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 $h^k = g$.
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Since $(n,k)= 1$, for some $a,b$, $na+kb=1$. Then $h=g^b$ works for any $g$. |
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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). |
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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 $a\cdot b^{-1} = 1$. |
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