Looking at the solution for part b), can someone explain what is meant by $\langle 2 \rangle = (\mathbb{Z}_{11} -{0},\cdot)$ and how come the choices for $f(2)$ are only 1,3,7 and 9? why arent any other elements choices for f(2)
im having alot of trouble understanding this question and solution.
-thanks
On
$\langle 2\rangle=(\mathbb{Z}_{11}-0,\cdot)$ means that the group is generated by $2$. This is shown in part (a): by taking all powers of $2$, you enumerate every element of the group before coming back to the identity element.
The image of a generator under an isomorphism $f$ must also generate the group. But the other unlisted powers of $2$ do not generate the group. For example, $\langle 2^5\rangle=\langle 10\rangle=\{1,2^5\}$, so we rule out $f(2)$ taking the value $2^5$ because it fails to generate the group.
On
For your first issue, the equality $\langle 2\rangle = (\Bbb Z_{11}-\{0\}, \cdot)$ is a statement that the subgroup generated by the element 2 (under multiplication is all of $\Bbb Z_{11}-\{0\}$. This is just a way of rephrasing the result of the first part of the problem, which is expressing every element of $\Bbb Z_{11}-\{0\}$ as an element of $\langle 2 \rangle$, which can also be expressed as the elements expressible in the form $2^i$ (Mostly-unnecessary technical detail: since this is a finite group here it suffices to only consider positive values of $i$.)
For the second part, the argument given in the solution is that since every element of the group is a power of 2, then every element of the group is a power of $f(2)$ under any isomorphism $f$. To show this, note that $f$ being an isomorphism implies that every element of the target (which is in this case is $\Bbb Z_{11}-\{0\}$) is of the form $f(x)$ where $x$ is some element in the domain (which is also $\Bbb Z_{11}-\{0\}$). Any such $x$ is of the form $2^i$ by the first part of the problem, and $f$ being a homomorphism implies $f(2^i) = (f(2))^i$.
Thus, the values possible for $f(2)$ are restricted to those values $y$ such that $\langle y \rangle = (\Bbb Z_{11}-\{0\}, \cdot)$, or in other words, those $y$ such that the values of $y^i$ contain every element of $\Bbb Z_{11}-\{0\}$. All such $y$ are of the form $2^j$ for $0\le j \le 9$ by previous work. Now we need to see directly that for any $j$ other than $1,3,7,9$ the values $y^i = 2^{ij}$ do not contain every element of the group. The shortest way to argue this is that the other values of $j$ are divisible by 2 or 5, which means that all values of $ij$ will similarly be divisible by 2 or 5. To see this concretely, for instance look at the powers of $4= 2^2$:
$$4^1 = 2^2 = 4, 4^2 = 2^4 = 8, 4^3 = 2^6 = 9, 4^4 = 2^8 = 3, 4^5 = 2^{10} = 1$$
Note that choosing higher powers of 4 will loop around these five values since $4^5 =1$ implies $4^{5k+i} = (4^5)^k4^i = (1)^k4^i = 4^i$. Therefore, 4 is not a valid choice for $f(2)$.
The problem is talking about the group of units of the ring $\Bbb Z_{11}$, the integers modulo $11$. The group operation in this group is multiplication modulo $11$, so, for example:
$[7]\cdot[8] = [1]$, because $56 \equiv 1$ (mod $11$), since $56 = 1 + 5\cdot 11$.
The problem asks you to find $\langle 2\rangle = \{[2^k]: k \in \Bbb Z\}$, the subgroup of $(\Bbb Z_{11})^{\times}$ generated by $[2]$ (it is common practice to drop the brackets and refer to $[2]$ as $2$, which can be confusing. We are not talking about the integer $2$, but rather, the equivalence class of $2$ modulo $11 = \{m \in \Bbb Z\mid m - 2 = 11t$, for some $t \in \Bbb Z\}$.
It is easy to calculate the first few powers:
$[2^1] = [2]\\ [2^2] = [4]\\ [2^3] = [8]$
After this, a little more thought is required:
$[2^4] = [16] = [5]$ (since $16-5$ is divisible by $11$)
$[2^5] = [32] = [10]$ (since $32 - 10$ is divisible by $11$)
$[2^6] = [64] = [9]\\ [2^7] = [128] = [7]\\ [2^8] = [256] = [3]$
(for example, for the last one:
$[256] = [220 + 36] = [220] + [36] = [0] + [36] = [36] = [33] + [3] = [3]$).
Finally:
$[2^9] = [512] = [72] = [6]$ and $[2^{10}] = [1024] = [34] = [1]$
This shows that $[2]$ has order $10$, and generates all non-zero elements of $\Bbb Z_{11}$.
From this, we can conclude that all ten elements of $\Bbb Z_{11} - \{[0]\}$ are thus units (for example the inverse of $[2^6] = [9]$ is clearly $[2^4] = [5]$, since $[1] = [2^{10}] = [2^6\cdot 2^4] = [2^6]\cdot[2^4]$, and indeed:
$[9]\cdot [5] = [45] = [1]$ because $45 - 1 = 4\cdot 11$).
Thus $\Bbb Z_{11} - \{[0]\}$ is cyclic, of order $10$.
It follows that any isomorphism $f:(\Bbb Z_{10} - \{[0]\}) \to (\Bbb Z_{10} - \{[0]\})$ is completely determined by $f([2])$, and since every element is of the form $[2^k]$, we must only find which $k = \{0,1,2,\dots,9\}$ will do.
This, in turns means that $[2^k]$ must also be a generator, or else our mapping $f$ is not injective (and also thus not surjective). Hopefully, you can see that if $\gcd(k,10) \neq 1$, say, $d$, for example, so that $10 = du$ and $k = dv$, then:
$[(2^k)^u] = [2^{ku}] = [2^{(dv)u}] = [2^{(du)v}] = [(2^{10})^v] = [2^{10}]^v = [1]^v = [1^v] = [1]$, so that $[2^k]$ has order at most $u < 10$, and cannot be a generator.
Thus we have only $4$ possible $k$: those for which $\gcd(k,10) = 1$, namely: $k = 1,3,7,9$. You can verify that each of these is, in fact, a generator, and we are done.