e Theorem 12.16. group with IGI p\" for some integer n 0. (We call such a group
ID: 3109930 • Letter: E
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e Theorem 12.16. group with IGI p" for some integer n 0. (We call such a group a finite p-group.) Let H be a subgroup of G with HI pm. Then there is a tower of subgroups Ho CH, CH with Hi m+i for all i, 0 Sis n. pm Proof We proceed by induction on k n m. The result is trivial i n m 0. Suppose then that the result is true for k n- m -1, and th IGI and pm. By Theorem 1 G has a normal subgroup N with INI p. If N is not contained in H, then H S NH with INH pm 1. Since by induction, the result is true for k n m 1 n- (m 1), it follows that there is a tower of subgroups with Hi m+1. Then taking H Ho, we are done. pm Hence, we may assume that N g H. Let G G/N. By induction, since (G: H) m-1, there is a tower of subgroups Ho H C H, CExplanation / Answer
It seems from the proof that the Induction procedure is not correct. It is the weak form of Induction. Where it is assumed that the result is true for n-m-1, therefore it shows the result for (n-m). But to show that the intermediatory results have been required and they have been used as result e.g all the intermediatory indexes are going well. But in the weal form of Induction it is only assumed that H_(n-m-1) is a subgroup, not the lower indexes. Therefore one need to apply here the strong form of Induction, which states that as the Induction hypothesis "Let the result is true for all 0,1,2,.....(n-m-1). Next the proof will start to show the result also holds for (n-m) .
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