Week 5 HW, Chapter 7, Problem 10, MA453, Spring 2008, Prof. Walther
Problem Statement
Can somebody please state the question
Discussion
Does anyone know how to do this or how to start it? --Lchinn 16:46, 11 February 2009 (UTC)
I think it has something to do with LaGrange's Theorem, but I'm not sure where to go from there.--Awika 18:15, 11 February 2009 (UTC)
$ \scriptstyle\mid G\mid\ =\ 155 $. $ \scriptstyle1\neq\mid a\mid\neq\mid b\mid\neq1 $. Consider a subgroup $ \scriptstyle H $ of $ \scriptstyle G $ that contains $ \scriptstyle a $ and $ \scriptstyle b $. Then, $ \scriptstyle\mid a\mid\textstyle\mid\scriptstyle\mid H\mid\textstyle\mid\scriptstyle\mid G\mid $ and $ \scriptstyle\mid b\mid\textstyle\mid\scriptstyle\mid H\mid\textstyle\mid\scriptstyle\mid G\mid $. The divisors of 155 are 1, 5, 31, and 155. So the possible values $ \scriptstyle\mid a\mid $ and $ \scriptstyle\mid b\mid $ could have are 5, 31, and 155. If either $ \scriptstyle\mid a\mid $ or $ \scriptstyle\mid b\mid $ is 155, then $ \scriptstyle\mid H\mid $ must be 155 and obviously $ \scriptstyle H $ would equal $ \scriptstyle G $. Disregarding 155, one of $ \scriptstyle\mid a\mid $ and $ \scriptstyle\mid b\mid $ would be 5 and the other 31. But since $ \scriptstyle\mid a\mid $ and $ \scriptstyle\mid b\mid $ both divide $ \scriptstyle\mid H\mid $ and $ \scriptstyle\mid H\mid $ must divide $ \scriptstyle\mid G\mid= $155, $ \scriptstyle\mid H\mid $ must be $ \scriptstyle lcm(5,31)\ = $ 155. Therefore in any possible combination, $ \scriptstyle\mid H\mid\ =\ \mid G\mid $, and so $ \scriptstyle H\ =\ G $. $ \scriptstyle\Box $
- --Nick Rupley 18:24, 11 February 2009 (UTC)
