I am looking for a (preferably modern) reference to the following old result of Magnus.

Let $F$ be a free group of finite rank and $$ F_1 = [F,F], F_2 = [F_1,F_1], \dots , F_{n+1} = [F_n,F_n], \dots $$ Let $T_2$ be the group of all affine transformations $x \mapsto ax+b$ of the complex line. Here, both $a$ and $b$ are complex numbers and $a \ne 0$. Of course, one may view $T_2$ as the group of invertible upper-triangular $2\times 2$ complex matrices with $1$ at the lower right corner.

Theorem of Magnus (Crelle, 170 (1934)?): There exists an embedding of the quotient $F/F_2$ into $T_2$.

I also wonder if there are any generalizations of this result which deal with embeddings of quotients $F/F_n$ into groups of complex (upper-triangular) matrices.

  • 2
    $\begingroup$ Magnus' theorem is not only that there exists an embedding, but that some given explicit homomorphism is an embedding (which is stronger and more useful). $\endgroup$
    – YCor
    May 14, 2015 at 11:50
  • $\begingroup$ @YCor I would appreciate a reference where such a homomorphism is explicitly described. $\endgroup$ May 14, 2015 at 17:07

2 Answers 2


Assuming $F$ non-abelian, there is no embedding of $F/\mathcal{D}^nF$ for $n\ge 3$ into a group of matrices over any field. (Here I write $\mathcal{D}G=[G,G]$, so $F_n=\mathcal{D}^nG$ in your notation.) One reason is that any solvable group of matrices is nilpotent-by-abelian-by-finite.

But $F/\mathcal{D}^nF$ is not nipotent-by-abelian-by-finite when $n\ge 3$. This is well-known, but here's a proof. Clearly it suffices to do the case when $F$ is free of rank 2. Indeed assume, by contradiction, otherwise. Then it follows that for some $k_0$, $F/\mathcal{D}^nF$ has a subgroup of index $k_0$ that is nilpotent-by-abelian; hence $F/\mathcal{D}^3F$ (being a quotient of $F/\mathcal{D}^nF$) has the same property. In turn, it follows that every 2-generated $3$-solvable group has a subgroup of index $\le k_0$ that is nilpotent-by-abelian.

Now in order to get a contradiction, consider three primes $p,q,r$, all $>k_0$, such that $q$ divides $r-1$, and fix a nontrivial action of the cyclic group $C_q$ on $C_r$. Consider the wreath product $G=(C_r\rtimes C_q)\rtimes C_p$. Let's show it's 2-generated. It clearly 3-generated: let $s,t,u$ be generators of $C_p$, $C_q$, $C_r$. If $g\in C_r\rtimes C_q$ and $k$ is integer, write $g_k=s^kgs^{-k}$ (so $g=g_0$). Then $G=\{s,t_0,u_0\}$. Now observe that $G=\{s,t_0u_1\}$ (use that $t_0$ and $u_1$- commute and have coprime orders), so $G$ is 2-generated. Also, the derived subgroup of $G$ contains $C_r^p$ (since $C_r$ is in the derived subgroup of $C_r\rtimes C_q$), and contains the element $t_Ot_1^{-1}$, which is a commutator of $s$ and $t_0$. In particular, the derived subgroup $\mathcal{D}G$ of $G$ contains an element of order $r$, say $u_0$ and an element of order $q$, say $t_0t_1^{-1}$, that do not commute; hence $\mathcal{D}G$ is not nilpotent; in other words, $G$ is not nilpotent-by-abelian.

Now turning back to the contradictory hypothesis, $G$ should have a nilpotent-by-abelian subgroup $H$ of index $\le k_0$. Since $G$ is not nilpotent-by-abelian, we have $H\neq G$. On the other hand, because $p,q,r>k_0$, $G$ has no proper index subgroup of index $\le k_0$. We get a contradiction, and hence $F/\mathcal{D}^nF$ is not nilpotent-by-abelian-by-finite (and nor is its profinite completion, by the way) and in particular $F/\mathcal{D}^nF$ is not isomorphic to a group of matrices over any field (or product of fields).

  • $\begingroup$ Thanks! It seems that what you denote by $F_n$ is $F/F_n$ in my notation. $\endgroup$ May 14, 2015 at 16:01

I believe you are talking about the "Magnus Embedding", which has been studied by Fox (via the Free differential calculus - see his '53 paper). The most recent reference seems to be Kapovich-Drutu (Misha will probably give a better answer).


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.