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Hello? I have some questions in the group theory. I know that the intersection of the lower central series of a finitely generate free group is trivial. So I wonder whether every nontrivial subgroup of the free group containsu a term or not. I've tried, but coudn't have shown or found a counter example. It may be false. Then, how about a finite index subgroup?

Please let me free from this discomfort.

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A finite-index subgroup of a finitely-generated free group is a finitely-generated free group. An infinite-index subgroup of a finitely-generated free group is an infinitely-generated free group. –  Ryan Budney Jun 11 '11 at 19:50
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Any normal subgroup where the quotient is perfect does not contain any term of the lower central series. –  Tyler Lawson Jun 11 '11 at 20:38
    
Ryan - your second sentence is only true if you insert the words 'normal' and 'non-trivial'. –  HJRW Jun 11 '11 at 21:38
    
Tyler - why don't you insert your comment as an answer? –  HJRW Jun 11 '11 at 21:40

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up vote 4 down vote accepted

The answer is "no" in both cases.

The terms of the lower central series of a group are verbal subgroups. If we let $\gamma_c(G)$ denote the $c$th term of the lower central series of $G$, then for any groups $G$ and $K$ and any group homomorphism $\varphi\colon G\to K$, we have $\varphi(\gamma_c(G))=\gamma_c(\varphi(G))\subseteq \gamma_c(K)$; and if $\varphi$ is onto, then $\varphi(\gamma_c(G))$ maps onto $\gamma_c(K)$, so that we have equality.

In particular, if $\pi\colon G\to G/N$ is a quotient map, then $\gamma_c(G/N)$ is trivial if and only if $\gamma_c(G)\subseteq N$. That is, the quotient $G/N$ is nilpotent of class at most $c-1$ if and only if the $c$th term of the lower central series of $G$ is contained in $N$.

So let $F$ be a free group, and let $N$ be a normal subgroup of $F$. Then $N$ contains $\gamma_c(F)$ if and only if $F/N$ is nilpotent of class at most $c-1$.

So, for example, if $F$ is the free group on two generators, then there is a normal subgroup of $F$ such that $F/N\cong S_3$; since $S_3$ is not nilpotent, $N$ does not contain any term of the lower central series of $F$, even though $N$ is of finite index.

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Thank you! Could I ask more? Then, how about only the case the subgroup H s.t. between H and F, there exist a subnormal series with finite abelian p-groups as factor groups. Here, F is a finitely generated free group. (Is the case equivalent to taking a subgroup of F of prime power index?) Also, I wonder how can I check the including question for not normal subgroups. –  qkqh Jun 13 '11 at 8:12
    
@qkqh: For nonnormal subgroups, look at the core (largest normal subgroup contained in $H$, the intersection of all conjugates of $H$), which is normal. Since the lower central series term are verbal, they are fully invariant, so $\gamma_c(F)$ is contained in $H$ if and only if it is contained in the core of $H$. Note that in my example, with $F/N\cong S_3$, there is a normal subgroup $N_1$, $N\triangleleft N_1\triangleleft F$ such that $F/N_1$ is cyclic of order $2$, and $N_1/N$ is cyclic of order $3$ (i.e., finite abelian $p$-groups). So the answer is still "no". –  Arturo Magidin Jun 13 '11 at 18:28
    
@qkqh: However, if all factors are $p$-groups for the same prime $p$, then $F/N$ is a $p$-group, hence nilpotent, so $N$ contains an appropriate term of the lower central series. (Again, you may as well restrict yourself to normal subgroups of $F$). –  Arturo Magidin Jun 13 '11 at 18:30
    
@Arturo Magidin: In your latest comment, even if all factors are $p$-groups for the same $p$, $N$ may not normal in F, so $F/N$ cannot be a $p$-group. By any chance, do you mean that the core of a subgroup of a prime power index is of index a power of the prime index (even in a f.g. free group)? –  qkqh Jun 13 '11 at 19:23
    
@qkqh: I was thinking of $N$ normal in $F$. In general, it is not true that the core of a subgroup of prime power index is of the same prime power: as a counterexample, go again to $S_3$; lift a nonnormal subgroup of index $3$ to get an $H\lt F_2$ of index $3$, whose core $N$ is of index $6$. So, if you are just trying to think about the nonnormal subgroups, the condition you give is not sufficient either. Just look for nonnilpotent finite groups that have appropriate subnormal chains to get counterexamples. –  Arturo Magidin Jun 13 '11 at 21:07

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