Show-that-a-group-order-100-is-not-simple- Tinku Tara June 4, 2023 Algebra 0 Comments FacebookTweetPin Question Number 85664 by Jidda28 last updated on 23/Mar/20 Showthatagrouporder100isnotsimple Commented by mind is power last updated on 24/Mar/20 100=22.52letseesylowTheoremletGgroupoforder100letn5numberof5−SylowGroupwehaven5∣4..1n5=1modd[5]]n5∈{1,2,4}by1n5=1by2⇒H:5−syllowgroupofGisunique⇒HisstablebyconjugategHg−=Hby[definitionofp−SylowGroup⇒Hisnormal⇒GisnotSimpleSinceHissubgroupnormalofG Commented by Jidda28 last updated on 27/Mar/20 Thankyousir Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: etude-de-la-monotonie-svp-u-n-k-1-n-1-k-ln-n-Next Next post: In-rectangle-ABCD-AB-8-BC-20-P-is-a-point-on-AD-so-that-BPC-90-If-r-1-r-2-r-3-are-the-radii-of-the-incircles-of-APB-BPC-and-CPD-find-r-1-r-2-r-3- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![100=2^2 .5^2 let see sylow Theorem let G group of order 100 let n_5 number of 5−Sylow Group we have n_5 ∣4..1 n_5 =1modd[5]] n_5 ∈{1,2,4} by 1 n_5 =1 by 2 ⇒H:5−syllow group of G is unique ⇒H is stable by conjugate gHg^− =H by[definition of p−Sylow Group ⇒H is normal ⇒G is not Simple Since H is subgroup normal of G](https://www.tinkutara.com/question/Q85768.png)
