Show that a group order 100 is not simple


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Question Number 85664 by Jidda28 last updated on 23/Mar/20

$S how that a group order 100 is not simple$
Commented by mind is power last updated on 24/Mar/20
$100=2^2 .5^2 let see sylow Theorem let G group of order 100 let n_5 number of 5−Sylow Group we have n_5 ∣4..1 n_5 =1modd[5]] n_5 ∈{1,2,4} by 1 n_5 =1 by 2 ⇒H:5−syllow group of G is unique ⇒H is stable by conjugate gHg^− =H by[definition of p−Sylow Group ⇒H is normal ⇒G is not Simple Since H is subgroup normal of G$
$100 = 2^{2} . 5^{2}$ $l e t s e e s y l o w T h e o r e m$ $l e t G g r o u p o f o r d e r 100$ $l e t n_{5} n u m b e r o f 5 - S y l o w G r o u p$ $w e h a v e n_{5} ∣ 4 . . 1$ $n_{5} = 1 m o d d [5]]$ $n_{5} \in {1, 2, 4} b y 1 n_{5} = 1 b y 2$ $\Rightarrow H : 5 - s y l l o w g r o u p o f G i s u n i q u e \Rightarrow H i s s t a b l e b y c o n j u g a t e$ ${g H g}^{-} = H b y [d e f i n i t i o n o f p - S y l o w G r o u p$ $\Rightarrow H i s n o r m a l \Rightarrow G i s n o t S i m p l e S i n c e H i s s u b g r o u p n o r m a l$ $o f G$
Commented by Jidda28 last updated on 27/Mar/20

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