# proving that N is a group under operation Δ

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

this is a question i have in homework, what should i do?. prove that N under Δ have 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 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 $a^{-1}$