# Let $G$ be any group and $x \in G$ , then $N_G(\langle x \rangle)/C_G(\langle x \rangle)$ is finite ?

Let $G$ be any group and $x \in G$ , then is it true that $N_G(\langle x \rangle)/C_G(\langle x \rangle)$ is finite ? I know that

$N_G(\langle x \rangle)/C_G(\langle x \rangle)$ is isomorphic to some subgroup of $Aut (\langle x \rangle)$ but cannot proceed further . Please help .

• Do you know how to describe the automorphism group of a cyclic group? – Tobias Kildetoft Dec 12 '14 at 8:49
• @TobiasKildetoft: I see , are you hinting at that the order of the Automorphism group of any infinite cyclic group is $2$ ? ( if $\langle x \rangle$ is finite then we are obviously safe ) – user123733 Dec 12 '14 at 10:53
• That is indeed what I am hinting at. – Tobias Kildetoft Dec 12 '14 at 10:54
• @TobiasKildetoft: Thanks very much – user123733 Dec 12 '14 at 10:56

As already noted in the question, for any subgroup $H\leq G$ we have that $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\mathrm{Aut}(H)$.

To see that this means that $N_G(\langle x\rangle)/C_G(\langle x\rangle)$ is finite, we just need to show that $\mathrm{Aut}(\langle x\rangle)$ is finite for any $x$.

If $|x|$ is finite, this is obvious, so we are left with the case where $\langle x\rangle \cong \mathbb{Z}$.

But an automorphism of a cyclic group is uniquely determined by its value on any generator. And since this value must again be a generator, we see that there are only two automorphisms of $\mathbb{Z}$ (one sending $1$ to $1$ and one sending $1$ to $-1$). Hence $\mathrm{Aut}(\mathbb{Z})$ is finite which finishes the proof.

This is false. Let $G = \operatorname{GL}_2\left(\mathbb{Q}\right)$ and $x = \left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right)$. Then, $x^i$ is conjugate to $x$ for every nonzero integer $i$. Hence, the set of $G$-conjugates of $x$ which belong to $\left<x\right>$ is infinite. But this set has the same cardinality as $N_G\left(\left<x\right>\right) / C_G\left(\left<x\right>\right)$, because there is a bijection from the latter set to the former given by sending the residue class of $f \in N_G\left(\left<x\right>\right)$ to $fxf^{-1}$.

• There must be some detail wrong here. The quotient is isomorphic to a subgroup of the automorphism group of $\langle x\rangle$ which is certainly finite for any $x$. – Tobias Kildetoft Dec 12 '14 at 8:49
• Oh! I forgot that $N_G\left(H\right)$ is the set of all $g \in G$ such that $gHg^{-1} = H$, not such that $gHg^{-1} \subseteq H$. Good to finally see an example where it matters. Can you make your comment into an answer so that mine doesn't look like it's true? – darij grinberg Dec 13 '14 at 15:14
• Done (that was actually a fairly subtle mistake. No wonder I did not see where it was at first). – Tobias Kildetoft Dec 15 '14 at 10:31