What does $x^n=y^n$ imply for $x=\pm y$ in a group? I am wondering if the following statement is true:

If $G$ is a group, then for all $x,y \in G, x^n=y^n$ implies $x=y$ or $x=-y$.

I have tried using an induction argument and it might work by using casework on whether $n$ is odd or even, but I'm not sure.

EDIT: If $G$ is a group, then if $x,y \in G$ and $ x^n=y^n$, then what information does this give us about $x$ and $y$?
 A: No. If the group has order $n$ then $x^n = e$ (the group identity) for every $x$.
A: No. Even under the assumption of an abelian group, it merely implies that $x=uy$ for some $u$ with $u^n=1$.
A: Here is another example of what can happen. Suppose $x^2=e$ then we have $x=x^3=x^5=x^7=\dots = x^{2n+1}=\dots$
Now suppose that there is an element with $y^{2n+1}=x$ (we can impose this by specifying the group by generators and relations), then $y^{2n+1}=x^{2n+1}$.
Here $x$ has order $2$ and $y$ can have arbitrarily large order.
A: In general, you have no good information. For example, if $G$ has order $n$, then $g^n = e = e^n$ for all $g \in G$. You can find the same kind of problem in infinite groups too. What's going wrong? You cannot control the orders!
Define $$\phi_n : G \to G : g \mapsto g^n$$ 
Is it a homomorphism? Well, in general no: there is no reason to $(gh)^n = g^h h^n$.
But if $G$ is abelian it's really easy to see than $\phi_n$ is a homomophism. 
In this case, if $g^n = h^n$, then $(gh^{-1})^n =e$ and $ord(g h^{-1}) | n$. If $G$ is finite, we have $ord(g h^{-1}) = lcm(ord(g), ord(h))$ so it's divide $n$. Well... ok, it's a result, but... let's try to add something.
Let $G$ be a finite group and $m = |G|$, Lagrange's theorem tells us that $ord(g) | m$ , for all $g \in G$. If $\gcd(n, m) = 1$ and $g \neq e$, then $ord(g)$ is not a divisor of $n$, so $\phi_n(g) = g^n \neq e$ and $\ker \phi_n = \{e\}$, i.e., $\phi_n$ is one-to one, then 
$$g^n = \phi_n(g) = \phi_n(h) = h^n \quad \Rightarrow \quad g = h $$
Since $G$ is finite, it's in fact a bijection.
So, if $G$ is a abelian finite group with order $m$ and $n \in \mathbb{Z}$ is such that $\gcd(m,n)=1$, than $\phi_n$ is a automorphism of $G$ and $g^n = h^n$ implies $g=h$ (we have more: all elements have one, and only one, n-th root).
Note: I used $G$ abelian to define this homomorphism and $\gcd(m,n)=1$ to control orders.
A: If $x^n = y^n$, then $x^n y^{-n} = 1$, and hence $(xy^{-1})^n = 1$. In other words, $x$ times the inverse of $y$ is idempotent.
