How i can show that group of order 250 is a solvable group ? |G| = 2*(5^3) But i count show that (5^3)/{e} is abelian group Thanks
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Originally Posted by s.lateralus How i can show that group of order 250 is a solvable group ? |G| = 2*(5^3) But i count show that (5^3)/{e} is abelian group Thanks the Sylow 5-subgroup of is normal because and it's solvable because it's nilpotent. also is abelian and hence solvable. thus is solvable. more generally and with the same argument we see that if are primes, then every group of order is solvable.
a more direct method: again let be the Sylow 5-subgroup of we know that contains subgroups such that now look at this subnormal series:
How do you know that 5^2 is normal in 5^3 ?
Code: How do you know that 5^2 is normal in 5^3 ? Let then thus is abelian. A group of p-order is cyclic There's a little theorem that states that any sub-group with index p is normal
Originally Posted by Dinkydoe Code: How do you know that 5^2 is normal in 5^3 ? Let then thus is abelian. A group of p-order is cyclic There's a little theorem that states that any sub-group with index p is normal Not exactly: the theorem states that if p is the least prime dividing the order of a finite group G, then any sbgp. of index p is normal in G. Tonio
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