Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $F$ is free group and $1 \neq x \in F$, then $C_F(x)$ is cyclic.

help me please!

share|cite|improve this question

The subgroup $\langle x,C_F(x)\rangle$ is a subgroup of a free group, hence is free. The only free groups with nontrivial center are the free groups of ranks $0$ or $1$, and $x\neq 1$ is clearly central in $\langle x, C_F(x)\rangle$. Thus, $\langle x, C_F(x)\rangle$ is cyclic, hence its subgroup $C_f(x)$ is cyclic.

share|cite|improve this answer

Here is a sketch of the proof:

If $F$ is a free group, then there exists a $S \subset F$ such that every element of $F$ can be written uniquely as a product of elements of $S$. Suppose $x \neq 1$. Then $x = (s_1...s_n)^k$ for some $n \geq 1$, $s_i \in S$, and $k$ chosen to be largest. Show that $C_F(x)$ is the cyclic group generated by $s_1...s_n$. Use the fact that everything in $F$ can be written uniquely as product of elements of $S$.

share|cite|improve this answer
Suppose $F=\mathbb{Z}$. Then wouldn't $S$ be the set of nonnegative prime integers? So consider $12=2^2\times 3$. Then $12$ cannot be written as $(2\times 3)^2$. Am I missing something here? – math-visitor Jun 30 '12 at 9:31
@math-visitor First of all $\mathbb{Z}$ is not even a group under multiplication. You are probably think of $\mathbb{Z}$ with addition which is a free group generated by $1$. – William Jun 30 '12 at 9:38
I think you have to pick $k$ maximal here (i.e. ensure that $s_1\dots s_n$ is not itself a power of anything). – Ben Millwood Jun 30 '12 at 9:48
Oh, I see. Thanks William! This example is from under free groups. I was wondering how would one know whether one is talking about free groups under multiplication or free groups under addition, and I'm guessing if the group's operation is not explicitly mentioned, one is assuming a group under multiplication... – math-visitor Jun 30 '12 at 9:48
You can also do it by using the result that a subgroup of a free group is free, but that might be using a sledgehammer to crack a nut! – Derek Holt Jun 30 '12 at 11:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.