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

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.


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 $\{0.5\}$?

Tags:

Number Theory

Abstract Algebra

Group Theory

Group Homomorphism

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