91Ó°ÊÓ

Our goal is to find and prove the existence of a subgroup \(K\) of \(G\) that is isomorphic to the quotient group \(G / H\). In other words, we want to show that there exists a bijective homomorphism between \(K\) and \(G / H\).

Since \(G\) is an abelian group, its elements can be written in the form of their orders, for example, $$G = \{g_1^{a_1}, g_2^{a_2}, \dots, g_n^{a_n} \}.$$ We will construct the subgroup \(K\) by selecting elements from \(G\) in a specific manner. For each element \(g_i\) in \(G\), we will include in \(K\) the element \(g_i^{h_i}\), where \(h_i\) divides \(a_i\), and \(h_i\) is the smallest positive integer so that \(g_i^{h_i} \in H\). In other words, $$ K = \{g_1^{h_1}, g_2^{h_2}, \dots, g_n^{h_n}\}. $$

To show that \(K\) is a subgroup of \(G\), we need to show that it satisfies the subgroup criterion: 1. Closure: For any two elements \(g_i^{h_i}, g_j^{h_j} \in K\), their product \(g_i^{h_i}g_j^{h_j}\) is also in \(K\). Since the group \(G\) is abelian, the product can be reordered, and the result can be expressed in terms of the orders \(a_i\) and \(a_j\). Hence, the product is also in \(K\). 2. Identity: The identity element \(e=\{g_1^0, g_2^0, \dots, g_n^0\}\) is also in \(K\). 3. Inverses: For any element \(g_i^{h_i} \in K\), its inverse \(g_i^{-h_i}\) is also in \(K\) since \(g_i^{-h_i}g_i^{h_i} = g_i^0 = e\), where \(e\) is the identity element. Therefore, \(K\) is a subgroup of \(G\).

We define a homomorphism \(\phi : K \to G / H\) by \(\phi(k) = kH = \{kh\;|\;h \in H\}\) for every \(k \in K\). We need to prove that \(\phi\) is both injective and surjective: 1. Injective: Suppose that \(\phi(k_1) = \phi(k_2)\) for some \(k_1, k_2 \in K\). Then \(k_1H = k_2H\), which means \(k_2^{-1}k_1 \in H\). Since \(k_1, k_2 \in K\), their orders are divisible by the orders of elements in \(H\), and therefore, \(k_2^{-1}k_1\) must be the identity element \(e\). Hence, \(k_1 = k_2\), and therefore \(\phi\) is injective. 2. Surjective: To show that \(\phi\) is surjective, we need to show that for every coset \(gH \in G / H\) there exists an element \(k \in K\) such that \(\phi(k) = gH\). Since \(g \in G\), it can be written in the form \(g = g_1^{b_1}g_2^{b_2}\dots g_n^{b_n}\). We can construct the element \(k = g_1^{b_1'}g_2^{b_2'}\dots g_n^{b_n'} \in K\), where \(b_i' = b_i \mod h_i\), then \(\phi(k) = kH = g_1^{b_1'}Hg_2^{b_2'}H\dots g_n^{b_n'}H = gH\). Therefore, \(\phi\) is surjective. Since the homomorphism \(\phi\) is both injective and surjective, it is bijective, and thus \(K\) and \(G / H\) are isomorphic.