I have a problem and I also have the solution to the problem. I just want to understand the solution.
Consider the group $(\mathbb Z_{11} - \{0\},.)$
(b) Find all isomorphisms between this group and itself. and Here is the solution:
Can someone please explain the solution, especially the highlighted part? I don't know why the question can just say it follows that such and such.
In general is there a 'good' way to approach questions relating 'finding isomorphisms to it self '??
The statement is that since $2$ generates the group $G=U(11)=(\Bbb F_{11}\setminus 0,\cdot)$, so does $f(2)$. Note that $U(11)\cong \Bbb Z/10\Bbb Z$, which has $\phi(10)=4$ generators.
Suppose you manage to find a generator $a$ for $G$, then $a,a^3,a^7,a^9$ will be generators as well, since $\{1,3,7,9\}$ are integers less than, and co-prime, to $10$. Now we know that we can take $a=2$.
I am guessing that with the notation $(\mathbb{Z}_{11} - \{0\})$ you are denoting the multiplicative group taken over the units of the equivalences classes of the integers mod 11. This is slightly unconventional, so please let me know if my assumption is wrong. I will just use $G$ in this case.
Remark of course that $G = \langle 2 \rangle$, and thus it is cyclic (precisely, it is isomorphic to $\mathbb{Z}_{10}$). We can represent every element $a \in G$ then as some power of $2$, i.e. $a = 2^m, m \in \mathbb{Z}$.
Here it is important to acknowledge a basic fact about isomorphisms, that is that they map generators to generators. We can show this relatively simply: assume $\theta : G \rightarrow H$ is an isomorphism, and let $h \in H$. As $\theta$ is a surjection then, we have:
$$h = \theta(a) \text{ for some } a \in G = \theta(2^m) = \theta(2)^m$$
As our selection of $h$ was arbitrary, we note then all elements of $H$ can be written as powers of what the generator is mapped to under isomorphism, and thus $H$ is of course both cyclic but also generated by all the generators of $G$ under this mapping.
We will now return our gaze to just automorphisms, $f : G \rightarrow G$. You can verify that $2^3, 2^7, 2^9$ are also generators here for $G$ (generally known as $G \cong \mathbb{Z}_{10}$, which has $4 = \phi(10)$ generators). Thus, as we have just shown, $f$ must map 2 to one of these powers of $2$ (or map to itself).
This essentially gives us just 4 automorphisms, as all that particularly matters is what generator we choose to send $2$ to. The remainder of the proof provided it just using the properties of the homomorphism to establish explicit definitions of these mappings.
