If $(G, ∗, I)$ is a group and $a, b ∈ G.$ Show that $a^3 = I$ if and only if $(b^{−1} ∗ a ∗ b)^3 = I$. If $(G, ∗, I)$ is a group and $a, b ∈ G.$ Show that $a^3 = I$ if and only if 
$(b^{−1} ∗ a ∗ b)^3 = I$.
 A: Write out your multiplication, we have
$$
(b^{-1}ab)^3= (b^{-1}ab)(b^{-1}ab)(b^{-1}ab)=b^{-1}a(bb^{-1})a(bb^{-1})ab=b^{-1}a^3b
$$
But then if $a^3=1$, it is clear the above gives the identity. Similarly, if $(b^{-1}ab)^3=1$, by the above calculation, $b^{-1}a^3b=1$. But then multiply by $b$ on the left and $b^{-1}$ on the right and we get $a^3=bb^{-1}=1$.
In general, we call $b^{-1}ab$ conjugation (of $a$) by $b$. This is a common operation in Algebra. It is easy to see it's importance if you think about conjugation in the following way: let $b$ be a translation in the plane and $a$ be a rotation. Then $b^{-1}ab$ is the operation that moves some a 'something' somewhere, rotates it, then moves it back. Generally, this is what a conjugation is like: it 'does something', does something of interest, then 'puts it back'. It is a very natural operation that arises quite often. 
Your problem is actually a specific case of a more general result:
Given a group $G$, if $x \in G$ has finite order, then $y^{-1}xy$ has the same order as $x$.
Proof: This is the argument above using induction. Observe $(y^{-1}xy)^n=y^{-1}x^ny$. If $n=1$ this is trivial. Assume the result is true for $n=1,2,\cdots,k$. Then we have
$$
(y^{-1}xy)^{k+1}=(y^{-1}xy)^k (y^{-1}xy)=y^{-1}x^ky y^{-1}xy=y^{-1}x^nxy=y^{-1}x^{k+1}y
$$
so that the claim holds by induction. But then simply observe
$$
1=(y^{-1}xy)^m \text{ if and only if } y^{-1}x^my=1 \text{ if and only if }x^m=1
$$
so that they must have the same order.
A: In general if $x$ and $y$ are conjugate then $x$ and $y$ have the same order.
Proof: Let $g\in G$ be the element such that $x=gyg^{-1}$ and $|x|, |y|$ denote the order of $x$ and $y$. We have $e=x^{|x|}=(gyg^{-1})(gyg^{-1})…(gyg^{-1})=gy^{|x|}g^{-1}\Rightarrow y^{|x|}=g^{-1}g=e \Rightarrow |y|\leq |x|$ (and $|y|$ divides $|x|$).
On the other hand $y=g^{-1}xg$, by similar arguments $|x|\leq |y|$, therefore $|x|=|y|$.
