MHB Understanding Aut(G) Isomorphism to Z2

[Deveno]

Suppose that a group $G$ is cyclic, then it is generated by a generator, say $g$.
Then every element of $G$ can be written in the form $g^i$ for some $i$.
Suppose that $x,y\in G$ then $x=g^k$ and $y=g^m$.
We have $$xy=g^kg^m=g^{k+m}=g^{m+k}=g^mg^k=yx$$
Therefore, a cyclic group is abelian.

Is this correct? (Wondering)

Yes. Notice that all the "action" happens in the exponents-the "$x$" just acts as a placeholder, of sorts. So a cyclic group is abelian BECAUSE the integers are abelian (under addition). In fact, EVERY cyclic group is a homomorphic image of the integers-if it is an isomorphic image (1-1 and onto) it is isomorphic to the images. If it is just an image (only onto), it is a FINITE cyclic group (with the smallest (positive) "repeat" of powers being the ORDER), isomorphic to $\Bbb Z_n$ for some $n$.

Ah ok... (Nerd)



I see... (Nerd)

I understand... (Smile)

Cool.

So, having shown that a group of $4$ elements is abelian, we have that $\text{Aut}(\mathbb{Z}_8)$, which has $4$ elements, is abelian.

Therefore, $$\text{id}\circ f_i=f_i\circ \text{id} \\ f_i\circ f_j =f_j\circ f_i$$ for $i,j=1,2,3$ and $i\neq j$.

We have the following:
$$(\text{id}\circ \text{id})(x)=\text{id}(\text{id}(x))=\text{id}(x)=x$$
So, $\text{id}\circ\text{id}$ is the function $\mathbb{Z}_8\rightarrow \mathbb{Z}_8$ that maps $x\mapsto x$, so it is the function $\text{id}$.
Therefore, we have $h(\text{id}\circ\text{id})=h(\text{id})=(0,0)=(0,0)+(0,0)=h(\text{id})+h(\text{id})$. $$(f_1\circ f_1)(x)=f_1(f_1(x))=f_1(3x)=3( 3 x)=9x\equiv x$$
So, $f_1\circ f_1$ is the function $\mathbb{Z}_8\rightarrow \mathbb{Z}_8$ that maps $x\mapsto x$, so it is the function $\text{id}$.
Therefore, we have $h(f_1\circ f_1)=h(\text{id})=(0,0)=(0,1)+(0,1)=h(f_1)+h(f_1)$. $$(f_2\circ f_2)(x)=f_2(f_2 (x))=f_2(-3x)=-3(-3x)=9x\equiv x$$
So, $f_2\circ f_2$ is the function $\mathbb{Z}_8\rightarrow \mathbb{Z}_8$ that maps $x\mapsto x$, so it is the function $\text{id}$.
Therefore, we have $h(f_2\circ f_2)=h(\text{id})=(0,0)=(1,0)+(1,0)=h(f_2)+h(f_2)$. $$(f_3\circ f_3)(x)=f_3(f_3(x))=f_3(-x)=-(-x)=x$$
So, $f_3\circ f_3$ is the function $\mathbb{Z}_8\rightarrow \mathbb{Z}_8$ that maps $x\mapsto x$, so it is the function $\text{id}$.
Therefore, we have $h(f_3\circ f_3)=h(\text{id})=(0,0)=(1,1)+(1,1)=h(f_3)+h(f_3)$. $$(f_1 \circ f_2)(x)=f_1(f_2 (x))=f_1(-3x)=3(-3x)=-9x\equiv -x$$
So, $f_1\circ f_2$ is the function $\mathbb{Z}_8\rightarrow \mathbb{Z}_8$ that maps $x\mapsto -x$, so it is the function $f_3$.
Therefore, we have $h(f_1\circ f_2)=h(f_3)=(1,1)=(0,1)+(1,0)=h(f_1)+h(f_2)$. $$(f_1\circ f_3)(x)=f_1(f_3(x))=f_1(-x)=3(-x)=-3x$$
So, $f_1\circ f_3$ is the function $\mathbb{Z}_8\rightarrow \mathbb{Z}_8$ that maps $x\mapsto -3x$, so it is the function $f_2$.
Therefore, we have $h(f_1\circ f_3)=h(f_2)=(1,0)=(0,1)+(1,1)=h(f_1)+h(f_3)$. $$(f_2\circ f_3)(x)=f_2(f_3(x))=f_2(-x)=-3(-x)=3x$$
So, $f_2\circ f_3$ is the function $\mathbb{Z}_8\rightarrow \mathbb{Z}_8$ that maps $x\mapsto 3x$, so it is the function $f_1$.
Therefore, we have $h(f_2\circ f_3)=h(f_1)=(0,1)=(1,0)+(1,1)=h(f_2)+h(f_3)$.
So, we have that $h$ is an homomorphism.

Is this correct? (Wondering)

Yes...verified in excruciating detail. You might look for short-cuts with large groups.

Is $h$ 1-1 and onto because we have defined the function so that each element of $\text{Aut}(\mathbb{Z}_8)$ is mapped to one element in $\mathbb{Z}_2\times\mathbb{Z}_2$ and each element of $\mathbb{Z}_2\times\mathbb{Z}_2$ has its original in $\text{Aut}(\mathbb{Z}_8)$? (Wondering)

Yes, this is clear to see.

 

[mathmari]

Yes. Notice that all the "action" happens in the exponents-the "$x$" just acts as a placeholder, of sorts. So a cyclic group is abelian BECAUSE the integers are abelian (under addition). In fact, EVERY cyclic group is a homomorphic image of the integers-if it is an isomorphic image (1-1 and onto) it is isomorphic to the images. If it is just an image (only onto), it is a FINITE cyclic group (with the smallest (positive) "repeat" of powers being the ORDER), isomorphic to $\Bbb Z_n$ for some $n$.

Ah ok... (Thinking)

You might look for short-cuts with large groups.

What do you mean? (Wondering)

 

[Deveno]

Ah ok... (Thinking)


What do you mean? (Wondering)

Well suppose you wanted to classify $\text{Aut}(\Bbb Z_{90})$. There's a LOT of products to check, right?

 

[mathmari]

Well suppose you wanted to classify $\text{Aut}(\Bbb Z_{90})$. There's a LOT of products to check, right?

Yes... What would we do in that case? (Wondering)

 

[Deveno]

maybe if $\text{Aut}(\Bbb Z_n) \cong (\Bbb Z_n)^{\times} = U(n)$ that would help?