Question: Let $G$ be a group. Suppose $x$ has finite order $n$. Show that $x^k=1$ if and only if $k$ is a multiple of $n$.

Intuitively, I think I understand this, but I am not sure how to prove it. Any help is appreciated.

Answer: 
In most mathematical proofs, $P\implies Q \space \equiv\space \neg[P\land \neg Q]$.

There is also no causal relationship assumed between the antecedent $P$ and consequent $Q$, i.e. we do not assume that $P$ causes $Q$.

From the above definition we have:

1. If we assume $P$ is true and can prove that $Q$ is also true, then
we can infer that $P\implies Q$.
2. If $P$ is true and $[P\implies Q]$ is true, then can infer that $Q$
is true.
3. $P \implies Q\space$ follows from any of the following:

	* $\neg Q \implies \neg P$
	* $\neg P \space \space$
	* $Q \space \space $
Fundamentally, mathematics is a guide to reducing the unknown. When we have a statement "If $P$ then $Q$", that means if we somehow already know that $P$ is true, then we now can be assured that $Q$ is true. Each theorem that proves an implication allows us to expand our knowledge of true facts by using chains of implications. We start with known supposed true facts and, using chains of implications, we can conclude many other true facts on the basis of the suppositions. And finally arriving at what we want to prove to be true also. The Achilles' heel of this method is that we are dependant of the starting true facts, and if any one of them turns out to be false, then the results of our implications are now not assured to be true. Another weakness is that we could make several kinds of errors in applying the implications because "to err is human" and "a chain is as strong as its weakest link".