To solve the given problem based on the group table, let us carefully analyze and compute the results.

(a) Express the elements in terms of $\{e, x_1, x_2, \dots, x_7\}$:

1. $(x_5)^2$:
   To find $(x_5)^2 = x_5 \cdot x_5$, locate $x_5$ in the row of $x_5$:

   • From the table: $x_5 \cdot x_5 = x_7$.
     So, $(x_5)^2 = x_7$.

2. $(x_4)^3$:
   To find $(x_4)^3 = x_4 \cdot x_4 \cdot x_4$:

   • From the table: $x_4 \cdot x_4 = x_6$.
   • Then, $x_6 \cdot x_4 = e$.
     So, $(x_4)^3 = e$.

3. $(x_5)(x_4)$:
   To find $x_5 \cdot x_4$:

   • From the table: $x_5 \cdot x_4 = x_3$.
     So, $(x_5)(x_4) = x_3$.

4. $(x_4)(x_5)$:
   To find $x_4 \cdot x_5$:

   • From the table: $x_4 \cdot x_5 = x_2$.
     So, $(x_4)(x_5) = x_2$.

5. $(x_5)^{-1}$:
   To find the inverse of $x_5$, locate the element in the row of $x_5$ that gives $e$ when multiplied by $x_5$:

   • From the table: $x_5 \cdot x_7 = e$.
     So, $(x_5)^{-1} = x_7$.

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(b) Find all elements $x$ such that $x^3 = e$:

For $x^3 = e$, compute cubes of all elements $x$ in the group:

• $e^3 = e$.
• $(x_1)^3 = x_1 \cdot x_2 \cdot x_3 = e$ (verify from the table).
• $(x_2)^3 = x_2 \cdot x_3 \cdot x_4 = e$.
• $(x_3)^3 = x_3 \cdot x_4 \cdot x_5 = e$.
• $(x_4)^3 = x_4 \cdot x_4 \cdot x_4 = e$ (calculated above).
• $(x_5)^3 = x_5 \cdot x_6 \cdot x_7 = e$.
• $(x_6)^3 = x_6 \cdot x_7 \cdot e = e$.
• $(x_7)^3 = x_7 \cdot e \cdot x_1 = e$.

All elements satisfy $x^3 = e$: $\{e, x_1, x_2, x_3, x_4, x_5, x_6, x_7\}$.

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(c) Find all elements $x$ such that $x^2 = x_5$:

For $x^2 = x_5$, compute squares of all elements $x$:

• $e^2 = e \neq x_5$.
• $(x_1)^2 = x_2 \neq x_5$.
• $(x_2)^2 = x_3 \neq x_5$.
• $(x_3)^2 = x_4 \neq x_5$.
• $(x_4)^2 = x_6 \neq x_5$.
• $(x_5)^2 = x_7 \neq x_5$.
• $(x_6)^2 = e \neq x_5$.
• $(x_7)^2 = x_1 \neq x_5$.

No elements satisfy $x^2 = x_5$: $N$.

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(d) Find all elements $x$ such that $x^3 = x_4$:

For $x^3 = x_4$, compute cubes of all elements $x$:

• $e^3 = e \neq x_4$.
• $(x_1)^3 = e \neq x_4$.
• $(x_2)^3 = e \neq x_4$.
• $(x_3)^3 = e \neq x_4$.
• $(x_4)^3 = e \neq x_4$.
• $(x_5)^3 = e \neq x_4$.
• $(x_6)^3 = e \neq x_4$.
• $(x_7)^3 = e \neq x_4$.

No elements satisfy $x^3 = x_4$: $N$.

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Final Answers:

(a)

1. $(x_5)^2 = x_7$.
2. $(x_4)^3 = e$.
3. $(x_5)(x_4) = x_3$.
4. $(x_4)(x_5) = x_2$.
5. $(x_5)^{-1} = x_7$.

(b) $\{e, x_1, x_2, x_3, x_4, x_5, x_6, x_7\}$.
(c) $N$.
(d) $N$.

Would you like me to verify any calculations or provide further clarification?

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1. What is the significance of $x^n = e$ in a group structure?
2. How do you verify the inverse of an element in a group?
3. Can there be a group where no element satisfies $x^3 = e$?
4. Why is the identity element important when interpreting group tables?
5. How does this group table compare to a cyclic group structure?

Tip: Always check the closure property and inverses to ensure a valid group operation.