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# ahw6 - x or y it remains in that form Using x 6 = y 4 = 1...

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Math 5125 Monday, October 10 Sixth Homework Solutions 1. Let F denote the free group on x , y , let a be a generator for K , and deﬁne a ho- momorphism θ : F G by θ x = a and θ y = b (where a = ( a , 1 ) and b = ( 1 , b ) ). Note that θ is onto, because G is generated by a and b , hence G = F / ker θ . Let P = h x , y | x 6 = y 4 = 1 , yxy - 1 = x - 1 i and let R = { x 6 , y 4 , yxy - 1 x } , so P = F / R F . Since θ x 6 = a 6 = 1, θ y 4 = b 4 = 1, and θ ( yxy - 1 x ) = bab - 1 a = (( φ b ) a ) a = a - 1 a = 1 , we see that R ker θ . We want R F = ker θ , because then we will be able to conclude that F / R F = G as required. Since | F / ker θ | = | G | = 24, it will be sufﬁcient to show that | P | ≤ 24. We will do this by establishing that every element of F can be written in the form x i y j mod R F , where 0 i 5 and 0 j 3. It will be sufﬁcient to show that when we multiply such an element on the left or right by
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Unformatted text preview: x or y , it remains in that form. Using x 6 = y 4 = 1, we need only consider multiplication on the left by y and on the right by x . Now the relation yxy-1 = x-1 yields yx i y-1 = x-i , consequently yx i y j = x-i y j + 1 , which is alright. Also yxy-1 = x-1 yields yx-1 y-1 = x and hence y j xy-j = x or x-1 . We deduce that x i y j x = x i ± 1 y j and again we have an expression in the requisite form. We now see that | R / R F | ≤ 6 * 4 as required....
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