Question

Determine whether the set $G$is a group under the operation$*$.

1. role="math" localid="1653713749808" $G=\left\{2,4,6,8\right\}$IN role="math" localid="1653713775661" ${\mathrm{\mathbb{Z}}}_{10}$;role="math" localid="1653713806423" $a*b=ab$
2. role="math" localid="1653713859257" $G=\mathrm{\mathbb{Z}}$; role="math" localid="1653713871380" $a*b=a-b$
3. role="math" localid="1653713892785" $G=\left\{n\in \mathrm{\mathbb{Z}}\left|n\text{is odd}\right.\right\}$; role="math" localid="1653713914832" $a*b=a+b$
4. role="math" localid="1653713928251" $G=\left\{2^x\left|x\in \mathrm{\mathbb{Q}}\right.\right\}$; role="math" localid="1653713940941" $a*b=ab$

Short answer

1. $\left(G,*\right)$is group.
2. $\left(G,*\right)$ is not a group.
3. $\left(G,*\right)$ is not a group.
4. $\left(G,*\right)$is group.

Step-by-step solution

Solution to Part (a)

Consider the set, $G=\left\{2,4,6,8\right\}$.

(i) As all the numbers in the given set $G$are even numbers and the products of the two elements which lies in the given set$G$. Also, the elements in the given set $G$are not divisible by $10$.

Therefore, the product of two elements in $G$is taken modulo $10$, which is an even number from $1$to $9$, that will be the element from $G$.

(ii) As the multiplication in integers is commutative, we have,

$a*\left(b*c\right)=\left(a*b\right)*c$

(iii) As we can see that,role="math" localid="1653717091900" $$\begin{gathered} 2.6=6.2=12=2 \\ 4.6=6.4=24=4 \\ 6.6=6.6=36=6 \\ 8.6=6.8=48=8 \end{gathered}$$

From the above observation, it can be seen that 6 is the neutral element in role="math" localid="1653715562628" $\left(G,*\right)$.

(iv) Now,

$$\begin{gathered} 2.8=8.2=16=6 \\ 4.4=4.4=16=6 \\ 6.6=6.6=36=6 \\ 8.2=2.8=16=6 \end{gathered}$$

From the above observation, it can be seen that every element in role="math" localid="1653715596806" $\left(G,*\right)$has an inverse.

Hence, $\left(G,*\right)$is group.

Solution to Part (b)

As axiom 2 of the group states that if in a group $G$, $x,y$, and $z$ are three elements such that $xy=zy$, then $x=z$.

Therefore,

$\left(1*1\right)*2=0*2=-2\ne 2=1*\left(-1\right)=1*\left(1*1\right)$

Since it does not satisfy axiom $2$, therefore, $\left(G,*\right)$ is not a group.

Hence, $\left(G,*\right)$ is not a group.

Solution to Part (c) 

As the axiom of closure states that if in a group $G$, $x$, and $y$ are two elements, then $xy$ lies in the a group $G$.

Therefore,

$3*5=5\notin G$ but $3\in G$ and$5\in G$.

Since it does not satisfy the axiom of closure, therefore, $\left(G,*\right)$ is not a group.

Hence, $\left(G,*\right)$is not a group.

Solution to Part (d)

(i) Assume $a=2^x$, $x\in \mathrm{\mathbb{Q}}$and $b=2^y,y\in \mathrm{\mathbb{Q}}$as an arbitrary element in the group$G$.

Then, we have,

role="math" localid="1653717435103" $$\begin{gathered} a*b=2^x.2^y \\ \sin cex+y\in \mathrm{\mathbb{Q}} \\ =2^{(x+y)}\in G \end{gathered}$$

(ii) As $2^x·2^y=2^{x+y}$from the associativity, it is true in $\left(\mathrm{\mathbb{Q}},+\right)$.

(iii) Since $2^0=1$is the neutral element.

(iv) Assume $a=2^x$, $x\in \mathrm{\mathbb{Q}}$as an arbitrary element in group $G$.

Then we have,

$$\begin{gathered} a.2^{-x}=2^{-x}.a \\ =2^0 \\ =1 \end{gathered}$$

Where,$2^{-x}\in G$as $-x\in G$since $\left(\mathrm{\mathbb{Q}},+\right)$is a group.

Hence, $\left(G,*\right)$is a group.