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 math, the fact that a statement $A$ implies a statement $B$ is written this way: $A \implies B$

The meaning of $A \implies B$ is defined by this truth table: 

$$
 \begin{matrix}
 A & B & | & A \implies B\\
 T & T & | &T\\ 
 T & F & | &F\\
 F & T & | &T\\
 F & F & | &T\\
 \end{matrix}
$$

That means that $A \implies $B is false only when $A$ is true and $B$ is false. In the other cases $A \implies $B is true.

So let's say I have a theorem that states "If $n$ is a multiple of $6$ then it must be a multiple of $2$" ($n$ multiple of $6 \implies n$ multiple of $2$). For this theorem to be true, if you encounter a multiple of $6$ then it must be also a multiple of $2$ because if not the implication would be false. But if you encounter a number that is not a multiple of $6$ then it can be a multiple of $2$ or not without invalidating the theorem, because the theorem says nothing about numbers that are not multiple of $6$.

You can also see what the implication means by looking at sets: 
[![enter image description here](https://i.stack.imgur.com/U9naZ.png)](https://i.stack.imgur.com/U9naZ.png)
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 $x = 0$?