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. $A = \phi$)


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 $A = f^{-1}({9}) = 9kerf = \{9,-9,63,-63\} = \{9,13,37,41\}$


I cannot see where you are wrong. Since $13=3^{17} \bmod 50$, you have $f(13)=f(3^{17})=31^{17}=11 \neq 9$. Moreover, $9=3^2 \bmod 50$ and so $f(9)=31^2 =11 \neq 9$. Can you check the other proposed solutions? Why must the other person be right?

(Incidentally, $f(37)=11$ and $f(41)=11$, too, so at least the other person is wrong.)

After your edits, your colleague is wrong. He is using the fact that the kernel has the same size as any pre-image of a point, which is correct AS LONG AS THE PRE-IMAGE IS NON-EMPTY. The correct statement of that theorem goes as follows:

Let $\phi: G \to K$ be a group homomorphism. Then the set $\phi^{-1}(\phi(a))$ is equal to the coset $a \mathrm{ker}(\phi)$. In other words, if $\phi(a)=b$ then $\phi^{-1}(b) = a \mathrm{ker}(\phi)$.

Your colleague is wrong, because there is no such $a$ for $b=9$.

For a concrete example of this mistake, the inclusion homomorphism $i: \mathbb{Z} \to \mathbb{R}$ has kernel with one element, but not every pre-image has one element: what's the pre-image of $\{0.5\}$?

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  • $\begingroup$ Mr @Randall, He who claiming I was wrong sent his solution. I eidted my post $\endgroup$ – se-hyuck yang Sep 18 '19 at 13:10
  • $\begingroup$ @se-hyuckyang I edited my answer to point out his mistake. $\endgroup$ – Randall Sep 18 '19 at 13:14
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    $\begingroup$ Terrific answer! $\endgroup$ – se-hyuck yang Sep 18 '19 at 13:23

Since you have identified $3^4=31$, have been given that $3$ is a primitive root and know that the order of the group is $20$ you know that $31$ will generate a group of order $5$ as the image of the homomorphism, consisting of the fourth powers of elements in the original group.

$9$ is not a fourth power, so isn't in the image, so the pre-image is empty as you have concluded.

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  • 1
    $\begingroup$ Excellent answer. $\endgroup$ – Randall Sep 18 '19 at 12:50

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