Recent content by tehme2
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Symmetric and Alternating Groups disjoint cycles
Homework Statement Let a = (a1a2..ak) and b = (c1c2..ck) be disjoint cycles in Sn. Prove that ab = ba. The Attempt at a Solution Sn consists of the permutations of the elements of T where T = {1,2,3,...,n} so assume we take an i from T. Then either i is in a, i is in b, or i is in...- tehme2
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- Cycles Groups Symmetric
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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Find the order of a k cycle in group Sn
Homework Statement Prove that a k-cycle in the group Sn has order k. Homework Equations The Attempt at a Solution I'm mostly confused on how to write this in math notation. I know it will have order k because a1 → a2 → a3 ... ak-1 → ak → a1 if we do the compositions K times. and so...- tehme2
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- Cycle Group
- Replies: 1
- Forum: Calculus and Beyond Homework Help
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Show Quotient Groups are isomorphic
I'm meant to show the quotient group Z18/M is isomorphic to Z6. The elements of Z18/M are M + 0 = {0, 6, 12} M + 1 = {1, 7, 13} M + 2 = {2, 8, 14} M + 3 = {3, 9, 14} M + 4 = {4, 10, 15} M + 5 = {5, 11, 16} so, I need a function f that takes z18/m to Z6 , so the domain would be the...- tehme2
- Post #3
- Forum: Calculus and Beyond Homework Help
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Show Quotient Groups are isomorphic
Homework Statement Show that Z18/M isomorphic to Z6 where m is the cyclic subgroup <6> operation is addition The Attempt at a Solution M = <6> , so M = {6, 12, 0} I figured I could show that Z18/M has 6 distinct right cosets if I wanted to do M + 0 = {6, 12, 0} M + 1 = {7, 13, 1}...- tehme2
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- Groups quotient Quotient groups
- Replies: 4
- Forum: Calculus and Beyond Homework Help