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[[Category:MA453Spring2009Walther]] | [[Category:MA453Spring2009Walther]] | ||
I think it has something to do with LaGrange's Theorem, but I'm not sure where to go from there.--[[User:Awika|Awika]] 18:15, 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.--[[User:Awika|Awika]] 18:15, 11 February 2009 (UTC) | ||
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| + | ---- | ||
| + | <math>\scriptstyle\mid G\mid\ =\ 155</math>. <math>\scriptstyle1\neq\mid a\mid\neq\mid b\mid\neq1</math>. Consider a subgroup <math>\scriptstyle H</math> of <math>\scriptstyle G</math> that contains <math>\scriptstyle a</math> and <math>\scriptstyle b</math>. Then, <math>\scriptstyle\mid a\mid\textstyle\mid\scriptstyle\mid H\mid\textstyle\mid\scriptstyle\mid G\mid</math> and <math>\scriptstyle\mid b\mid\textstyle\mid\scriptstyle\mid H\mid\textstyle\mid\scriptstyle\mid G\mid</math>. The divisors of 155 are 1, 5, 31, and 155. So the possible values <math>\scriptstyle\mid a\mid</math> and <math>\scriptstyle\mid b\mid</math> could have are 5, 31, and 155. If either <math>\scriptstyle\mid a\mid</math> or <math>\scriptstyle\mid b\mid</math> is 155, then <math>\scriptstyle\mid H\mid</math> must be 155 and obviously <math>\scriptstyle H</math> would equal <math>\scriptstyle G</math>. Disregarding 155, one of <math>\scriptstyle\mid a\mid</math> and <math>\scriptstyle\mid b\mid</math> would be 5 and the other 31. But since <math>\scriptstyle\mid a\mid</math> and <math>\scriptstyle\mid b\mid</math> both divide <math>\scriptstyle\mid H\mid</math> and <math>\scriptstyle\mid H\mid</math> must divide <math>\scriptstyle\mid G\mid=</math>155, <math>\scriptstyle\mid H\mid</math> must be <math>\scriptstyle lcm(5,31)\ =</math> 155. Therefore in any possible combination, <math>\scriptstyle\mid H\mid\ =\ \mid G\mid</math>, and so <math>\scriptstyle H\ =\ G</math>. <math>\scriptstyle\Box</math> | ||
| + | :--[[User:Narupley|Nick Rupley]] 18:24, 11 February 2009 (UTC) | ||
Revision as of 13:24, 11 February 2009
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)
