Let $\ast$ defined on $\mathbb{Z}$ by $a\ast b=a+b-1$ $\forall a,b \in \mathbb{Z}$. Show that $(\mathbb{N},\ast)$ is not a subgroup of $(\mathbb{Z},\ast)$,where $\mathbb{N}$ is the set of natural numbers.
I showed it but I want to know whether it is correct or not?
Firstly I found inverse of element $b \in \mathbb{N}$ and then I found $ab^{-1}$ which is $a-b+1$ but it isn't $\in \mathbb{N}$ then $ab^{-1}$ isn't $\in \mathbb{N}$ so then I think I proved it?
If I am wrong please correct me.
On
You're on the right track by proving that the inverse operation is not well defined, or not internal. It can be easier to take an element in particular that is in $\mathbb{N}$, but which doesn't have an inverse in $\mathbb{N}$. Do you know about such an element?
On
To disprove that $(\mathbb{N}, *)$ is a subgroup of $(\mathbb{Z}, *)$ you need to give a counterexample to one of the axioms. For example, every element in the group needs an inverse that is also in the group.
Let $3=a \in \mathbb{N}$. Now we have to look for the number $a^{-1} \in \mathbb{N}$ and $a*a^{-1} = 1$. Now we know that $a*a^{-1} = a + a^{-1} -1 = 1 \implies a^{-1} = -1$. This is a contradiction to our assumption that $a^{-1} \in \mathbb{N}$. Therefore, $(\mathbb{N}, *)$ cannot be a subgroup of $(\mathbb{Z}, *)$.
Well, it is hard to follow your line of reasoning. Is $a-b+1$ not in $\mathbb{N}$ for all choices of $a$ and $b$ or only some? To establish that a subset of a group is not a subgroup of that group, what is it exactly that you need to show?
To show that $(\mathbb{N}, *)$ is not a subgroup it suffices to show that there exists an $a \in (\mathbb{N}, *)$ such that $a^{-1} \not \in (\mathbb{N}, *)$. And indeed $a^{-1} = 2-a$, which is not in $(\mathbb{N}, *)$ if $a \ge 2$ say $a=3.$