Deveno said:
Yes, all cyclic groups are abelian. Can you prove this?
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)
Deveno said:
If $a$ has order 2, then the only powers of $a$ that are distinct are $e = a^0$ and $a = a^1$, since:
$a^2 = e$
$a^3 = a(a^2) = a$
$a^4 = (a^2)(a^2) = ee = e$
$a^5 = (a^3)(a^2) = ae = a$...and so on (so only two non-negative powers).
But if $a^2 = e$ then $a^{-1} = a$, and this is the unique inverse of $a$. So for any negative power:
$a^{-k} = (a^{-1})^k = a^k$, so we don't get any more powers here.
Since $G$ has four elements, and the subgroup generated by $a$ only has two elements, we must have something else in $G$, right?
Ah ok... (Nerd)
Deveno said:
I would argue as follows:
$ba = e \implies bba = b \implies a = b$, contradiction.
$ba = a \implies baa = aa \implies b = e$, contradiction.
$ba = b \implies bba = bb \implies a = e$, contradiction.
Hence $ba = ab$.
(proving this is *possible* is shown by the example $\Bbb Z_2 \times \Bbb Z_2$ with $a = (1,0)$ and $b = (0,1)$).
I see... (Nerd)
Deveno said:
Well it suffices to show that "everything commutes with everything", right?
$e$ clearly commutes with anything.
$a$ commutes with $a$: because $aa = aa$.
$a$ commutes with $b$: because $ab = ba$.
$a$ commutes with $ab$: because $a(ab) = a(ba) = (ab)a$.
Now we have already shown that $b$ commutes with $e$ and $a$, so we just need to show that $b$ commutes with itself (which is obvious) and $ab$:
$b(ab) = (ba)b = (ab)b$, done.
Finally, above we see that $ab$ commutes with $e,a$ and $b$, so the only thing we haven't checked is that $ab$ commutes with itself, which is again obvious:
$(ab)(ab) = (ab)(ab)$.
I understand... (Smile)
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)
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)