Given $H \leq G$, prove a group $C(x) \leq G$ and $N_G(H) \leq G$

Question

Let $G$ be a group and $H \leq G$
a) Prove for any $a \in G$ , $aHa^{-1} \leq G$
b) Let $x\in G$ and $C(x) = \{ a \in G : ax = xa \}$. Prove $C(x ) \leq G$.
c)Let $N_G (H) = \{a \in G : aH = Ha\}$. Show that $N_G (H) \leq G$ and $H \leq N_G(H)$.

Note: my professor uses $H \leq G$ to denote H is a subgroup of G. As I am still very new to the subject, forgive this quiet naive and obvious question. I have read about the idea of normal subgroup in textbook, but we have not formally discussed it in class, so bare with if I miss any obviously available lemma.
I proved all three with the fact that H is a subgroup of G iff H nonempty, and $\forall x, y\in H, xy^{-1} \in H$.
But it feels like a rather redundant way of proving it, as my guess is there is some way of using a) for b), and a) b) for c), but I cannot formulate how exactly.
(Additional small question: I realize that $N_G(H)$ is a normal subgroup; is $C(x)$ any commonly seen group with a name?)
I am giving some of my attempts below but I do understand it is quiet messy and naive.

Attempt for c):
$\forall a \in N_G (H)$, we also have $a \in G$, i.e. $C(x) = \{a \in G: ax = xa \}$ for some $x \in G$. We have proved $C(x) \leq G$.
As a result of a), we also have $a C(x) a^{-1} \leq G$.
Let $y \in C(x)$, then $y = xyx^{-1}$. we now have $axyx^{-1}a^{-1} \in G$. And for which I am stuck now...

Any suggestion and help is appreciated.

Answer 1

Here’s a simple proof, and for simplicity I’ll show the centraliser and normalisers are subgroups of your group $G$.

Theorem: $C_G(A),N_G(A) \leq G$ are subgroups.

Proof:

Let $G$ be a group. We show $C_G(A) \le G$ where $A \subset G$. Let $x,y \in C_G(A)$, then for all $a \in A$ we have $$x a = a x$$ and $$y a = a y$$ Then \begin{align*} x y a &= x a y\\ &=a x y \end{align*} Forcing $x y \in C_G(A)$ and we have closure. For closure of inverses, let $x \in C_G(A)$. Then $$x a = a x.$$ Hitting both sides by $x^{-1}$ we obtain $$x^{-1} x a x^{-1} = x^{-1} a x x^{-1}.$$ This forces $$a x^{-1} = x^{-1} a,$$ forcing $x^{-1} \in C_G(A)$. Thus $C_G(A) \leq G$ is a subgroup.

Let $G$ be a group. We show $N_G(A) \le G$. Let $x,y \in N_G(A)$. We show $xy \in N_G(A)$ and $x^{-1} \in N_G(A)$. For closure, if $x,y \in N_G(A)$, then $$x A=A x$$ and $$y A=A y.$$ Thus we can compute \begin{align*} (x y) A &= x (y A) \\ &=x (A y)\\ &= (x A) y \\ &=(A x) y\\ &=A (x y) \end{align*} forcing $x y \in N_G(A)$ and we have closure. For closure of inverse, let $x \in N_G(A)$, then $$x A=A x,$$ thus we can compute by hitting the left sides by $x^{-1}$: \begin{align*} x^{-1} A x&=x^{-1} x A\\ &=A \end{align*} thus it follows that $x^{-1} A=A x^{-1}$ forcing $x^{-1} \in N_G(A)$ thus $N_G(A) \leq G$ as was desired. $\blacksquare$

Added: Moreover, we have that $C_G(A) \trianglelefteq N_G(A)$ is a normal subgroup. Could you prove that fact?

Also, please upvote if you deem the proof valid for future student references.

Answer 2

Firstly, $e\in aHa^{-1}, C(a), N(H)$, so that they are all nonempty subsets of $G$.

a) Let $x,y\in aHa^{-1}$; then, there are $h,h'\in H$ such that $x=aha^{-1}$ and $y=ah'a^{-1}$; therefore, $xy^{-1}=$ $aha^{-1}(ah'a^{-1})^{-1}=$ $aha^{-1}ah'^{-1}a^{-1}=$ $ahh'^{-1}a^{-1}\in aHa^{-1}$.

b) Let $x,y\in C(a)$; then, $xa=ax$ and $ya=ay$; from this latter, by taking the inverses, $a^{-1}y^{-1}=y^{-1}a^{-1}$, whence $y^{-1}a=ay^{-1}$; therefore, $xy^{-1}a=xay^{-1}=axy^{-1}$, and hence $xy^{-1}\in C(a)$.

c) Let $x,y\in N(H)$; then $xH=Hx$ and $yH=Hy$; from this latter, $yh=h'y$ for some $h,h'\in H$, whence, by taking the inverses, $h^{-1}y^{-1}=y^{-1}h'^{-1}$ and finally $y^{-1}H=Hy^{-1}$; therefore, $xy^{-1}H=$ $xHy^{-1}=$ $Hxy^{-1}$, and hence $xy^{-1}\in N(H)$. Note that $h\in H\Longrightarrow hH=H=Hh$, whence $h\in N(H)$, and hence $H\le N(H)$.