I've got G, a cyclic group, and x and y two element from G. Let's suppose that o(x)|o(y). Can we make sure that x is a power of y? (I mean, can we make sure that we can write x=y^(something) ?) I need to prove this, it's not enough with an example.
I know that as G is a cyclic group, it is generated by only one element: G = < g > and the elements from G are the powers of g:
G = < g > = {1, g, g^2, g^3, . . . ,g^n} being |G|=n < infinite.
So if I call g=y, so that we have G = < y > = {1, y, y^2, y^3, . . . ,y^n}, I can say that if any of those elements is x, I can write x as a power of y.
But if y is for exaple y=g^2 or y=g^3… can I always write x as a power of y? Taking into account that o(x)|o(y)
Best Answer
Let $G$ be a cyclic group and $y,x\in G$ be elements of the group such that $o(x)|o(y)$. First thing to observe is that $H=\langle y \rangle \subset G$ and $K=\langle x \rangle \subset G$, and that both $H$ and $K$ are the unique subgroups of $G$ with orders $o(y)$ and $o(x)$ respectively. Since $H$ is a cyclic group itself, and $o(x)|o(y)$ there's a subgroup $L$ of $H$ with order $o(x)$, which is also a cyclic subgroup of $G$ of order $o(x)$. This tells us that $L=K$, and so $x\in \langle y \rangle$ which in turn implies that $y^k=x$ for some $k\in \mathbb{Z}$