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: 
A statement $A$ *implies* another statement $B$ (written as $A\Rightarrow B$), if from the truth of the former, it necessarily follows the truth of the latter.

**Example**. If I am in London, I am necessarily in England. So the statement "I am in London" *implies* the statement "I am in England".

On the other hand, if it is possible for $B$ to be true, while simultaneously $A$ is false, then $A$ is *not implying* $B$.

**Example**. If I am in England, I am not necessarily in London (I could be in Oxford). Hence, the statement "I am in England" does *not imply* the statement "I am in London".
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 $x = 0$?