Define $a\ast b=a+b+5$, and show $(\Bbb Z,\ast)$ is a group.

Question

Let the set $\mathbb Z$ have the operation $*$ defined by $a * b = a + b + 5$ for all $a,b \in\mathbb Z$. Show this is a group.

I understand how to prove closure and associativity.

For identity, I have:

Suppose $-5$ is an identity. Then $a * -5 = a$. So $a + 5 + (-5) = a$. Thus $a = a$.

But I know I have to prove $-5 * a = a$ because I cannot assume commutativity. So $-5 * a = a$. Then $-5 + a + 5 = a$. Thus $a = a$. Hence $\mathbb Z$ has an identity.

Am I allowed to assume the last part of my proof although I do not know if this set is commutative?

For the inverse part, I am confused about where to start.

Answer 1

You have identified the identity element $e=-5$. To show that you have inverses, you need to prove that for every $a \in \mathbb{Z}$ there is some $b \in \mathbb{Z}$ such that $a*b = b*a = e$. Assuming you proved $*$ is commutative as mentioned in the comments, then it suffices to show there is a $b$ such that $a*b=e$.

I claim that $b=-a-10$ fits the bill. If $a \in \mathbb{Z}$, clearly $-a-10\in \mathbb{Z}$. Then $a*(-a-10) = a + (-a-10) + 5 = a - a - 10 + 5 = -5 = e$. Thus $-a-10$ is the inverse of $a$.

Answer 2

Here's a restatement of what's going on here: we have a function $\varphi(n)=n+5$ from $\mathbb{Z}$ to itself and want to check that $n*m=\varphi(n+m)$ is also a group structure. So, it may leave the waters less muddy if we state the problem more generally: given a set $S$ with a group structure $(*,e)$ and a function $\varphi:S\to S,$ when is the function $*':(s,t)\mapsto \varphi(s*t)$ also a group operation on $S$?

Well, a group needs an identity. An identity for our possible operation $*'$ would be $e'\in S$ such that $s*e'=e'*s=\varphi^{-1}(s)$ for every $s\in S$. So $\varphi$ had better be a bijection. Considering $s=e$ shows $e'$ could only be $\varphi^{-1}(e)$. We've also shown that $e'$ has to be in the center of $S$, that is, commute with all $s\in S$. To narrow down what $\varphi$ looks like even further, compute $\varphi(a)=\varphi(ae'^{-1}e')=\varphi(\varphi^{-1}(ae'^{-1}))=ae'^{-1}$. So $\varphi$ is just right multiplication by $e'^{-1}$ (equivalently, left multiplication, since $e'$ is in the center of $S$).

So we've seen it's necessary for $*'$ to be a group operation that $\varphi$ be multiplication by an element of the center of $S$. That's actually all we need: let's check. We already required $e'$ was an identity for $*'$ when we introduced it. How about inverses? Given $s$ we want $t$ so that $s*'t=s*t*e'^{-1}=e'=t*s*e'^{-1}=t*'s$. This yields $t=s^{-1}*e'*e'=e'*s^{-1}*e'$. These two expressions for $t$ are equal since $e'$ is in the center of $S$ and determine $t$ uniquely as the inverse of $s$ for $*'$. Finally for associativity we must have $$(r*'s)*'t=(r*s*e'^{-1})*'t=r*s*e'^{-1}*t*e'^{-1}=r*s*t*e'^{-1}*e'^{-1}=r*'(s*t*e'^{-1})=r*'(s*'t)$$

And we're saved yet again by the condition that $e'$ be central, which makes the third equality hold.

So we've seen that the functions $\varphi:S\to S$ such that $\varphi(s*t)$ is a group multiplication are exactly those given by multiplication by elements of the center of $S$.

Answer 3

$$ \text{Given e = -5, only show right inverse}\\ \text{let b is the inverse of a}\\ \Rightarrow\text{a*b=e}\\ \Rightarrow a + b + 5 = e = -5\\ \Rightarrow a + b = -5 - 5\\ \Rightarrow a + b + (-a) = -5-5+(-a)\quad(\because \text{I always can add an Integer in both side})\\ \Rightarrow b = -a -10\quad(\because \text{Addition is commutative and a+(-a)=0})\\ \Rightarrow \text{b is the right inverse of a} $$