# proving that $\mathbb{N}$ is a group under operation $Δ$

On the set $$\mathbb{N}$$ of natural numbers there is Commutative binary operation that is marked as $$*$$. it is known that $$\mathbb{N}$$ is a group under $$*$$, $$2$$ is an Identity element and that $$3$$ is an inverse of $$1$$ in the group. For $$\mathbb{N}$$ we define another binary operation called $$Δ$$ which is defined for $$a,b\in \mathbb{N}$$ as $$aΔb=(a*1)*b$$.
prove that $$\mathbb{N}$$ is a group under the operation $$Δ$$.

This is a question I have in homework, what should I do?. prove that $$\mathbb{N}$$ under $$Δ$$ has Closure, Associativity, Identity element, and Inverse element?
I think I found the identity element of $$Δ(3)$$ but I don't sure about how to prove it, I need to write something like "$$3Δb=(3*1)* b=e*b=b$$ and since $$\mathbb{N}$$ is associative under $$*$$, $$(a*1)* b=a*(1*b)$$ so $$aΔ3=(a*1)* 3=a*(1*3)=a*e=a$$"?.

• – Shaun Apr 30 '18 at 0:58

You have the plan correct. You need to prove the group axioms for your new operation. For the identity being $3$ you want to prove $a \Delta 3 = a = 3\Delta a$. Your proof is almost correct, but it should go $3\Delta a=(3*1)*a=2*a=a$. It is good to keep the operation signs when there is more than one operation around. To prove $a \Delta 3=a$ takes the commutativity and associativity of $*$. Closure is easy because you inherit it from $*$.
To get associativity you again go to the definition. Write out $(a \Delta b) \Delta c$. With the commutativity and associativity of $*$ you should be able to get that to match $a\Delta(b\Delta c)$.
Finally for inverses you will need to use that $a$ has an inverse under $*$, which you might as well call $a^{-1}$. You need to find in inverse for $a$ under $\Delta$, which will be based on $a^{-1}$