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Chapter 16: Problem 3

Let \(G\) be a group where \(a^{2}=e\) for all \(a \in G\). Prove that \(G\) is abelian.

Short Answer

Expert verified
By using the properties of a group, associativity and existence of identity and inverse, we proved that for any two elements \(a\) and \(b\) of \(G\), \(ab = ba\), hence the group is abelian.

Step by step solution

01

Choose two arbitrary elements

To prove that all elements in the group are commutative, pick any two arbitrary elements. Let \(a\) and \(b\) be two elements in \(G\).
02

Express the product in terms of identity

As given, the square of any element is the identity, i.e. \(a^{2} = e\) and \(b^{2} = e\). We can write the product \(ab\) in terms of \(e\) as \(ab = e(ab)^{2}e\). By substituting the square identity, we get \(ab = e(a^{2})(b^{2})e\).
03

Use Associativity

We know that group \(G\) is associative. This property allows us to remove or insert brackets without changing the result. This gives us \(ab = ((ea)b)(be)\). Given that \(e\) is an identity and multiplying any element by \(e\) leaves the element unchanged, we simplify this to \(ab = abbe\).
04

Find the inverse

We also know that for each element in a group, an inverse exists which when multiplied by the element, yields the identity. So, \(bb = e\). Substituting this, we get \(ab = ae = a\). By a similar process, we can prove \(ba = a\). Since \(a\) was an arbitrary element, and \(a = ab = ba\), then \(ab = ba\) is true for all \(a,b \in G\). Therefore, group \(G\) is abelian.

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Most popular questions from this chapter

For each of the following encoding functions, find the minimum distance between the code words. Discuss the errordetecting and error-correcting capabilities of each code. a) \(\begin{array}{rl}E: \mathbf{Z}_{2}^{2} \rightarrow \mathbf{Z}_{2}^{5} \\\ 00 \rightarrow 00001 & 01 \rightarrow 01010 \\ 10 \rightarrow 10100 & 11 \rightarrow 11111\end{array}\) b) \(E: \mathbf{Z}_{2}^{2} \rightarrow \mathbf{Z}_{2}^{10}\) \(\begin{array}{cl}00 \rightarrow 0000000000 & 01 \rightarrow 0000011111 \\ 10 \rightarrow 1111100000 & 11 \rightarrow 1111111111\end{array}\) \begin{array}{ll} \text { c) } E: \mathbf{Z}_{2}^{3} \rightarrow \mathbf{Z}_{2}^{6} & \\ 000 \rightarrow 000111 & 001 \rightarrow 001001 \\ 010 \rightarrow 010010 & 011 \rightarrow 011100 \\ 100 \rightarrow 100100 & 101 \rightarrow 101010 \\ 110 \rightarrow 110001 & 111 \rightarrow 111000 \\ \text { d) } E: \mathbf{Z}_{2}^{3} \rightarrow \mathbf{Z}_{2}^{8} & \\ 000 \rightarrow 00011111 & 001 \rightarrow 00111010 \\ 010 \rightarrow 01010101 & 011 \rightarrow 01110000 \\ 100 \rightarrow 10001101 & 101 \rightarrow 10101000 \\ 110 \rightarrow 11000100 & 111 \rightarrow 11100011 \end{array}

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