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 $aΔ3=(a*1)* 3=a*(1*3)=a*e=a$"?.


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 $a^{-1}$

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