# What did I wrong in this question?(Group homomorphism and primitive roots)

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.

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 $$\{0.5\}$$?

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

• Excellent answer. – Randall Sep 18 '19 at 12:50