prove a group of certain order is never simple

Asked Mar 18 '17 at 21:02

Active Mar 18 '17 at 21:02

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There is no simple group of order $400$.

$400=2^{4}\times 5^{2}$,so by sylow theorem, there is a 2-sylow group $A$ or 5-sylow group $B$,then by group action(left multiplication) we have a group homorphism from $G$ to symmtry groups $S_{5^{2}}$ or $S_{2^{4}}$.Kernel of these two morphisms are all normal subgroup of $G$ if they are nontrivial.It's enough for us to prove that one of these two homorphisms have nontrivial kernel.But I don't know how to prove it. Since there is an answer in mathstackexchange by using sylow theorem not group action, I think this should not be duplicated.

asked Mar 18 '17 at 21:02

Jack

2,017

Together with a clarification also here. With the question for other methods, see here. – Dietrich Burde Mar 18 '17 at 21:16
@DietrichBurde They didn't use group action to prove the theorem.My question is that how to use group action to solve this problem. – Jack Mar 18 '17 at 21:19
I see. The other links suggest, that one should rather use Sylow (or at least, there is no reason not to use Sylow. Also, Sylow is in the same spirit as group actions, because that's the way to prove Sylow). – Dietrich Burde Mar 18 '17 at 21:20
@DietrichBurde about the previous answer, I don't know why $P\cap Q$ has at least 125 elements, can you explain why? – Jack Mar 18 '17 at 21:26
@DietrichBurde Made a mistake I mean PQ has at least 125 elements – Jack Mar 18 '17 at 21:28
Which previous answer? There are at least three answers given on MSE. – Dietrich Burde Mar 18 '17 at 21:28
Let us continue this discussion in chat. – Jack Mar 18 '17 at 21:30

prove a group of certain order is never simple

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