Which is the difference between the two notations?

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



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?



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?

• In case you don't already know, $N_G(H)$ is referred to as the "normalizer of $H$ inside $G$", and $C_G(H)$ is the "centralizer of $H$ inside $G$". Feb 24 '16 at 16:22
• I'd use a slightly different notation. For $g \in G$, define the automorphism $\phi_g : H \to H$ of $H$ by $\phi_g(h) = g^{-1}hg$. The homomorphism $g \mapsto \phi_g$ is tricky, because it's from a group with product $\cdot$ (regular multiplication) to a group with product $\circ$ (function composition). This should help to see that you want to show $\phi_{g_1g_2}(h) = \phi_{g_1}(\phi_{g_2}(h)) = (\phi_{g_1} \circ \phi_{g_2})(h)$. It's probably best to ask a new question though, as you already have a few good answers to the original question. Feb 28 '16 at 20:20

Let me give you an example.

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

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

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.

• I think I got it... Thanks for explaining it to me!! Could you explain also to me the meaning of $N_G(H)/C_G(H)$ ? Feb 24 '16 at 21:32
• That's a quotient group, as the second subgroup is a normal subgroup of the first one. Feb 24 '16 at 22:10
• Ah ok... To show that $N_G(H)/C_G(H)$ is isomorphic to $\text{Aut}(H)$ we have to define a function that maps an element of $N_G(H)/C_G(H)$ to an element of $\text{Aut}(H)$ and show that it is bijective and an homomorphism, right? How does an element of $N_G(H)/C_G(H)$ look like? Feb 24 '16 at 22:25
• Use the first isomorphism theorem. Start with the map $N_{G}(H) \to \mathrm{Aut}(G)$ given by $x \mapsto (h \mapsto x^{-1} h x)$. Show that this is a homomorphism, with kernel $C_{G}(H)$. Feb 24 '16 at 22:28
• Yes, sorry, a misprint. Feb 24 '16 at 22:36

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 $N_G(G)\not= C_G(G)$.

• Perfect answer, allow me also to rephrase some of the thing you have written above (sometimes rephrasing helps understand better). We have that $g \in N_G(H)$ if and only if the restriction of $\phi_g$ to $H$ give an endomorphism, actually an automorphism, of $H$ (i.e. an ${\phi_g}_{|H} \colon H \to H$). Instead $g \in C_G(H)$ if and only if not only ${\phi_g}_{|H}$ is an $H$-automorphism but in particular ${\phi_g}_{|H}=\text{id}_H$. This is just a different way to say what anomaly already said. Feb 24 '16 at 16:57
• I think I got it... Thanks for explaining it to me!! Could you explain also to me the meaning of $N_G(H)/C_G(H)$ ? Feb 24 '16 at 21:32
• It's a quotient group. What are you confused by? Feb 24 '16 at 22:11
• Ah ok... To show that $N_G(H)/C_G(H)$ is isomorphic to $\text{Aut}(H)$ we have to define a function that maps an element of $N_G(H)/C_G(H)$ to an element of $\text{Aut}(H)$ and show that it is bijective and an homomorphism, right? How does an element of $N_G(H)/C_G(H)$ look like? Feb 25 '16 at 21:30
• The quotient is not isomorphic to $\operatorname{Aut}(H)$ in general. For example, take again $G = H$. Feb 25 '16 at 21:36

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 $gh=h'g$.

• Ah ok... To show that $N_G(H)/C_G(H)$ is isomorphic to $\text{Aut}(H)$ we have to define a function that maps an element of $N_G(H)/C_G(H)$ to an element of $\text{Aut}(H)$ and show that it is bijective and an homomorphism, right? How does an element of $N_G(H)/C_G(H)$ look like? Feb 25 '16 at 21:30