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Chapter 16: Problem 3

Why is the set \(\mathbf{Z}\) not a group under subtraction?

Short Answer

Expert verified
The set of integers ( \mathbf{Z} ) is not a group under subtraction because it fails to meet the properties of associativity and identity.

Step by step solution

01

Check for Closure

Closure means that if a and b are elements in the set, then a - b (the result of the operation) is also in the set. Considering that the operation is subtraction and the set is all integers, given any two integers by subtracting one from the other, the result will still be an integer. Therefore, the set \( \mathbf{Z} \) is closed under subtraction.
02

Check for Associativity

Associativity means that for all a, b, and c in the set, the equation (a - b) - c equals a - (b - c). Subtraction is not an associative operation because generally (a - b) - c \neq a - (b - c) for integers a, b, and c. Therefore, the set \( \mathbf{Z} \) is not associative under subtraction.
03

Check for Identity

An identity element is an element e in the set, for all a in the set, the equation a - e = a and e - a = a hold. However, there is no such integer e can satisfy the equations under subtraction. Thus, the set \( \mathbf{Z} \) does not have an identity under subtraction.
04

Check for Invertibility

Invertibility is ignored in this exercise because even the identity element does not exist in the set \( \mathbf{Z} \) under subtraction. Hence there is no need to further check for an inverse.

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Most popular questions from this chapter

Express each of the following elements of \(S_{7}\) as a product of disjoint cycles. $$ \begin{aligned} \alpha &=\left(\begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 4 & 6 & 7 & 1 & 5 & 3 \end{array}\right) \\ \beta &=\left(\begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 3 & 6 & 5 & 2 & 1 & 7 & 4 \end{array}\right) \\ \gamma &=\left(\begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 3 & 1 & 7 & 5 & 4 & 6 \end{array}\right) \\ \delta &=\left(\begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 4 & 2 & 7 & 1 & 3 & 6 & 5 \end{array}\right) \end{aligned} $$

Let $$ H=\left[\begin{array}{lllllll} 1 & 1 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right] $$ be the parity-check matrix for a Hamming \((7,4)\) code. a) Encode the following messages: \(\begin{array}{llllll}1000 & 1100 & 1011 & 1110 & 1001 & 1111 .\end{array}\) b) Decode the following received words: \(\begin{array}{cccc}1100001 & 1110111 & 0010001 & 0011100\end{array}\) c) Construct a decoding table consisting of the syndromes and coset leaders for this code. d) Use the result in part (c) to decode the received words given in part (b).

Let \(H\) and \(K\) be subgroups of a group \(G\), where \(e\) is the identity of \(G\). a) Prove that if \(|H|=10\) and \(|K|=21\), then \(H \cap K=\\{e\\}\). b) If \(|H|=m\) and \(|K|=n\), with \(\operatorname{gcd}(m, n)=1\), prove that \(H \cap K=\\{e\\}\)

Let \(G=\\{q \in \mathbf{Q} \mid q \neq-1\\}\). Define the binary operation \(\circ\) on G by \(x \circ y=x+y+x y\). Prove that \((G, \circ)\) is an abelian group

a) Construct a decoding table (with syndromes) for the group code given by the generator matrix $$ G=\left[\begin{array}{lllll} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \end{array}\right] $$ b) Use the table from part (a) to decode the following received words. \(\begin{array}{cccc}11110 & 11101 & 11011 & 10100 \\ 10011 & 10101 & 11111 & 01100\end{array}\) c) Does this code correct single errors in transmission?
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