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Unformatted text preview: Math 103A Homework Solutions HW4 Jacek Nowacki February 27, 2007 Chapter 6 Problem 4. Show that U (8) is not isomorphic to U (10). Proof. First, we note that U (8) = { 1 , 3 , 5 , 7 } and U (10) = { 1 , 3 , 7 , 9 } . Hence, we can’t use orders as they are the same. Instead, we notice that each element of U (8) is of order two. But, 3 ∈ U (10) is of order four. As we know, isomorphisms preserve order. So, the two groups can’t be isomorphic. Problem 6. Prove that the notion of group isomorphism is transitive. That is, if G , H , and K are groups and G ≈ H and H ≈ K , then G ≈ K . Proof. Let us start by noting that G ≈ H means that there is at least one isomophism from G to H (also, at least one isomorphism exists the other way). There are situations in which there is more than one isomorphism between the groups. Let φ : G→ H be an isomorphism between G and H . Let ψ : H→ K be an isomorphism between H and K . Now, we define a map δ : G→ K by δ = φ ◦ ψ . We will show that δ is an isomorphism. For that, we need it to be bijective. That is clear, as it is a composition of two bijective functionsbe bijective....
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This note was uploaded on 01/18/2012 for the course INFORMATIK 2011 taught by Professor Phanthuongcang during the Winter '11 term at Cornell.
 Winter '11
 PhanThuongCang

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