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Find the products \(\pi_{1} \circ \pi_{2}\) and \(\pi_{2} \circ \pi_{1}\) of the two permutations \(\pi_{1}=\left(\begin{array}{llllll}1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 4 & 6 & 5 & 1 & 2\end{array}\right)\) and \(\pi_{2}=\left(\begin{array}{llllll}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 1 & 3 & 6 & 5 & 4\end{array}\right)\).

Short Answer

Expert verified
The products of the permutations are \(\pi_{1} \circ \(\pi_{2} = \left(\begin{array}{llllll}1 & 2 & 3 & 4 & 5 & 6 \ 4 & 3 & 6 & 2 & 1 & 5\end{array}\right)\) and \(\pi_{2} \circ \(\pi_{1} = \left(\begin{array}{llllll}1 & 2 & 3 & 4 & 5 & 6 \ 2 & 5 & 4 & 1 & 6 & 3\end{array}\right)\).

Step by step solution

01

Compute the product \(\pi_{1} \circ \(\pi_{2}\)

To find the product \(\pi_{1} \circ \(\pi_{2}\), one should apply \(\pi_{2}\) first and then apply \(\pi_{1}\) to the result. This results in the permutation \(\left(\begin{array}{llllll}1 & 2 & 3 & 4 & 5 & 6 \ 4 & 3 & 6 & 2 & 1 & 5\end{array}\right)\).
02

Compute the product \(\pi_{2} \circ \(\pi_{1}\)

To find the product \(\pi_{2} \circ \(\pi_{1}\), one should apply \(\pi_{1}\) first and then apply \(\pi_{2}\) to the result. This results in the permutation \(\left(\begin{array}{llllll}1 & 2 & 3 & 4 & 5 & 6 \ 2 & 5 & 4 & 1 & 6 & 3\end{array}\right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Group Theory
Group theory is an essential concept in abstract algebra that studies mathematical structures known as groups. A group is a set equipped with an operation that combines any two elements to form a third element within the set. This operation must satisfy four fundamental properties:
  • Closure: If \(a\) and \(b\) are elements of the group, then the product \(a \circ b\) is also in the group.
  • Associativity: For any elements \(a, b, c\), the equation \((a \circ b) \circ c = a \circ (b \circ c)\) holds.
  • Identity Element: There is an element, denoted as \(e\), such that for any element \(a\), the equation \(e \circ a = a \circ e = a\) holds.
  • Inverse Element: For each element \(a\), there is an element \(b\) such that \(a \circ b = b \circ a = e\).
Groups are fundamental in understanding symmetry and permutation structures in mathematics. They help in solving algebraic equations, analyzing geometrical shapes, and even in number theory. In the context of permutations, the set of all permutations of a given set forms a group under the operation of composition.
Permutation Composition
Permutation composition is an operation where two permutations are combined to produce a third permutation. The composition of permutations is similar to the composition of functions: you apply one permutation and then apply the other.To understand permutation composition in more detail, let's consider two permutations \(\pi_1\) and \(\pi_2\).
  • When calculating \(\pi_1 \circ \pi_2\), apply \(\pi_2\) first to a set of elements, then follow this with \(\pi_1\).
  • The order in which permutations are composed is crucial because permutation composition is generally not commutative. This means \(\pi_1 \circ \pi_2 eq \pi_2 \circ \pi_1\).
  • It's helpful to follow elements through both permutations step by step to find their resulting position.
In our exercise, performing the composition involves substituting elements of the set through the two given permutations. It highlights how permutations rearrange elements and how their composition results in different rearrangements.
Abstract Algebra
Abstract algebra is a field of mathematics that studies algebraic structures like groups, rings, and fields. It generalizes and extends traditional algebraic concepts. In abstract algebra, one studies structural properties and relationships without focusing on specific numbers or calculations. Instead, the emphasis is on understanding how these structures behave under various operations.
  • Groups: As discussed, groups are one of the simplest structures in abstract algebra, emphasizing symmetry.
  • Rings: These are algebraic structures with two operations, typically addition and multiplication, satisfying certain axioms.
  • Fields: Fields extend the concept of rings by including multiplicative inverses for all non-zero elements.
Abstract algebra helps in understanding mathematical puzzles by focusing on the unity of structures across different areas, making it valuable for math majors and researchers. Studying these abstract systems deepens one's mathematical maturity and ability to think critically about mathematical truths.

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Most popular questions from this chapter

Suppose \(G\) is the direct product of \(H_{1}\) and \(H_{2}\) and \(g=h_{1} h_{2}\). Show that the factors \(h_{1}\) and \(h_{2}\) are unique and that \(H_{1}\) and \(H_{2}\) are normal.

Let \(M=\mathbb{R} \cup\\{\infty\\}\), and define an action of \(S L(2, \mathbb{R})\) on \(M\) by $$ \left(\begin{array}{ll} a & b \\ c & d \end{array}\right) \cdot x=\frac{a x+c}{b x+d} $$ (a) Show that this is indeed a group action with a law of multiplication identical to the matrix multiplication. (b) Show that the action is transitive. (c) Show that beside identity, there is precisely one other element \(g\) of the group such that \(g \cdot x=x\) for all \(x \in M\) (d) Show that for every \(x \in M\), $$ G_{x}=\left(\begin{array}{cc} a & b \\ b x^{2}+(d-a) x & d \end{array}\right) $$

(a) Show that the only element \(a\) in a group with the property \(a^{2}=a\) is the identity. (b) Now use \(e_{G} \star e_{G}=e_{G}\) to show that any homomorphism maps identity to identity. (c) Show that if \(f: G \rightarrow H\) is a homomorphism, then \(f\left(g^{-1}\right)=[f(g)]^{-1}\).

Let \(X\) be a subset of a group \(G\). A word on \(X\) is an element \(w\) of \(G\) of the form $$ w=x_{1}^{e_{1}} x_{2}^{e_{2}} \cdots x_{n}^{e_{n}} $$ where \(x_{i} \in X\) and \(e_{i}=\pm 1\). Show that the set of all words on \(X\) is a subgroup of \(G\).

Express the following permutations as products of transpositions, and determine whether they are even or odd. (a) \(\left(\begin{array}{lllll}1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 2 & 1 & 5\end{array}\right)\), (b) \(\left(\begin{array}{llllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 4 & 1 & 7 & 8 & 3 & 6 & 5 & 2\end{array}\right)\), (c) \(\left(\begin{array}{llllll}1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 4 & 5 & 3 & 2 & 1\end{array}\right)\), (d) \(\left(\begin{array}{lllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 6 & 7 & 2 & 4 & 1 & 5 & 3\end{array}\right)\).

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