Let $G$ be a group and $u \in G$ be a fixed element. By the following, prove that $(G,\bullet)$ is a group. Let $G$ be a group and $u \in G$ be a fixed element. Define the operation $\bullet$ on G as $\forall a,b \in G, a \bullet b=au^{-1}b.$ Prove that $(G,\bullet)$ is a group.
So, I know that in order for it to be a Group, it must have closure, associativity, identity, and inverse. 
Closure is obvious. 
Associativity: Let $a ,b ,c \in G$, 
then $(a \bullet b) \bullet c=(au^{-1}b \bullet c)$
$au^{-1}bu^{-1}c=au^{-1}(bu^{-1}c)=au^{-1}(b \bullet c)=a \bullet (b \bullet c)$
Identity: If $a,b \in G$ and $e$ is the identity in G, then $a \bullet u=au^{-1}u=ae=a$ and $u \bullet a=uu^{-1}a=ea=a$
Inverse: Now, I believe that the above is correct but Im a little unsure about the inverse. 
 A: Arthur and Squirtle's answers are 100% correct, but just in case you want something more conceptual:
This is an example of transport of structure. In detail, if $(G,\cdot)$ is a group, and $f$ is a bijection from $G$ onto a set $X$, then we can define an operation $\bullet$ on $X$ by setting
$$ f(g) \bullet f(h) = f(g \cdot h) $$
for all $g,h \in G$. Note that every element of $X$ has the form $f(g)$ for some $g \in G$, so this operation is well defined on all of $X$.
All of the group properties for $(X,\bullet)$ now follow straight from the properties for $(G,\cdot)$. For instance, $(X,\bullet)$ is associative because
$$
( f(g) \bullet f(h) ) \bullet f(k) = f(g \cdot h) \bullet f(k)\\
= f((g \cdot h) \cdot k)\\
= f(g \cdot (h \cdot k))\\
= f(g) \bullet f(h \cdot k)\\
= f(g) \bullet (f(h) \bullet f(k)).\\
$$
I'll leave you to prove that the identity of $(X,\bullet)$ is $f(\text{id}_{G})$, and the inverse of $f(g)$ is $f(g^{-1})$.
Your example is then $X = G$ (as sets) and $f(g) = ug$, since
$$f(ab) = uab = (ua)u^{-1}(ub) = f(a)\bullet f(b).$$
Now write $g^{-1(\bullet)}$ for the inverse of $g$ with respect to $\bullet$, and assume we've proved that $f(e) = u$ is the identity of $(G,\bullet)$. If $g \in G$, then $g = f(u^{-1}g).$ So the inverse property becomes
$$ g^{-1(\bullet)} = f(u^{-1}g)^{-1(\bullet)} = f((u^{-1}g)^{-1}) = f(g^{-1}u) = ug^{-1}u, $$
as in Arthur and Squirtle's answers.
A: As for the inverse, given an $a \in G$, we want to find some $b\in G$ such that $a\bullet b = u$. We have
$$
a\bullet b = u\\
au^{-1}b = u\\
ua^{-1}au^{-1}b = ua^{-1}u\\
b = ua^{-1}u
$$
So the inverse of $a$ in $(G, \bullet)$ is $ua^{-1}u$ (it remains, of course, to show that it is a left inverse as well).
A: What you have above is correct so far.  The answer to your last question is a little funny, but The inverse of $b$ is $ub^{-1}u$ where the inverse here is in the first group structure.  Explicitly, $(ub^{-1}u)\bullet b=(ub^{-1}u)u^{-1}b=u$; similarly, $b\bullet (ub^{-1}u)=bu^{-1}ub^{-1}u=u$
