# If $n$ the order of a group and $\gcd(k,n) = 1$ then there exists an element in the group

Let $G$ be a finite group of order $n$ and $k$ a positive integer relatively prime to $n$. Prove that for each $g$ in $G$ there is an $h$ in $G$ such that $h^k = g$.

-
@John, please do not do this again to any ofyour questions. –  Mariano Suárez-Alvarez Dec 12 '12 at 7:15
If you submit a full confession to cheating in writing along with the take home final, I will take that into consideration when determining the scope of the penalty. Enclose the URL to either this page or your edit history. –  user52748 Dec 12 '12 at 8:08
Further evidence there should be a "temporary lock" feature for such suspicious questions, where answers cannot be submitted in the first 3 or so hours after posting. –  user641 Dec 12 '12 at 16:08
@Steve: Absolutely not. That would affect a great many entirely legitimate questions. –  Brian M. Scott Dec 12 '12 at 23:34
@Steve: I doubt that the questions will suffer, but as one who answers quite a few questions, I’d be somewhat inconvenienced. The real point, however, is that there’s no reason for such an embargo apart from the rather insulting presumption that questions that appear simply to be quoted are attempts to cheat. –  Brian M. Scott Dec 13 '12 at 7:05

Since $(n,k)= 1$, for some $a,b$, $na+kb=1$. Then $h=g^b$ works for any $g$.

-

Hint $\rm\,\ g^N = 1\: \Rightarrow\: g^J = g^{J\ mod\ N}.\:$ Your case is $\rm\: J = \dfrac{1}K.\,$ It exists mod $\rm N$ by $\rm\:(K,N)=1$ and Bezout.

Remark $\$ The idea becomes clearer if we use additive notation for the cyclic group $\rm\langle g\rangle,\:$ viz.

$\rm\qquad n\cdot g = 0\:\Rightarrow\: j\cdot g = (j\ mod\ n)\cdot g,\$ so the "scalar" multipliers may be considered mod $\rm\:n,$

since $\rm\ j\ mod\ n = r\:\Rightarrow\: j = kn\!+\!r\:\Rightarrow\: j\cdot g = (kn\!+\!r)\cdot g = k\cdot (n\cdot g) + r\cdot g = r\cdot g\:$ by $\rm\:n\cdot g = 0.$

Here, in this additive linear form, the problem of taking a $\rm\:k$'th root, i.e. raising to exponent $\rm\:1/k,\:$ translates into taking a $\rm\:k$'th part, i.e. multiplying by the scalar $\rm\:1/k,\:$ for which there exists a unique solution by Bezout's identity since $\rm\:(k,n) = 1.\:$ The implicit coefficient and linear structure here is a generalization of the notion of a vector space where one allows the "coefficients" to come from any ring, not only fields. It is known as a module and plays a fundamental role in number theory and algebra by abstracting ubiquitous linear structure in ring theory. See this answer for some nice examples and further discussion (which may prove enlightening even to more advanced readers).

-

Let $G$ be a group and $k$ relatively prime to $|G|$ then $\{g^k \in G | g \in G \} = G$ because $a^k = b^k \implies a=b$ since $(a \cdot b^{-1})^k = 1$ iff $a\cdot b^{-1} = 1$.

-
Is this homework? If so, you should tag it as such, and also let everybody know what you have done, or where you are stuck. –  Martin Argerami Dec 12 '12 at 21:14
This works only if $a^kb^{-k}=(ab^{-1})^k$, i.e. if $G$ is abelian. –  Stefan Hamcke Dec 14 '12 at 13:15