Is the centralizer of a group equal to the intersection of the centralizers of its generators? Let $G$ be a finite group, and $H\leq G$ such that $H=\langle x,y\rangle$. Is the following true:
$C_G (H) = C_G (x) \cap C_G (y)$
It seems to me the answer is yes. Given $c\in C_G (H)$ then $ch=hc$ for all $h\in H$, so certainly $cx=xc$ and $cy=yc$ so $c\in C_G (x) \cap C_G (y)$. Now given some $c\in C_G (x) \cap C_G (y)$, then $c$ must commute with any word $x^{k_1}y^{l_1}...x^{k_n}y^{l_n}$, that is, any element generated by $x$ and $y$ and therefore any element of $H$. Thus $c\in C_G (H)$. In almost the same way I show the following:
$C_G (H) = C_{C_G (x)}(y)$
My only concern is that where I'm reading a similar result, it specifies that (in their case) we can only do this because $x$ and $y$ commute. I can't quite see why this is necessary, since my proof above does not require that $x$ and $y$ commute. Is there a flaw in my argument?
Thank you in advance
 A: Yes, we always have $G_G(H) = C_G(x) \cap C_G(y)$ if $H = \langle x,y\rangle$, regardless of whether $G$ is finite or not. Your argument to show the equality is correct.
For $G$ infinite, one should also mention that the argument uses the - easy to verify - fact that $C_G(x^{-1}) = C_G(x)$ (and $C_G(y^{-1}) = C_G(y)$), since then a word in $\langle x,y\rangle$ may contain negative exponents. If $x$ and $y$ have finite order, as must be if $G$ is finite, then by reducing the exponent modulo the pertinent order one can assume all exponents non-negative.
The second equality,
$$C_G(H) = C_{C_G(x)}(y)\tag{$\ast$}$$
is where the question whether $x$ and $y$ commute may be relevant. It is clear what the notation $C_K(z)$ means if $K$ is a group with $z \in K$, but what if $K$ is a subgroup of $G$ and $z \in G \setminus K$? We can of course say that then
$$C_K(z) := \{ k \in K : k\cdot z = z \cdot k\},$$
i.e. $C_K(z) = C_G(z) \cap K$. In that case, $(\ast)$ is the very same as $C_G(H) = C_G(x) \cap C_G(y)$ and true. But it may be that the author only assigns meaning to $C_K(z)$ when $z \in K$, and then the expression $C_{C_G(x)}(y)$ only makes sense if $y \in C_G(x)$, i.e. if $x$ and $y$ commute.
I think that the latter is the case, the author only assigns meaning to $C_K(z)$ if $z\in K$, and then they can only write $C_G(H) = C_{C_G(x)}(y)$ when $x$ and $y$ commute.
