There is no simple group of order 400

Question

There is no simple group of order 400.

My attempt:

Firstly, I started supposing that $G$ was not simple, and using the Sylow theorems I got that $G$ must have $25$ $2$-Sylow and $16$ $5$-Sylow.

The thing is I don't know how to end in a contradiction at this point.

As the number of Sylows is quite considerable, I cannot use the fact that $G$ divides $n!/2$ because obviously does.

Finally, I tried to count on elements but I cannot afirm that the intersection of Sylows is trivial and even get a minimum value

Does this answer your question? No group of order 400 is simple — ndhanson3, May 14 '21 at 19:34
You made a big mistake: you started by assuming what you wanted to prove. — , May 14 '21 at 21:34

score 0 · Answer 1 · answered May 14 '21 at 20:42

0

By Burnside's theorem it would be solvable. But a non-abelian simple group can't be solvable. Meanwhile if it were abelian, its order would have to be prime.

answered May 14 '21 at 20:42

There is no simple group of order 400

1 Answers1