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?