proof about the order of subgroup

Here is a simple question and I have the answer for it but I have trouble understanding the answer.

Question.

Suppose that G is a group with |G|=27. Prove that G contains a subgroup H satisfying |H|=3.

Answer.

Clear so far: applying Lagrange theorem, possible orders of subgroups are 1, 3, 9, 27. Possible orders of elements are the same (1, 3, 9, 27). There can only be one element of order 1, namely the identity element, so all other elements msut have order 3, 9 or 27.

*Then it gets muddy. *

"Let x be any non-identity element of the group. If x has order 3, we can take H=<x>." *- then |H|=3?? because H={x^1, x^2, x^3}?*

"If x has order 9, we can take H=<x^3> and this will satisfy |H|=3 since x^3 will have order 3." *- is that because H={x^1, x^2, x^3} again? so that we can multiply 9 by 1, 2 or 3 to get to a number less than 27.*

"If x has order 27, we may take H=<x^9>" *- same logic as in the previous statement, I guess.*

Thanks for your comments!