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We have $F$ a free group.
If we have a chain $F = C_0 \gt C_1 \gt \ldots$ such that each $C_{i+1}$ is a proper characteristic subgroup of $C_i$. We want to prove $\cap_i C_i = 1$.

Now, I have a feeling that we must use the fact that a free group is residually finite. I.e: $\cap_{H\lt_{f}F}H=1$. With $\lt_f$ meaning that $H$ is of finite index.
My main issue is that we do not know if the $C_i$'s are of finite index.

Another idea was to take $w \in K = \cap_i C_i $ and prove that $w=1$ but I do not see how to make this happen.

Robert Shore
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Cactus
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1 Answers1

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Alright, so, using Corollary 9 in the chapter on Nielsen's method in D.L Johnson's book, as Derek Holt suggested, I got a pretty clear idea on how to solve that problem. My only issue is that I am not supposed to know about the concepts such as the Nielsen transformation and being N-reduced. However, I believe I can form a pretty concrete proof, please let me know if there are any mistakes.

Take $B_i$, a free basis of $C_i$. Suppose that we have $w \in X_i$ such that $w \in C_{i+1}$, as the homomorphisms switching the elements of $X_i$ with eachother are automorphisms, $X_i \subset C_{i+1}$, since $C_{i+1}$ is characteristic. However, then $C_i \leq C_{i+1} $, which is impossible as they are proper subgroups.
Now take $w_0 \in X_i$ such that $l(w_0)\leq l(w), w\in X_i$, with $l$ the length using the basis of F.
We can assume $l(w_0) \leq l(w), w\in C_i$. Or we transform the basis to have $w_0$ in it which gives an automorphism.
Now, we have that $l_i :=l(w_0)=min\{l(w), w \in C_i\}$ and for all $w \in C_i$ such that $l(w)=l_i$, we can once again create a new basis in which $w$ belong, and we can create an automorphism between the basis so $w \notin C_{i+1}$.
Now, we have proven that the shortest $w \in C_{i+1}$ has length $l(w)>l_i$.
Using this, we have $1 = l_1 < l_2 < l_3 < ...$. So if we take $v \in F, l(v)=k$, then $v \notin C_{k+1}$.
So $\cap C_i = 1$

Cactus
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  • Presumably $B_i=X_i$? The claim that $l(w_0) \le l(w)$ for all $w \in C_i$ is precisely Corollary 8 in Johnson's book, and is the non-trivial part of the proof. So that still needs justifying. Your sentence "Or we transform the basis to have $w_0$ in it which gives an automorphism" doesn't make much sense because $w_ 0$ is already in the basis. – Derek Holt Dec 11 '25 at 09:16
  • @DerekHolt then I am most likely expected to use another method as we are not familiar with the Johnson's methods in my course. – Cactus Dec 12 '25 at 07:43