MHB Proving the Normality of Inner Automorphism Group in Group Theory

[mathmari]

Hey!

I want to show that $\text{Inn}(G)\trianglelefteq\text{Aut}(G)$ for each group $G$.

We have that the inner automorphisms of $G$ is the following set $\text{Inn}(G)=\{\phi_g\mid g\in G\}$ where $\phi_g$ is an automorphism of $G$ and it is defined as follows: $$\phi_g : G\rightarrow G \\ x\mapsto x^g=g^{-1}xg$$

To show that $\text{Inn}(G)$ is an normal subgroup of $\text{Aut}(G)$ we have to show that $$h\phi_gh^{-1}=\phi_g\in \text{Inn}(G), \forall h\in \text{ Aut}(G) \text{ and } \forall \phi_g\in \text{Inn}(G)$$

We have the following:
$$h\phi_gh^{-1}(x)=h(\phi_g (h^{-1}(x)))=h(g^{-1}h^{-1}(x)g)$$
Since $h$ is an homomorphism we have that $$h(g^{-1}h^{-1}(x)g)=h(g^{-1})h(h^{-1}(x))h(g)=h(g^{-1})xh(g)$$

Is everything correct so far? (Wondering)

How could we continue? (Wondering)

 

[Deveno]

Hey!

I want to show that $\text{Inn}(G)\trianglelefteq\text{Aut}(G)$ for each group $G$.

We have that the inner automorphisms of $G$ is the following set $\text{Inn}(G)=\{\phi_g\mid g\in G\}$ where $\phi_g$ is an automorphism of $G$ and it is defined as follows: $$\phi_g : G\rightarrow G \\ x\mapsto x^g=g^{-1}xg$$

To show that $\text{Inn}(G)$ is an normal subgroup of $\text{Aut}(G)$ we have to show that $$h\phi_gh^{-1}=\phi_g\in \text{Inn}(G), \forall h\in \text{ Aut}(G) \text{ and } \forall \phi_g\in \text{Inn}(G)$$

No, that is incorrect. In general, to show a group $N$ is normal in a group $G$, we need to show that:

$gng^{-1} \in N$ for any $g \in G$, and $n \in N$.

This is *not* the same as saying $gng^{-1} = n$. Your are confusing *normalizing* with *centralizing*. We only require conjugates to again be in $N$, we do not require conjugation FIXES elements of $N$.

We have the following:
$$h\phi_gh^{-1}(x)=h(\phi_g (h^{-1}(x)))=h(g^{-1}h^{-1}(x)g)$$
Since $h$ is an homomorphism we have that $$h(g^{-1}h^{-1}(x)g)=h(g^{-1})h(h^{-1}(x))h(g)=h(g^{-1})xh(g)$$

Is everything correct so far? (Wondering)

How could we continue? (Wondering)

Personally, I wouldn't use $h$ to denote an automorphism since it might get confused with an ELEMENT of $G$.

So let $\psi \in \text{Aut}(G)$.

We want to show that $\psi\phi_g\psi^{-1}$ is an inner automorphism, in other words we need to produce some $a \in G$ such that:

$\psi\phi_g\psi^{-1} = \phi_a$.

Starting with some arbitrary $x \in G$ is a good idea, however:

$\psi\phi_g\psi^{-1}(x) = \psi(\phi_g(\psi^{-1}(x))) = \psi(g^{-1}(\psi^{-1}(x))g)$

$= \psi(g^{-1})\psi(\psi^{-1}(x))\psi(g)$

Here, we catch a break-the middle factor in our 3-fold product simplifies:

$\psi(\psi^{-1}(x)) = x$, so we have:

$= \psi(g^{-1})x\psi(g)$

Now, since $\psi$ is a homomorphism, $\psi(g^{-1}) = [\psi(g)]^{-1}$, and thus we have:

$= [\psi(g)]^{-1}x\psi(g)$, which suggests we choose $a = \psi(g)$, that is:

$\psi\phi_g\psi^{-1} = \phi_a$, and $\phi_a \in \text{Inn}(G)$.

 

[mathmari]

In general, to show a group $N$ is normal in a group $G$, we need to show that:

$gng^{-1} \in N$ for any $g \in G$, and $n \in N$.

This is *not* the same as saying $gng^{-1} = n$. Your are confusing *normalizing* with *centralizing*. We only require conjugates to again be in $N$, we do not require conjugation FIXES elements of $N$.
Personally, I wouldn't use $h$ to denote an automorphism since it might get confused with an ELEMENT of $G$.

So let $\psi \in \text{Aut}(G)$.

We want to show that $\psi\phi_g\psi^{-1}$ is an inner automorphism, in other words we need to produce some $a \in G$ such that:

$\psi\phi_g\psi^{-1} = \phi_a$.

Ah ok... I see... (Thinking)

since $\psi$ is a homomorphism, $\psi(g^{-1}) = [\psi(g)]^{-1}$

Why does this stand? (Wondering)

 

[Deveno]

It suffices to show that:

$\psi(g^{-1})\psi(g) = e_G$.

But $\psi$ is a homomorphism, so:

$\psi(g^{-1})\psi(g) = \psi(g^{-1}g) = \psi(e_G)$.

Now, we just have to show that $\psi(e_G) = e_G$.

$\psi$ is bijective, so given any $x \in G$, we have $x = \psi(y)$ for some $y \in G$.

Hence $x\psi(e_G) = \psi(y)\psi(e_G) = \psi(ye_G) = \psi(y) = x$, and:

$\psi(e_G)x =\psi(e_G)\psi(y) = \psi(e_Gy) = \psi(y) = x$.

Since $e_G$ is the UNIQUE element of $G$ such that:

$xe_G = e_Gx = x$, for all $x \in G$, we conclude $\psi(e_G) = e_G$, QED.

 

[mathmari]

It suffices to show that:

$\psi(g^{-1})\psi(g) = e_G$.

But $\psi$ is a homomorphism, so:

$\psi(g^{-1})\psi(g) = \psi(g^{-1}g) = \psi(e_G)$.

Now, we just have to show that $\psi(e_G) = e_G$.

$\psi$ is bijective, so given any $x \in G$, we have $x = \psi(y)$ for some $y \in G$.

Hence $x\psi(e_G) = \psi(y)\psi(e_G) = \psi(ye_G) = \psi(y) = x$, and:

$\psi(e_G)x =\psi(e_G)\psi(y) = \psi(e_Gy) = \psi(y) = x$.

Since $e_G$ is the UNIQUE element of $G$ such that:

$xe_G = e_Gx = x$, for all $x \in G$, we conclude $\psi(e_G) = e_G$, QED.

I understand! (Nerd)

Starting with some arbitrary $x \in G$ is a good idea, however:

$\psi\phi_g\psi^{-1}(x) = \psi(\phi_g(\psi^{-1}(x))) = \psi(g^{-1}(\psi^{-1}(x))g)$

$= \psi(g^{-1})\psi(\psi^{-1}(x))\psi(g)$

Here, we catch a break-the middle factor in our 3-fold product simplifies:

$\psi(\psi^{-1}(x)) = x$, so we have:

$= \psi(g^{-1})x\psi(g)$

Now, since $\psi$ is a homomorphism, $\psi(g^{-1}) = [\psi(g)]^{-1}$, and thus we have:

$= [\psi(g)]^{-1}x\psi(g)$, which suggests we choose $a = \psi(g)$, that is:

$\psi\phi_g\psi^{-1} = \phi_a$, and $\phi_a \in \text{Inn}(G)$.

We have shown that
$$\psi\phi_g\psi^{-1}(x)=[\psi(g)]^{-1}x\psi(g)=\phi_{\psi (g)}$$

Is this equivalent to $$\psi\phi_g\psi^{-1} = \phi_{\psi (g)}$$ ? (Wondering)

Does it stand because $x$ is arbitrary? (Wondering)

 

[Deveno]

I understand! (Nerd)


We have shown that
$$\psi\phi_g\psi^{-1}(x)=[\psi(g)]^{-1}x\psi(g)=\phi_{\psi (g)}$$

Is this equivalent to $$\psi\phi_g\psi^{-1} = \phi_{\psi (g)}$$ ? (Wondering)

Does it stand because $x$ is arbitrary? (Wondering)

Yes, and yes.

 

[mathmari]

Yes, and yes.

Nice... Thank you very much! (Smile)