Which is the difference between the two notations?

Question

I want to show that if $H\leq G$ then $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\text{Aut}(H)$.

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We have the following:

$$N_G(H)=\{g\in G\mid gH=Hg\} \\ C_G(H)=\{g\in G\mid gh=hg, \forall h\in H\}$$

Could you explain to me the difference between these two notations?

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EDIT:

We have to show that the map $N_{G}(H) \to \mathrm{Aut}(H)$ is an homomorphism and the kernel is $C_G(H)$, right?

Do we maybe show that the map $g\mapsto (h\mapsto g^{-1}hg)$ is an homomorphism as follows?

$$h\mapsto (g_1g_2)^{-1}h(g_1g_2) \\ \Rightarrow h\mapsto (g_1g_2)^{-1}h(g_1)h(g_2) \\ \Rightarrow h \mapsto g_2^{-1}g_1^{-1}h(g_1)h(g_2)$$

But how could we continue?

Answer 1

Let me give you an example.

Let $G = S_{3}$, $H = A_{3} = \{ 1, (1 2 3), (1 3 2) \}$.

Then $N_{G}(H) = G > H = C_{G}(H)$.

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The point being that if you take $h = (1 2 3) \in H$, and $g = (1 2) \in G$, then $$ g h = (1 3) \ne (2 3) = h g, $$ while $$ g H = \{ (12), (13), (23) \} = \{ (12), (23), (13) \} = H g, $$ the products being taken in order.

Answer 2

For each $g\in G$, we have the inner automorphism $\phi_g(x) = g^{-1} x g$. For a subgroup $H\subset G$, the normalizer $N_G(H)$ is the set of $g$ such that $\phi_g$ maps $H$ to itself (that is, $H$ is closed under conjugation by $g$); the centralizer $C_G(H)$ is the set of $g$ such that $\phi_g$ fixes $H$ pointwise. For example, if $H = G$, then $N_G(G) = G$, but $C_G(G)$ is the center $Z(G)$. In particular, if $G$ is nonabelian, then $N_G(G)\not= C_G(G)$.

Answer 3

The key thing to realize is that the notation $gH=Hg$ is not saying that $gh=hg$ for all $h\in H$. Rather, it says that given $h\in H$ there exists $h'\in H$ such that $gh=h'g$.