Say you have a set G with a subgroup H,
s.t.
$G=HC_G(H)$
($C_G(h)$ being the centraliser of the group).
My first question :
What is the structure of this group ?
My attempt:
$C_G(H)=\{g\in G |hg=gh, \forall h \in H\}$
i.e. $hgh^{-1}=g,\forall h \in H$
So $G=HC_G(H)=\{hg| h\in H, g \in C_G(H)\}$
My second question:
This would mean that H is normal in G correct ?
My attempt:
To be normal implies that $xHx^{-1}=H,\forall x \in G$
Well we already know what elements of G look like $hg \in G$. These g's being exactly those which commute with all the h's in H.
So then
$(hg)H(hg)^{-1}=\{hghh^{-1}g^{-1}|h \in H, g \in G\}$
But $hghh^{-1}g^{-1}=hgg^{-1}=h$
So $(hg)H(hg)^{-1}=H,\forall g\in G$
And so H is normal in G
It seems like you have a problem with the second part. Let $x\in G$. We want to show that $xHx^{-1} = H$, which is equivalent to $xHx^{-1}\subset H$. It is correct that for some $g\in C_g(H)$ and $a\in H$, $x=ag$. As such, $xHx^{-1}= (ag)H(ag)^{-1}=\{ag hg^{-1}a^{-1}\mid h\in H\}$. However, $h$ doesn't necessarily equal $a$, which is implicitly assumed in your first step. However, you do know that $a$ and $a^{-1}$ commute with $g$, $g^{-1}$, and $h$ as $a,a^{-1},h\in H$ and $a,a^{-1}\in C_G(H)$. As such, $xHx^{-1}= (ga)H(ga)^{-1}=\{ga ha^{-1}g^{-1}\mid h\in H\} = \{gaha^{-1}gg^{-1}\mid h\in H\} = \{aha^{-1}\mid h\in H\}$. However, $a,h,a^{-1}\in H$, so $aha^{-1}\in H$ and so $xHx^{-1}\subset H$.