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 $
If you take two statements $P$ and $Q$ then saying "$P$ implies $Q$" or equivalently $P \implies Q$ means that if $P$ holds then $Q$ holds. For instance, the following are some factual implications.

1. "$x$ is a real number greater than zero" implies "$(-1)\cdot x$ is a real number less than zero." This is true because if the first statement holds then one may conclude the second holds as well.
2. "x is an odd number" implies "There exists a natural number $k$ such that $x = 2k + 1$." Again, from the first statement one can conclude the second.

Here is an example of an **incorrect** implication:

1. "x is a real number" implies "$x$ is less than zero." Here, one cannot conclude the second statement from the first, so it is incorrect to write this. What if $x = 0$?