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Chapter 16: Problem 2

If \(G, H\), and \(K\) are groups and \(G=H \times K\), prove that \(G\) contains subgroups that are isomorphic to \(H\) and \(K\).

Short Answer

Expert verified
The subgroups \(H' = H \times \{e_K\}\) and \(K' = \{e_H\} \times K\) are indeed subgroups of \(G\) and are isomorphic to \(H\) and \(K\) respectively.

Step by step solution

01

Defining the subgroups

First step is to define the nessecary subgroups for \(H\) and \(K\). The natural choice for these would be \(H' = H \times \{e_K\}\) and \(K' = \{e_H\} \times K\), where \(e_H\) and \(e_K\) are the identity elements of \(H\) and \(K\) respectively.
02

Prove the subgroups

The second step is to prove that \(H'\) and \(K'\) are indeed subgroups of \(G\). For \(H'\), it must be shown that: \n1. Closure: For any \(h1, h2 \in H\), \((h1, e_K)(h2, e_K) = (h1h2, e_Ke_K) = (h1h2, e_K) \in H'\).\n2. Identity: The pair \((e_H, e_K) \in H'\).\n3. Inverses: For every \(h \in H\), \((h^{-1}, e_K) \in H'\) is the inverse of \((h, e_K)\). By similar argumentation the same holds for \(K'\).\n4. Associativity is inherited from \(H\) and \(K\), thus also holds in \(H'\) and \(K'\).
03

Proving Isomorphisms

The third step will be proving that \(H'\) is isomorphic to \(H\) and \(K'\) is isomorphic to \(K\). The map \(phi: H -> H'\), \(h -> (h, e_K)\) is an isomorphism. Check:\n1. Well-defined: \(phi\) does map \(H\) to \(H'\) since for each \(h \in H\), \((h, e_K) \in H'\).\n2. Bijective: \(phi\) is one-to-one and onto.\n3. Homomorphism: \(\phi(h1*h2) = (h1*h2, e_K) = (h1, e_K) * (h2, e_K) = phi(h1) * phi(h2)\). Same reasoning can be used for the isomorphism from \(K\) to \(K'\).

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