How to show these groups are isomorphic through finding an isomorphism Show, by finding a suitable isomorphism, that the two groups $(\Bbb Z,+)$ and $(\Bbb Z,∗)$, where $a∗b=a+b−1$, are isomorphic.
The first step should be to take a closer look on the group $(\mathbb{Z},\ast)$. In particular what is the neutral element?
A group homomorphism has to send the neutral element to the neutral element. That is a good starting point.
Turns out the neutral element of $(\mathbb{Z},\ast)$ is $1$.
As $a\ast 1=a+1-1=a$ and $1\ast a=1+a-1=a$.
So $\varphi(0)=1$.
The easiest function that obeys this rule, is $x\mapsto x+1$.
So now we have to show that $\varphi(x)=x+1$ is an isomorphism.
That is a bijection (easiest shown by giving an inverse, but showing injective and surjective should be easy enough too), and that it is a homomorphism.
So we have to show that $\varphi(a+b)=\varphi(a)\ast\varphi(b)$.
I leave the rest to you.
The first step should be to take a closer look on the group $(\mathbb{Z},\ast)$. In particular what is the neutral element?
A group homomorphism has to send the neutral element to the neutral element. That is a good starting point.
Turns out the neutral element of $(\mathbb{Z},\ast)$ is $1$.
As $a\ast 1=a+1-1=a$ and $1\ast a=1+a-1=a$.
So $\varphi(0)=1$.
The easiest function that obeys this rule, is $x\mapsto x+1$.
So now we have to show that $\varphi(x)=x+1$ is an isomorphism.
That is a bijection (easiest shown by giving an inverse, but showing injective and surjective should be easy enough too), and that it is a homomorphism.
So we have to show that $\varphi(a+b)=\varphi(a)\ast\varphi(b)$.
I leave the rest to you.