Cyclic group of order 8

Question

In a cyclic group of order 8 show that element has a cube root. So for some $a\in G$ there is an element $x \in G$ with $x^3=a.$

Also show in general that if $g=<a>$ is a cyclic group of order m and $(k,m)=1$ then each element in G has a $k$th root. What element will $a^k$ generate? Use this to express any element as a $k$th powers.

Where do I begin? For the first one is it just through closure essentially? And the second one Im stuck on. Where do I begin? I know that gcd between k & m is 1 so $kx+my=1$ with $x,y\in Z$. Thank you.

Answer 1

For the second part of your question, $a$ is a generator of $G$ ( i.e. $G= <a>$ ).

Since ($k$ , $m$) $=1$ , so $a^k$ is also a generator of $G$ ( $< a^k > $ = $ <a^{gcd(m,k)}>$ ).

Therefore, for $x\in G$ , $x= a^{sk}$ $= (a^s)^k$ , for some $s$.

Answer 2

Since $\;\text{gcd}\;(3,8)=1\;$ , if $\;G=\langle z\rangle\;$ then also $\;G=\langle z^3\rangle\;$ , and from here for each $\;a\in G\;$ there exists $\;k\in\Bbb Z\;$ so that we have

$$a=(z^3)^k= (z^k)^3$$

Take just $\;x=z^k\;$

Answer 3

Perhaps simpler approach consider $f:G\to G$ defind as $x\to x^3$.Now by using the fact that G is abelian and does not have any element of order $3$,show that $G$ is automorphism and hence done.Also i think abelian is sufficient condition!