This is the question that don't have any idea Which point I was wrong
Q) Let group homomorphism $f : Z_{50}^* \to Z_{50}^* $ $by$ $f(3) = 31$
(Here, the $Z_{50}^* = \{a \in Z_{50} \vert gcd(a,50)=1\}$ )
$3$ is a primitive root for $mod$ $50$
Find all the element of the $A = \{ x \in Z_{50}^* \vert f(x)=9\}$
My attempt) $31 = 81 = 3^4$ $(mod50)$ So, $f(3) = 3^4$
Plus Owing to the $3$ is a primitive roots of the $Z_{50}^*$, $3$ is a generator of the group $Z_{50}^*$
Then All we have to do is just find the $3^a s.t.$ $f(3^a) = 3^{4a} =3^2(mod50)$
Hence Find the $a$ satisfying $4a = 2(mod \phi(50))$
(Here the $a \in \{x \vert 1 \leq x <50, gcd(x,50)=1\}$)
But $\phi(50)$ = $20$, There aren't exist the "$a$ ". (I.E. $A = \phi$)
p.s.)
But the someone who gave me this question said the answer is $A = \{9,13,37,41\}$
I totally couldn't understand Which point I was wrong.
Please help me. Thanks.
Additional post) Here is that person's solution who claiming the $A = \{9,13,37,41\}$
He might be the suggesting the incorrect answer, Surely there are two possibility that Should be incorrect either mine or his.
For the integer set $Z$, since $3^4 = 31$, $imf=\{f(3^a) \vert a \in Z\} = \{(3^4)^a \vert a \in Z\}= <3^4>$
Then $\vert imf \vert =5$ and $\vert Z_{50}^* / kerf \vert = \vert imf \vert = 5$
Hence $\vert kerf \vert =4$
Also, $f(-7) = f(3^5) = f(3)^5 = 3^{20} =1$
Plus, $f(-1) = f(3^{10} ) = f(3)^{10} = 3^{40} =1$ Therefore $\{-1, -7\} \in kerf$ So, $kerf = \{1,-1,7,-7\} $
We can conclude the $A = f^{-1}({9}) = 9kerf = \{9,-9,63,-63\} = \{9,13,37,41\}$