Question

The set of all integral multiples of 5 is a subgroup of
a) The set of all rational numbers under multiplication
b) The set of all integers under multiplication
c) The set of all non-zero rational numbers under multiplication
d) The set of all integers under addition

Answer 1

Hint: A subgroup must satisfy all the properties of a group: Closure, Associativity, Identity, and Inverse. A sufficient condition is that it must satisfy the closure property. Note that “The set of all integral multiples of 5” is obtained when we multiply 5 with any number ‘n’.

Complete step-by-step solution:
Example: When we multiply 5 with $n=1$, we get 5. When we multiply 5 with $n=2$, we get 10 and so on.
A group is a set G together with an operation (.), which satisfies the following properties: If for $a\in G,$ $b\in G$ and $c\in G$, we have that
Closure: For $a\in G,$ $b\in G$ , we have $\text{a }\text{. b}\in G$. This says that for any two elements in G if we apply the operation (.) then the resulting element will also be in G
Associativity: For $a\in G,$ $b\in G$ and $c\in G$ , we have $\text{a }\text{.(b }\text{. c) = (a }\text{. b)}\text{.c}$ . This says that for any three elements in G, associativity between them holds when the operation (.) is applied.
Identity: For $a\in G,$ there exists an element $e\in G$ such that $\text{a }\text{. e = e }\text{. a = a}$ , where e is the identity element.
Inverse: For $a\in G,$ there exists an element ${{a}^{-1}}\text{ }\in \text{ G}$ such that $\text{a }\text{. }{{\text{a}}^{-1}}=\text{ }{{\text{a}}^{-1}}\text{ }\text{. a = e}$ . Here ${{a}^{-1}}$ is the inverse element.
A subgroup must satisfy all the properties of a group. The sufficient condition for a subgroup is that it must satisfy the closure condition.
“The set of all integral multiples of 5” is obtained when we multiply 5 with any number ‘n’. Let $H=\left\{ 5n,\text{ }where\text{ }n\in I \right\}$ . Here ‘I’ is a set with all ‘n’ values. Here H is a non-empty set with at least one element.
Let $a=5x$ and $b=5y$ . Now we apply an addition operator for a and b. So, $a+b=5x+5y=5(x+y)$ . Since $5x\in G$ and $5y\in G$ , $5(x+y)\in G$ . Hence, closure is satisfied.
Hence, the correct answer is
The set of all integral multiples of 5 is a subgroup of
d)The set of all integers under addition.

Note: We can also find the correct answer by applying all the properties of a group to the set of all integral multiples of 5. Example: let $a=5$, $b=10$ and $c=15$ . Now $a+b\text{ = }5+10=15=5(3)=5(1+2)\in G$ , $a+(b+c)=5+(10+15)=30\text{ and }(a+b)+c=(5+10)+15=30$ ,$a+e=5+0=5=0+5=e+a$ , where $e=0$ , $a+{{a}^{-1}}=5+(-5)=0=e$ and ${{a}^{-1}}+a=(-5)+5=0=e$ , where ${{a}^{-1}}=-5$
Hence, the set of all integral multiples of 5 satisfies all the properties of a group. This verifies our answer that the set of all integral multiples of 5 is a subgroup of the set of all integers under addition.