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

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\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\cdot e=n $ for all n in S, and a right identity if $ e\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\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
 $ \begin{align}
  & 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{align} $
By using the left- identity $ n\cdot e=n $ and right-identity $ e\cdot n=n $ , we can solve the product as
 $ \begin{align}
  & =a\cdot a\cdot a\cdot b\cdot b \\
 & =e\cdot a\cdot e \\
 & =a\cdot e \\
 & =a \\
\end{align} $
Expanding the product and applying identity element in case (2), we get
 $ \begin{align}
  & 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{align} $
By using the left- identity $ n\cdot e=n $ and right-identity $ e\cdot n=n $ , we can solve the product as
 $ \begin{align}
  & =a\cdot a\cdot a\cdot e\cdot b\cdot b \\
 & =a\cdot a\cdot e\cdot b \\
 & =a\cdot e \\
 & =a \\
\end{align} $
Therefore, on considering both the cases, the evaluated result in both the cases comes out to be $ a $ .
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.