Centralizer, Normalizer and Conjugate I am looking at Group Theory notes on Centralizer and Normalizer for next semester and come up with this question:
Let $H$ be a subgroup of $G$, and let $g$ be an element in $G$. Show that
$$(a)\ C_G(H^g) = C_G(H)^g$$
$$(b)\  N_G(H^g) = N_G(H)^g.$$
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For (a), I sure the LHS is as follow but correct me if I am wrong:
$C_G(H^g) = \{x \in G : xg^{-1}hg = g^{-1}hgx = xghg^{-1} = ghg^{-1}x, \forall h^g \in H^g$ and where $ g, g^{-1} \in G\} $
But I am not sure for the RHS of (a). Does it literally mean $[C_G(H)]^g$?
For the LHS of (b), correct me if I am wrong here:
$$N_G(H^g) = \{x \in G : xH^g = H^gx\}$$
$$ = \{x \in G : xg^{-1}Hg = g^{-1}Hgx = xgHg^{-1} = gHg^{-1}x\}$$
Again for the RHS, does it mean literally $[N_G(H)]^g$?
Thank you very much for your time and help. Happy holidays to you all!
 A: Yes the right hand side of both equations literally means what it says. 
Here is a hint for both problems: write down precisely what it means that some element $x$ is a member of the left hand side. Then write down precisely what it means that some element $x$ is a member of the right hand side. Then manipulate both expressions so make them look more like each other. If you can manipulate them into the same expression then you're done and you can write down the proof nicely.
Here is something to get you started:
$$ \begin{aligned} x \in C_G(H^g) &\iff \dots \\ &\iff \dots \\ &\iff x^{g^{-1}}\in C_G(H) \\ &\iff x \in C_G(H)^g \end{aligned}$$
Spoiler: solution for (a)

 $$ \begin{aligned} x \in C_G(H^g) &\iff (\forall h\in H)( (h^g)^x = h^g) \\ &\iff (\forall h\in H)(h^{gxg^{-1}}=h) \\ &\iff x^{g^{-1}}\in C_G(H) \\ &\iff x \in C_G(H)^g \end{aligned}$$

Spoiler: solution for (b)

 $$ \begin{aligned} x \in N_G(H^g) &\iff  (H^g)^x = H^g \\ &\iff H^{gxg^{-1}}=H \\ &\iff x^{g^{-1}}\in N_G(H) \\ &\iff x \in N_G(H)^g \end{aligned}$$

