Prove $G$ is a group (unusual star operation). I've come across another rather simple question, but I'm having trouble with $G$'s defined operation.
Let $(G,\circ)$ be a group and $x_0 \in G$. Consider the operation $*$ defined in $G$ by $a * b = a \circ x_0 \circ b, a,b\in G$.
a) Knowing $*$ is associative, show $(G,*)$ is a group.
Well, so far I've got this:


*

*Since $x_0 \in G$, G is non-empty.

*Closed under multiplication: Let $a,b \in G$. $a * b = a \circ x_0 \circ b$. Since all $a, b,$ and $x_0$ are in $G$, then its product is in $G$ as well.

*Identity: $a * 1_G = a$ (I don't think I've got this one quite right, because I'm messing with inverses, and that's probably not required)


$a * 1_G = a == a \circ x_0 \circ 1_G = a == a^-1 \circ a \circ x_0^-1 \circ 1_G = a^-1 \circ a == x_0 \circ 1_G = 1_G == x_0 = 1_G$ 
I'm left with proving the inverse element, which should probably be done before the identity. Or maybe I should re-think my identity strategy. I'm a bit lost here. Any help? Thanks.
EDIT: What I mean is: Clearly, we need the concept of inverses to work out the group's identity element. Thing is, I proved it before proving the very existence of the inverse element. Is this not a problem?
 A: The problem you are having with the identity element is that $1_{(G,\circ)}$ is not the identity element in $(G,\ast)$, $x_0^{-1}$(with inverse taken with respect to $(G,\circ)$) is since $$a\ast x_0^{-1}=a\circ x_0\circ x_0^{-1}=a\circ 1_{(G,\circ)}=a$$ and $x_0^{-1}\ast a=a$ follows almost identically.
With this in mind, can you figure out what the inverses are?
Also, you mention that you think you should find inverses before finding the identity. This is ill-advised because inverses are defined in terms of the identity, so without identity, the concept of inverses doesn't mean anything.
Edit: to show that $(G,\ast)$ is closed under $\ast$, we can use the fact that $(G,\circ)$ is closed under $\circ$. Explicitly, for any $a,b\in G$, $$a\ast b=a\circ x_0 \circ b\in G$$ since $\circ$ is a binary operation on $G$.
A: As you can see, $a *1_G=a\iff x_0 \circ 1_{G_2}=1_{G_1}$ where $G_1=(G,\circ)$ and $G_2=(G,*)$. Hence you get $1_{G_2}={x_0^{-1}}_{G_1}$. That would mean the inverse of $x_0$ under the operation "$\circ$" is the identity element in $G_2$, which is $G$ under $*$ operation.
As for finding the inverse, for a given $a\in G$, we want to find $b$ such that $a*b=1_{G_2}={x_0^{-1}}_{G_1}$. So, $a*b=a \circ x_0 \circ b={x_0^{-1}}_{G_1}$.
Noticing that the last line is simply an expression taking place in $G_1$, under that operation "$\circ$", we can proceed, noting ${y}^{-1}_{G_1}={y^{-1}}$, assuming we are already making steps in $G_1$: $a \circ x_0 \circ b={x_0^{-1}}_{G_1}\Rightarrow  x_0 \circ b=a^{-1}{x_0^{-1}}\Rightarrow  b={x_0^{-1}}a^{-1}{x_0^{-1}}$, which, getting back to $G_2$, will be denoted: $a^{-1}_{G_2}=b=({x_0^{-1}}_{G_1})(a^{-1}_{G_1})({x_0^{-1}}_{G_1})$. It is very important to notice that $a^{-1}_{G_2}$ and $a^{-1}_{G_1}$ are not necessarily the same.
A: As already noted, Identity element in $(G,*)$ is $x_0^{-1}$.
The inverse of an element $a \in (G,*)$ is $x_0^{-1} \circ a^{-1}\circ x_0^{-1}$:
$$
a \ast (x_0^{-1} \circ a^{-1}\circ x_0^{-1})=\\
= a \circ x_0 \circ (x_0^{-1} \circ a^{-1}\circ x_0^{-1})=\\
x_0^{-1}=(x_0^{-1} \circ a^{-1}\circ x_0^{-1}) \ast a.
$$
A: I don't think you have anything yet.  
1) You say G is non-empty because $x_0 \in G$.  Well yes, but you know that as a set $<G, \circ> = <G, *>$ so $<G, *>$ is non-empty.
What we have to prove instead is that ever $g \in G$ can be written as $h \circ x_0 \circ i$.  Which it can as $g = g \circ x \circ c_0^{-1}$
2) Before you could make the claim you had to show all $a = a' \circ x_0 \circ a~$, which we did above.  So as $a,b$ in $<G, \circ$> then $a,b$ in $<G, *>$ so $a \circ x_0 \circ b \in <G, \circ>$ which as a set is $<G, *>$.  But we did have to show that all a $in G$ were expressible as $a' \circ x_0 \circ a~$ first
3)  $a * I_G = a \circ x_0 \circ I_G = a \circ x_0 \ne a$
So that won't do it!  we need to find $e$ such that $c \circ x_0 e = c = e \circ x_0$ for all $c \in G$.
If $c \circ x_0  e = c$ then $e = x_0^{-1} \circ c^{-1} \circ c = x_0^{-1}$.
Meanwhile $c*x^{-1} = c \circ x_0 \circ x_0^{-1} = c  = x_0^{-1} \circ x_0 \circ c $.
So $x^{-1}$ is the indentity element of $<G , *>$
3) Finally inverse: For $a  \in <G, *>$ what $a \circ x_0 \circ b = b \circ x_0 \circ a = x^{-1}$?
That would be $x_0^{-1} \circ a^{-1} \circ x_0^{-1}$.  Try it. 
