Given that the Cayley table for set *S* = {*a*, *b*, *c*, *d*, *e*, *f*, *g*, *h*} with operation * is as follows, complete parts B1 and B2:
 

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g

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***b***

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***c***

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***d***

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***e***

c

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***f***

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***g***

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***h***

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1.  State the identity element of the set *S* with operation * and explain why it satisfies the definition of the identity property.

2.  Explain why the set *S* with operation * is not a group by giving a relevant counterexample.

Let's analyze the Cayley table for the set \( S = \{a, b, c, d, e, f, g, h\} \) with the binary operation \( * \).

### Part B1: Identity Element

The identity element \( e \) in a set \( S \) with a binary operation \( * \) is defined as the element that, when combined with any element \( x \in S \) using the operation \( * \), returns \( x \) itself. In other words, if \( e \) is the identity element, then:
\[ e * x = x * e = x \quad \text{for all } x \in S \]

Looking at the Cayley table:

- We see that for the row corresponding to \( g \), the element \( g \) returns the element itself when combined with any element:
  \[
  \begin{align*}
  g * a &= a \\
  g * b &= b \\
  g * c &= c \\
  g * d &= d \\
  g * e &= e \\
  g * f &= f \\
  g * g &= g \\
  g * h &= h \\
  \end{align*}
  \]

- Similarly, for the column corresponding to \( g \), the element \( g \) is returned when combined with any element:
  \[
  \begin{align*}
  a * g &= a \\
  b * g &= b \\
  c * g &= c \\
  d * g &= d \\
  e * g &= e \\
  f * g &= f \\
  g * g &= g \\
  h * g &= h \\
  \end{align*}
  \]

Thus, \( g \) is the identity element for this set \( S \) with operation \( * \).

### Part B2: Why \( S \) is Not a Group

For a set with a binary operation to be a group, it must satisfy the following properties:

1. **Closure**: The operation must be closed on the set.
2. **Associativity**: The operation must be associative.
3. **Identity Element**: There must be an identity element (as found in B1).
4. **Inverses**: Every element must have an inverse within the set.

Let’s check if every element in \( S \) has an inverse under the operation \( * \).

For example, consider element \( a \). Its row and column in the table give us:
\[
\begin{aligned}
a * a & = g \\
a * b & = h \\
a * c & = e \\
a * d & = f \\
a * e & = c \\
a * f & = d \\
a * g & = a \\
a * h & = b \\
\end{aligned}
\]

To find an inverse of \( a \), we need an element \( x \in S \) such that:
\[
a * x = x * a = g
\]

However, looking at the table:

- \( a * g = a \), not \( g \).
- \( g * a = a \), not \( g \).

Since \( g \) is the identity, and there's no \( x \) such that both \( a * x = g \) and \( x * a = g \), \( a \) does not have an inverse. This violates the requirement for every element to have an inverse in the group.

### Conclusion:
Since not every element in \( S \) has an inverse, the set \( S \) with the operation \( * \) is **not** a group.

---

Would you like further details or have any questions?

Here are some related questions:

1. Can a set have more than one identity element?
2. How can we verify the associativity property in a Cayley table?
3. What would happen if there was more than one element in the set without an inverse?
4. Is it possible for a set with no identity element to be a group?
5. How can we determine whether the operation is commutative?

**Tip:** When working with Cayley tables, always start by identifying the identity element, as it simplifies checking for other group properties.

Explore Cayley tables and group properties with detailed examples. Learn how to identify identity elements and analyze why certain sets with operations are not groups. Includes a specific analysis of set S with operation *.

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