Question Number 85664 by Jidda28 last updated on 23/Mar/20
$$\boldsymbol{{S}\mathrm{how}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{group}}\:\boldsymbol{\mathrm{order}}\:\mathrm{100}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{not}}\:\boldsymbol{\mathrm{simple}} \\ $$
Commented by mind is power last updated on 24/Mar/20
![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](Q85768.png)
$$\mathrm{100}=\mathrm{2}^{\mathrm{2}} .\mathrm{5}^{\mathrm{2}} \\ $$$${let}\:{see}\:{sylow}\:{Theorem} \\ $$$${let}\:{G}\:{group}\:{of}\:{order}\:\mathrm{100} \\ $$$${let}\:{n}_{\mathrm{5}} \:\:{number}\:{of}\:\mathrm{5}−{Sylow}\:{Group} \\ $$$${we}\:{have}\:{n}_{\mathrm{5}} \mid\mathrm{4}..\mathrm{1} \\ $$$$\left.{n}_{\mathrm{5}} =\mathrm{1}{modd}\left[\mathrm{5}\right]\right] \\ $$$${n}_{\mathrm{5}} \in\left\{\mathrm{1},\mathrm{2},\mathrm{4}\right\}\:{by}\:\mathrm{1}\:{n}_{\mathrm{5}} =\mathrm{1}\:{by}\:\mathrm{2}\: \\ $$$$\Rightarrow{H}:\mathrm{5}−{syllow}\:{group}\:{of}\:\:{G}\:{is}\:{unique}\:\Rightarrow{H}\:{is}\:{stable}\:{by}\:{conjugate} \\ $$$${gHg}^{−} ={H}\:{by}\left[{definition}\:{of}\:{p}−{Sylow}\:{Group}\right. \\ $$$$\Rightarrow{H}\:{is}\:{normal}\:\Rightarrow{G}\:{is}\:{not}\:{Simple}\:{Since}\:{H}\:{is}\:{subgroup}\:\:{normal} \\ $$$${of}\:{G} \\ $$
Commented by Jidda28 last updated on 27/Mar/20
$$\boldsymbol{{T}}{hank}\:{you}\:{sir} \\ $$