Which function do we define to show the homomorphism?

[mathmari]

Hey!

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:

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

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

Which function do we have to define to show that the map $N_{G}(H) \to \mathrm{Aut}(G)$ is an homomorphism? (Wondering)

 

[Opalg]

Hey!

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:

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

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

Which function do we have to define to show that the map $N_{G}(H) \to \mathrm{Aut}(G)$ is an homomorphism? (Wondering)

The function should go from $N_{G}(H)$ to $\mathrm{Aut}(H)$, not to $\mathrm{Aut}(G)$.

For a hint, notice that if $gH = Hg$ then $g^{-1}Hg = H.$

 

[mathmari]

The function should go from $N_{G}(H)$ to $\mathrm{Aut}(H)$, not to $\mathrm{Aut}(G)$.

Oh yes... (Blush)

For a hint, notice that if $gH = Hg$ then $g^{-1}Hg = H.$

A $g$ that satisfies that condition is an element of $N_G(H)$, right? To what can we map this element so that we have an automorphism? (Wondering)
I do not really have an idea...

 

[Opalg]

A $g$ that satisfies that condition is an element of $N_G(H)$, right? To what can we map this element so that we have an automorphism? (Wondering)
I do not really have an idea...

If $g^{-1}Hg = H$ then the map $h \mapsto g^{-1}hg$ ($h\in H$) takes $H$ to itself (and in fact is an automorphism of $H$).

 

[mathmari]

If $g^{-1}Hg = H$ then the map $h \mapsto g^{-1}hg$ ($h\in H$) takes $H$ to itself (and in fact is an automorphism of $H$).

So, do we define the map $$g\mapsto (h\mapsto g^{-1}hg)$$ ? (Wondering)

 

[Opalg]

So, do we define the map $$g\mapsto (h\mapsto g^{-1}hg)$$ ? (Wondering)

Yes, exactly. (Clapping) (Of course, you need to check that this does indeed give an isomorphism from $N_G(H)/C_G(H)$ to a subgroup of $\text{Aut}(H)$.)

 

[mathmari]

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

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

(Wondering)

But how could we continue? (Wondering)

 

[Deveno]

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

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

(Wondering)

But how could we continue? (Wondering)

With all due respect to Opalg, I believe there is an error, here.

The map $g \mapsto \phi_g$ where $\phi_g(x) = g^{-1}xg$ is not even a homomorphism, but rather an anti-homomorphism.

The map you want is rather:

$g \mapsto \psi_g$ where $\psi_g(x) = gxg^{-1}$. THIS is a homomorphism.

To see this, we must verify that:

$g_1g_2 \mapsto \psi_{g_1}\circ\psi_{g_2}$

Now under our mapping, we have $g_1g_2 \mapsto \psi_{g_1g_2}$, so we need to show that:

$\psi_{g_1g_2} = \psi_{g_1} \circ \psi_{g_2}$

We do this by comparing values on an arbitrary $x \in G$:

$\psi_{g_1g_2}(x) = (g_1g_2)x(g_1g_2)^{-1} = (g_1g_2)x(g_2^{-1}g_1^{-1}) = g_1(g_2xg_2^{-1})g_1^{-1}$

while:

$(\psi_{g_1}\circ \psi_{g_2})(x) = \psi_{g_1}(\psi_{g_2}(x)) = \psi_{g_1}(g_2xg_2^{-1}) = g_1(g_2xg_2^{-1})g_1^{-1}$

Now, this is a homomorphism $G \to \text{Aut}(G)$, but that's not what we want. We want a homomorphism from:

$N_G(H) \to \text{Aut}(H)$.

So, we can just RESTRICT our homomorphism to the domain $N_G(H)$, but in so doing, we must check that the image $\psi_n$ for $n \in N_G(H)$ is an automorphism of $H$.

Now $\psi_n$ is an inner automorphism of $G$, and is bijective, and will REMAIN bijective when restricted to $H$. It will also still be a homomorphism, so it is an isomorphism between $H$ and $\psi_n(H)$.

So to show $\psi_n$ is an automorphism of $H$, we must show the image $\psi_n(H) = H$.

But this image is $nHn^{-1} = H$, since for any $n \in N_G(H)$, we have $nH = Hn$.

 

[mathmari]

The map $g \mapsto \phi_g$ where $\phi_g(x) = g^{-1}xg$ is not even a homomorphism, but rather an anti-homomorphism.

What exactly is an anti-homomorphism? (Wondering)

Having the map $g \mapsto \phi_g$ where $\phi_g(x) = g^{-1}xg$, we have the following:

$$\phi_{g_1g_2}(x)=(g_1g_2)^{-1}x(g_1g_2)=g_2^{-1}g_1^{-1}xg_1g_2$$
$$(\phi_{g_1}\circ\phi_{g_2})(x)=\phi_{g_1}(\phi_{g_2}(x))=\phi_{g_1}(g_2^{-1}xg_2)=g_1^{-1}g_2^{-1}xg_2g_1$$

So, we have that $ \phi_{g_1g_2}(x)\neq (\phi_{g_1}\circ\phi_{g_2})(x)$. That's why it is not an homomorphism, right? (Wondering)

The map you want is rather:

$g \mapsto \psi_g$ where $\psi_g(x) = gxg^{-1}$. THIS is a homomorphism.

To see this, we must verify that:

$g_1g_2 \mapsto \psi_{g_1}\circ\psi_{g_2}$

Now under our mapping, we have $g_1g_2 \mapsto \psi_{g_1g_2}$, so we need to show that:

$\psi_{g_1g_2} = \psi_{g_1} \circ \psi_{g_2}$

We do this by comparing values on an arbitrary $x \in G$:

$\psi_{g_1g_2}(x) = (g_1g_2)x(g_1g_2)^{-1} = (g_1g_2)x(g_2^{-1}g_1^{-1}) = g_1(g_2xg_2^{-1})g_1^{-1}$

while:

$(\psi_{g_1}\circ \psi_{g_2})(x) = \psi_{g_1}(\psi_{g_2}(x)) = \psi_{g_1}(g_2xg_2^{-1}) = g_1(g_2xg_2^{-1})g_1^{-1}$

Now, this is a homomorphism $G \to \text{Aut}(G)$

Ah ok... I see...

We want a homomorphism from:

$N_G(H) \to \text{Aut}(H)$.

So, we can just RESTRICT our homomorphism to the domain $N_G(H)$, but in so doing, we must check that the image $\psi_n$ for $n \in N_G(H)$ is an automorphism of $H$.

Now $\psi_n$ is an inner automorphism of $G$, and is bijective, and will REMAIN bijective when restricted to $H$. It will also still be a homomorphism, so it is an isomorphism between $H$ and $\psi_n(H)$.

Why will $\psi_n$ still be bijective and still be an homomorphism when restricted to $H$ ? And why do we want to restict it to $H$ ? (Wondering)

So to show $\psi_n$ is an automorphism of $H$, we must show the image $\psi_n(H) = H$.

Why do we have to show that? (Wondering)

 

[Deveno]

What exactly is an anti-homomorphism? (Wondering)

A map $f: G \to G'$ between two groups where $f(a\ast b) = f(b)\ast f(a)$. An example is the inversion map $g \mapsto g^{-1}$ in any non-abelian group.

Having the map $g \mapsto \phi_g$ where $\phi_g(x) = g^{-1}xg$, we have the following:

$$\phi_{g_1g_2}(x)=(g_1g_2)^{-1}x(g_1g_2)=g_2^{-1}g_1^{-1}xg_1g_2$$
$$(\phi_{g_1}\circ\phi_{g_2})(x)=\phi_{g_1}(\phi_{g_2}(x))=\phi_{g_1}(g_2^{-1}xg_2)=g_1^{-1}g_2^{-1}xg_2g_1$$

So, we have that $ \phi_{g_1g_2}(x)\neq (\phi_{g_1}\circ\phi_{g_2})(x)$. That's why it is not an homomorphism, right? (Wondering)

Yes, it reverses the order of the composition.

Ah ok... I see...

sehr gut

Why will $\psi_n$ still be bijective and still be an homomorphism when restricted to $H$ ?

Strictly speaking, I should have also said "to its image". Look, suppose we have an injective function $f:A \to B$.

the function $f:A \to f(A)$ is surjective. So if $f$ is injective, then $f$ is bijective from $A$ to its image $f(A)$.

Now suppose $X$ is a subset of $A$. We define the restriction $f|_X:X \to f(A)$ by:

$f|_X(x) = f(x)$ (this makes sense because $x \in X$, and since $X \subseteq A$, we have $x \in A$, and $f$ is defined on $A$).

I claim $f|_X$ is injective, too, if $f$ is.

Suppose $f|_X(x_1) = f|_X(x_2)$. Then, by definition, $f(x_1) = f(x_2)$. Since $f$ is injective, we have $x_1 = x_2$, and so $f|_X$ is injective.

Since $f|_X: X \to f(X)$ is surjective, this makes $f|_X$ bijective onto its image $(f|_X(X) = f(X)$).

Homomorphism are, after all, functions (with a special property).

Recall that the homomorphism property holds for ANY two elements of our homomorphism domain group, including those that happen to lie in some subgroup.

And why do we want to restrict it to $H$ ? (Wondering)

We want to show that $\psi_n$ is an automorphism of $H$. Normally, the domain of definition of an inner automorphism of $G$ is $G$, so we must "restrict" our inner automorphism to be defined "just on $H$". However, we must also show it maps $H \to H$ (or else it wouldn't be an endomorphism, much less an automorphism).

Why do we have to show that? (Wondering)

See above. The map $\psi_n$, just by virtue of being a bijective homomorphism, takes $H$ to some subgroup $\psi_n(H)$ of $G$. Here is where we use the "special" property of $n$ (namely, that it normalizes $H$) to conclude $\psi_n(H) = nHn^{-1} = H$.

 

[mathmari]

With all due respect to Opalg, I believe there is an error, here.

The map $g \mapsto \phi_g$ where $\phi_g(x) = g^{-1}xg$ is not even a homomorphism, but rather an anti-homomorphism.

The map you want is rather:

$g \mapsto \psi_g$ where $\psi_g(x) = gxg^{-1}$. THIS is a homomorphism.

To see this, we must verify that:

$g_1g_2 \mapsto \psi_{g_1}\circ\psi_{g_2}$

Now under our mapping, we have $g_1g_2 \mapsto \psi_{g_1g_2}$, so we need to show that:

$\psi_{g_1g_2} = \psi_{g_1} \circ \psi_{g_2}$

We do this by comparing values on an arbitrary $x \in G$:

$\psi_{g_1g_2}(x) = (g_1g_2)x(g_1g_2)^{-1} = (g_1g_2)x(g_2^{-1}g_1^{-1}) = g_1(g_2xg_2^{-1})g_1^{-1}$

while:

$(\psi_{g_1}\circ \psi_{g_2})(x) = \psi_{g_1}(\psi_{g_2}(x)) = \psi_{g_1}(g_2xg_2^{-1}) = g_1(g_2xg_2^{-1})g_1^{-1}$

Now, this is a homomorphism $G \to \text{Aut}(G)$

In my notes I found the following:

When we have $g\in G$, we define $\phi_g:G\rightarrow G$ with $x\mapsto x^g=g^{-1}xg$. It is an automorphism of $G$.
$\phi_g$ is an hommomorphism and bijective.
The set $\{\phi_g \mid g\in G\}=\text{Inn}(G)$ is called the inner automorphisms of $G$.

Is this definition wrong? (Wondering)

Since $f|_X: X \to f(X)$ is surjective, this makes $f|_X$ bijective onto its image $(f|_X(X) = f(X)$).

What do you mean by "bijective onto its image" ? (Wondering)
That the map form $X$ to its image is bijective? (Wondering)

 

[mathmari]

To clarify...

We define the map $$\psi_g : G\rightarrow G \\ x\mapsto gxg^{-1}, \text{ with } g\in G$$ which is an automorphism of $G$.

The map $g\mapsto \psi_g$ is an epimorphism $G\rightarrow \text{Aut}(G)$.

We are looking for an epimorphism $N_G(H)\rightarrow \text{Aut}(H)$.

We can restrict the above epimorphism to the domain $N_G(H)$.
If we restrict the domain to $N_G(H)$, we have to check that the image $\psi_n$ for $n\in N_G(H)$ is an automorphism of $H$.

We have that $\psi_n$ is an inner automorphism of $G$, so it is an homomorphism, 1-1 and onto.
Since $H\leq G$, we have that $\psi_n$ remains 1-1 and onto when we restrict it to $H$, so $\psi_n\rightarrow \psi_n(H)$ is 1-1 and onto.

Since $n\in N_G(H)$ we have that $nH=Hn\Rightarrow nHn^{-1}=H \Rightarrow \psi_n(H)=H$.

That means that $\psi_n\rightarrow H$ is 1-1 and onto.

Since $\psi_n:G \rightarrow G$ is an homomorphism, that means that this identity holds for any two elements of the domain, so also for these that belong to $H$.
So, $\psi_n\rightarrow H$ is also an homomorphism.

So, we have that $\psi_n\rightarrow H$ is 1-1, onto and an homomorphism, that means that $\psi_n\rightarrow H$ is an automorphism of $H$. The map $$N_G(H)\rightarrow \text{Aut}(H) \\ n\mapsto \psi_n$$ is an homomorphism and onto, because the map $$G\rightarrow \text{Aut}(G) \\ g\mapsto \psi_g$$ is an homomorphism and onto and $N_G(H)\subseteq G$, or not? (Wondering) Now it is left to show that the kernel of the map is $C_G(H)$, right? (Wondering)

 

[Deveno]

In my notes I found the following:

When we have $g\in G$, we define $\phi_g:G\rightarrow G$ with $x\mapsto x^g=g^{-1}xg$. It is an automorphism of $G$.
$\phi_g$ is an homomorphism and bijective.
The set $\{\phi_g \mid g\in G\}=\text{Inn}(G)$ is called the inner automorphisms of $G$.

Is this definition wrong? (Wondering)

No, both maps:

$x \mapsto gxg^{-1}$
$x \mapsto g^{-1}xg$

One is conjugation by $g^{-1}$ (also written ${}^gx)$, and one is conjugation by $g$ (also written $x^g$). If one composes "right-to-left" (that is $(f_1\circ f_2)(s) = f_1(f_2(s))$, in other words first do $g$, then do $f$), to get a HOMOMORPHISM:

$G \to \text{Aut}(G)$ one needs to use the first mapping. Basically it's a "left-versus-right" thing that crops up so often in algebra.

Both kinds of conjugation are inner automorphisms (if $g \in G$, so is $g^{-1}$).

What do you mean by "bijective onto its image" ? (Wondering)
That the map from $X$ to its image is bijective? (Wondering)

Yes.

 

[Deveno]

To clarify...

We define the map $$\psi_g : G\rightarrow G \\ x\mapsto gxg^{-1}, \text{ with } g\in G$$ which is an automorphism of $G$.

The map $g\mapsto \psi_g$ is an epimorphism $G\rightarrow \text{Aut}(G)$.

No, it is not onto. It's image is $\text{Inn}(G)$, the subgroup of $\text{Aut}(G)$ consisting of *inner* automorphisms (there are also automorphisms which are not inner, these are called outer automorphisms. For example, the map:

$(1,0) \mapsto (1,1)$
$(0,1) \mapsto (0,1)$

is an outer automorphism of $\Bbb Z_2 \times \Bbb Z_2$ - this group is abelian, and all inner automorphisms of it are thus the identity).

We are looking for an epimorphism $N_G(H)\rightarrow \text{Aut}(H)$.

Again, no, we do not need the homomorphism to be onto. We just need its image to be a SUBGROUP of $\text{Aut}(H)$.

We can restrict the above epimorphism to the domain $N_G(H)$.
If we restrict the domain to $N_G(H)$, we have to check that the image $\psi_n$ for $n\in N_G(H)$ is an automorphism of $H$.

Yes.

We have that $\psi_n$ is an inner automorphism of $G$, so it is an homomorphism, 1-1 and onto.

Yes, but only from $G$ to $G$.

Since $H\leq G$, we have that $\psi_n$ remains 1-1 and onto when we restrict it to $H$, so $\psi_n\rightarrow \psi_n(H)$ is 1-1 and onto.

Still 1-1, automatically. Only onto if we consider $\psi_n$ as a map $H \to \psi_n(H)$.

Since $n\in N_G(H)$ we have that $nH=Hn\Rightarrow nHn^{-1}=H \Rightarrow \psi_n(H)=H$.

That is what a normalizer does. It normalizes (by definition, for any $n \in N_G(H)$, we have $nHn^{-1} = H$. This is equivalent to $nH = Hn$).

That means that $\psi_n\rightarrow H$ is 1-1 and onto.

"...from $H$ to $H$".

Since $\psi_n:G \rightarrow G$ is an homomorphism, that means that this identity holds for any two elements of the domain, so also for these that belong to $H$.
So, $\psi_n\rightarrow H$ is also an homomorphism.

Yes.

So, we have that $\psi_n\rightarrow H$ is 1-1, onto and an homomorphism, that means that $\psi_n\rightarrow H$ is an automorphism of $H$.

"...onto $H$". This is important.

The map $$N_G(H)\rightarrow \text{Aut}(H) \\ n\mapsto \psi_n$$ is an homomorphism and onto, because the map $$G\rightarrow \text{Aut}(G) \\ g\mapsto \psi_g$$ is an homomorphism and onto and $N_G(H)\subseteq G$, or not? (Wondering)

It's a homomorphism. It does not HAVE to be onto, we are only trying to prove a SUBGROUP of $\text{Aut}(H)$ is isomorphic to $N_G(H)/C_G(H)$.

Now it is left to show that the kernel of the map is $C_G(H)$, right? (Wondering)

Correct.