. It's quite a careful choice of minimal representation. Another way to describe it is to ask for minimum length and then lexicographically minimal. – Colin McQuillan Feb 12 '12 at 23:46
@ColinMcQuillan This is embarrassing but I still don't understand. I think we want to obtain a representation of $g$ of the form $c_1^{p_1}...c_k^{p_k}$ for fixed $c_i.$ I.e., we must know that $c_1$ is the base of the first factor, $c_2$ of the second, etc. They must be fixed or the bound won't be $(n+1)^{n^2}.$ I think we are considering our min/max conditions in the set of strings beginning with $c_1^{p_1}...c_k^{p_k},$ again with $p_i$ variable but $c_i$ fixed. So if we allow our terms to change to anything, we will wander outside this set, where the conditions don't apply. – user23211 Feb 13 '12 at 0:25

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