I'm stuck at the final stage of a problem and could use some help.
Let the binary operation * on the naturals N be an Abelian group (i.e also complies with ab = ba for every a and b of N) It is known that 2 is the identity element and that 3 is inverse to 1.
N has a new binary operation Δ defined on it as such: aΔb = (a * 1)*b for every a,b ∈ N prove that N with Δ is a group.
I have been successful in showing closure, associativity, and the existence of the identity element which turns out to be e = 3.
I am stuck at the last requirement of proving the existence of an inverse element for each element.
Basically I need to prove that for every a ∈ N there is a b ∈ N such that a Δ b = 3 so then (a * 1)* b = 3 but I have no idea how to isolate b
I appreciate your help in advance.
The element $(a*1)$ has an inverse under $*$; let's call it $c$. Then $b=c*3$ works.
This is a special case of a general set-up. Take a group $G$ (not necessarily Abelian) with operation $*$. Fix $g_0\in G$ and define $\circ$ on $G$ by $$a\circ b=a*g_0*b.$$ Then $G$ is also a group under the operation $\circ$.