# Show that $\left( Z,* \right)$ is an infinite abelian group, where ‘*’ is defined as:

$a*b=a+b+2$ , and $Z$ is the set of all integers.

Last updated date: 25th Mar 2023

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Answer

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Hint: First check all the four fundamental properties of a group. Which are closure property, associativity, the identity property, and the inverse property. Then check if the operation is commutative or not.

Complete step-by-step answer:

We can say that $\left( Z,* \right)$ is a group if it satisfies the following four properties:

Closure property: If we take any two elements from the set, which is $Z$ here, then the composition of those two elements also should belong from that set. That is, $a,b\in Z\Rightarrow a*b\in Z$.

Associativity: The defined operation should be associative. That is, $\left( a*b \right)*c=a*\left( b*c \right)$.

The identity property: There exist an element, say $e$ , such that: $a*e=e*a$ , for all $a\in Z$.

The inverse property: For each $a\in Z$, there exist an element ${{a}^{-1}}$ in $Z$ such that: $a*{{a}^{-1}}={{a}^{-1}}*a=e$

$\left( Z,* \right)$ is said to be an abelian group if for all $a,b\in Z$, $a*b=b*a$.

Now let us check the first four properties one by one.

Let, $a,b\in Z$.

$a*b=a+b+2$ also belongs to the set of integers. Because if a and b are integers then addition of this two integers will give us an integer and if we add 2 with them it will again give us an integer. Therefore the closure property holds.

Let $a,b,c\in Z$

$\begin{align}

& a*b=a+b+2 \\

& \left( a*b \right)*c=\left( a*b \right)+c+2=a+b+2+c+2=a+b+c+4 \\

\end{align}$

$\begin{align}

& b*c=b+c+2 \\

& a*\left( b*c \right)=a+\left( b*c \right)+2=a+b+c+2+2=a+b+c+4 \\

\end{align}$

Therefore:

$a*\left( b*c \right)=\left( a*b \right)*c$, so the associativity holds.

Let us assume that for every element that belongs to the set, there exist an element e such that:

$\begin{align}

& a*e=a \\

& \Rightarrow a+e+2=a \\

& \Rightarrow e+2=a-a \\

& \Rightarrow e+2=0 \\

& \Rightarrow e=-2 \\

\end{align}$

So the identity element exists and that is -2.

Let us assume that for each element $a\in Z$ there exist an element ${{a}^{-1}}$ such that:

$\begin{align}

& a*{{a}^{-1}}=e \\

& \Rightarrow a+{{a}^{-1}}+2=-2 \\

& \Rightarrow a+{{a}^{-1}}=-2-2 \\

& \Rightarrow {{a}^{-1}}=-4-a \\

\end{align}$

So for each element the inverse element exists.

Therefore $\left( Z,* \right)$ satisfies all the four properties. Hence $\left( Z,* \right)$ is a group.

Now we have to check if it is abelian or not.

$a*b=a+b+2=b+a+2=b*a$

Therefore for all $a,b\in Z$

$a*b=b*a$

Hence $\left( Z,* \right)$ is an abelian group.

We know that there are infinite numbers of elements in the set of all integers.

Hence, $\left( Z,* \right)$ is an infinite abelian group.

Note: We generally make mistakes while checking the four properties. To be a group all the four properties have to be satisfied. Then we need to check the abelian property separately. A group is not always abelian.

Complete step-by-step answer:

We can say that $\left( Z,* \right)$ is a group if it satisfies the following four properties:

Closure property: If we take any two elements from the set, which is $Z$ here, then the composition of those two elements also should belong from that set. That is, $a,b\in Z\Rightarrow a*b\in Z$.

Associativity: The defined operation should be associative. That is, $\left( a*b \right)*c=a*\left( b*c \right)$.

The identity property: There exist an element, say $e$ , such that: $a*e=e*a$ , for all $a\in Z$.

The inverse property: For each $a\in Z$, there exist an element ${{a}^{-1}}$ in $Z$ such that: $a*{{a}^{-1}}={{a}^{-1}}*a=e$

$\left( Z,* \right)$ is said to be an abelian group if for all $a,b\in Z$, $a*b=b*a$.

Now let us check the first four properties one by one.

Let, $a,b\in Z$.

$a*b=a+b+2$ also belongs to the set of integers. Because if a and b are integers then addition of this two integers will give us an integer and if we add 2 with them it will again give us an integer. Therefore the closure property holds.

Let $a,b,c\in Z$

$\begin{align}

& a*b=a+b+2 \\

& \left( a*b \right)*c=\left( a*b \right)+c+2=a+b+2+c+2=a+b+c+4 \\

\end{align}$

$\begin{align}

& b*c=b+c+2 \\

& a*\left( b*c \right)=a+\left( b*c \right)+2=a+b+c+2+2=a+b+c+4 \\

\end{align}$

Therefore:

$a*\left( b*c \right)=\left( a*b \right)*c$, so the associativity holds.

Let us assume that for every element that belongs to the set, there exist an element e such that:

$\begin{align}

& a*e=a \\

& \Rightarrow a+e+2=a \\

& \Rightarrow e+2=a-a \\

& \Rightarrow e+2=0 \\

& \Rightarrow e=-2 \\

\end{align}$

So the identity element exists and that is -2.

Let us assume that for each element $a\in Z$ there exist an element ${{a}^{-1}}$ such that:

$\begin{align}

& a*{{a}^{-1}}=e \\

& \Rightarrow a+{{a}^{-1}}+2=-2 \\

& \Rightarrow a+{{a}^{-1}}=-2-2 \\

& \Rightarrow {{a}^{-1}}=-4-a \\

\end{align}$

So for each element the inverse element exists.

Therefore $\left( Z,* \right)$ satisfies all the four properties. Hence $\left( Z,* \right)$ is a group.

Now we have to check if it is abelian or not.

$a*b=a+b+2=b+a+2=b*a$

Therefore for all $a,b\in Z$

$a*b=b*a$

Hence $\left( Z,* \right)$ is an abelian group.

We know that there are infinite numbers of elements in the set of all integers.

Hence, $\left( Z,* \right)$ is an infinite abelian group.

Note: We generally make mistakes while checking the four properties. To be a group all the four properties have to be satisfied. Then we need to check the abelian property separately. A group is not always abelian.

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