Suppose $G$ is a cyclic group with generator $a$ of order $n$. Let integer $k > 1$.

Prove: $a$ has a $k^{th}$ root iff $\gcd(n,k) = 1$.

What I have so far:

Suppose $\gcd(n,k) = 1$. Then there are integers $p,q$ such that $$np + kq = 1$$ So we have $$a^{np+kq} = a^{np}a^{kq} = (a^n)^p a^{kq} = ea^{kq} = a^{kq}$$ where $e \in G$ is the identity. Then $$a^{kq} = a \Rightarrow (a^q)^k = a$$ so there exists $b = a^q \in G$ such that $b^k = a$. This proves one part.

Now suppose there is a $b \in G$ such that $b^k = a$. How would we show that $\gcd(n,k) = 1$?



Suppose that $a= b^k$. We must have $b = a^r$ for some integer $r$ since $a$ generates $G$. Therefore $a = b^k = a^{rk}$. Hence $rk \equiv 1 \thinspace (mod \thinspace n)$ and so $gcd(k,n) = 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.