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Unformatted text preview: Math 4107 Midterm 1 Solutions, Fall 2009 October 18, 2009 1. You can look up these definitions yourself. 2. There are many ways to solve this problem. We present two here: Way 1. Let H be the subgroup of order 35. By Lagrange, we know that every element of H has order either 1 , 5 , 7 or 35. If some element has order 5 or 35, we are done (if g has order 35, then g 7 has order 5). So, let us assume every element has order 1 or 7. So, since the identity is the only element of order 1, every nonidentity element has order 7. If g 1 and g 2 are two elements of order 7, then either the cyclic group ( g 1 ) = ( g 2 ) or ( g 1 ) ( g 2 ) = { e } . What this means is that the subgroups of order 7 are essentially disjoint, apart from intersecting in the identity; so, letting A 1 , ..., A k be all these different subgroups of order 7, together they contribute 6 k elements of order 7, and 1 element of order 1. But this implies 35 =  H  = 1 + 6 k, which cannot hold because 35 1 = 34 is not divisible by 3.1 = 34 is not divisible by 3....
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This note was uploaded on 10/23/2011 for the course MATH 4107 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.
 Fall '08
 Staff
 Math, Algebra

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