HW4 - Math 103A Homework Solutions HW4 Jacek Nowacki...

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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 functions

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