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Let $G$ be an infinite profinite group.

Then $G$ is an inverse limit of an inverse limit system $\{G_i\}_{i=1}^{\infty}$ where $G_{i+1} \to G_i$ is surjective for all $i$.

I want to show that $G$ is conjugacy separable, i.e., given any two elements $a,b \in G$ that are not conjugate, there exists a finite quotient group where their images are also not conjugate.

To do this, it suffices to show that if the images of $a$ and $b$ are conjugate in $G_i$ for each $i$, they are conjugate in whole group $G$.

Let $a_i$ and $b_i$ be the image of $a$ and $b$ in $G_i$.

Suppose that there exists a $t_i \in G_i$ such that $t_ig_{1,i} \equiv g_{2,i}t_i$ in $G_i$.

Thus the equations $xa_i \equiv b_ix$ in $G_i$ are solvable for $x$.

Let $S_i$ be the set of solutions of each equation $xa_i \equiv b_ix$ in $G_i$ for each $i$.

Then $\{S_i\}^{\infty}_{i=1}$ forms an inverse system. Since each $S_i$ is nonempty and finite, there exists an inverse limit $S = \lim_{\leftarrow}S_i$.

For any element $t \in S$, $t$ is to be a solution of the above equations. Thus $G$ is conjugacy separable.

Is my proof correct?

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Yes, profinite groups are conjugacy separable for the reason you describe.

Another way of seeing it:

In a compact group, conjugacy classes are closed (because they are continuous images of the group), so if $x$ and $y$ are not conjugate, then $x$ cannot be approximated by elements of the conjugacy class $Y$ of $y$. In a profinite group, there is a base for the topology consisting of finite index normal subgroups, so saying that $x$ cannot be approximated by elements of $Y$ is the same as saying that there is a finite index open normal subgroup $H$ such that $xH$ is disjoint from $Y$. Now $YH/H$ is the conjugacy class of $yH$ in $G/H$, and we have $xH \not\in YH/H$, so $xH$ and $yH$ are not conjugate in $G/H$.

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