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Question: In three element group \[\left\\{ e,a,b \right\\}\] where e is the identity, \( {{a}^{5}}{{b}^{4}} \...

In three element group \left\\{ e,a,b \right\\} where e is the identity, a5b4{{a}^{5}}{{b}^{4}} is equal to
A.a
B.e
C.ab
D.b

Explanation

Solution

Hint: For solving this question, we consider a given group as G whose one of the elements is an identity. So, we write all the possibilities, that is, a⋅a=e and b⋅b=e or a⋅b=b⋅a=ea\cdot a=e\text{ and b}\cdot \text{b=e or }a\cdot b=b\cdot a=e . Now after writing all the possibilities we have to obtain the common value by considering each individual case.

Complete step-by-step answer:
In mathematics, an identity element is a special type of element of a set with respect to a binary operation on that set, which leaves any element of the set unchanged when combined with it. An element of set S is called the left identity if n⋅e=nn\cdot e=n for all n in S, and a right identity if e⋅n=ne\cdot n=n for all n in S.
According to the problem statement, we are given a group G = {e, a, b}, whose element e is an identity. By using the other two elements, we have two possible cases: a⋅a=e and b⋅b=e or a⋅b=b⋅a=ea\cdot a=e\text{ and b}\cdot \text{b=e or }a\cdot b=b\cdot a=e .
Expanding the product and applying identity element in case (1), we get
a⋅a=e and b⋅b=e a5b4=a⋅a⋅a⋅a⋅a⋅b⋅b⋅b⋅b =e⋅a⋅a⋅a⋅b⋅b⋅e \begin{aligned} & a\cdot a=e\text{ and b}\cdot \text{b=e} \\\ & {{a}^{5}}{{b}^{4}}=a\cdot a\cdot a\cdot a\cdot a\cdot b\cdot b\cdot b\cdot b \\\ & =e\cdot a\cdot a\cdot a\cdot b\cdot b\cdot e \\\ \end{aligned}
By using the left- identity n⋅e=nn\cdot e=n and right-identity e⋅n=ne\cdot n=n , we can solve the product as
=a⋅a⋅a⋅b⋅b =e⋅a⋅e =a⋅e =a \begin{aligned} & =a\cdot a\cdot a\cdot b\cdot b \\\ & =e\cdot a\cdot e \\\ & =a\cdot e \\\ & =a \\\ \end{aligned}
Expanding the product and applying identity element in case (2), we get
a⋅b=b⋅a=e a5b4=a⋅a⋅a⋅a⋅a⋅b⋅b⋅b⋅b =a⋅a⋅a⋅a⋅e⋅b⋅b⋅b \begin{aligned} & a\cdot b=b\cdot a=e \\\ & {{a}^{5}}{{b}^{4}}=a\cdot a\cdot a\cdot a\cdot a\cdot b\cdot b\cdot b\cdot b \\\ & =a\cdot a\cdot a\cdot a\cdot e\cdot b\cdot b\cdot b \\\ \end{aligned}
By using the left- identity n⋅e=nn\cdot e=n and right-identity e⋅n=ne\cdot n=n , we can solve the product as
=a⋅a⋅a⋅e⋅b⋅b =a⋅a⋅e⋅b =a⋅e =a \begin{aligned} & =a\cdot a\cdot a\cdot e\cdot b\cdot b \\\ & =a\cdot a\cdot e\cdot b \\\ & =a\cdot e \\\ & =a \\\ \end{aligned}
Therefore, on considering both the cases, the evaluated result in both the cases comes out to be aa .
Therefore, option (a) is correct.
Note: Students must remember the left identity and right identity to solve and obtain a particular answer. If in both the cases the obtained value is different than the methodology is wrong. In this way, answers can be verified.