/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Let \(H_{1}\) be a subgroup of a... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 11

Let \(H_{1}\) be a subgroup of an abelian group \(G_{1}\) and \(H_{2}\) a subgroup of an abelian group \(G_{2}\). Show that \(H_{1} \times H_{2}\) is a subgroup of \(G_{1} \times G_{2}\).

Short Answer

Expert verified
Question: Show that \(H_1 \times H_2\) is a subgroup of \(G_1 \times G_2\) if \(H_1\) is a subgroup of \(G_1\) and \(H_2\) is a subgroup of \(G_2\), where \(G_1\) and \(G_2\) are abelian groups. Answer: To prove that \(H_1 \times H_2\) is a subgroup of \(G_1 \times G_2\), we showed that the following conditions hold: the existence of an identity element in \(H_1 \times H_2\), closure under the group operation, and the existence of inverses for all elements in \(H_1 \times H_2\). Since all these conditions are satisfied, we can conclude that \(H_1 \times H_2\) is a subgroup of \(G_1 \times G_2\).

Step by step solution

01

Identity element

First, let's find the identity element for the groups \(H_1\) and \(H_2\). Since \(H_1\) is a subgroup of \(G_1\), and \(H_2\) is a subgroup of \(G_2\), we know that the identity element of \(H_1\) is also the identity element of \(G_1\), denoted by \(e_1\), and similarly, the identity element of \(H_2\) is \(e_2\). Now, let's verify that the identity element of \(H_1 \times H_2\) is \((e_1, e_2)\). For any element \((h_1, h_2) \in H_1 \times H_2\), \((h_1, h_2)(e_1, e_2) = (h_1e_1, h_2e_2) = (h_1, h_2)\) and also \((e_1, e_2)(h_1, h_2) = (e_1h_1, e_2h_2) = (h_1, h_2)\). Since the identity element doesn't change the value of the element it's multiplied with, we can conclude that \((e_1, e_2)\) is the identity element of \(H_1 \times H_2\).
02

Closure

Now we need to show that the product of any two elements in \(H_1 \times H_2\) is also in \(H_1 \times H_2\). Let \((h_1, h_2)\) and \((h'_1, h'_2)\) be any two elements in \(H_1 \times H_2\). Then: \((h_1, h_2)(h'_1, h'_2) = (h_1h'_1, h_2h'_2)\). Since \(H_1\) is a subgroup of \(G_1\), and \(H_2\) is a subgroup of \(G_2\), the product \(h_1h'_1\) is in \(H_1\), and the product \(h_2h'_2\) is in \(H_2\). Therefore, the product \((h_1h'_1, h_2h'_2)\) is in \(H_1 \times H_2\), proving closure.
03

Inverses

Lastly, we need to show that if \((h_1, h_2)\) is an element of \(H_1 \times H_2\), then its inverse \((h_1^{-1}, h_2^{-1})\) is also in \(H_1 \times H_2\). Since \(H_1\) is a subgroup of \(G_1\), we know that the inverse of \(h_1\), denoted as \(h_1^{-1}\), exists in \(H_1\), and similarly, \(h_2^{-1}\) exists in \(H_2\). Therefore, the inverse \((h_1^{-1}, h_2^{-1})\) of the element \((h_1, h_2)\) exists in \(H_1 \times H_2\). Since we have shown that the identity element exists, closure holds, and inverses exist for all elements in \(H_1 \times H_2\), we can conclude that \(H_1 \times H_2\) is a subgroup of \(G_1 \times G_2\).

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