Math Help - solvable group

1. solvable group

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

2. 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
$P,$ the Sylow 5-subgroup of $G,$ is normal because $[G:P]=2$ and it's solvable because it's nilpotent. also $G/P$ is abelian and hence solvable. thus $G$ is solvable.

more generally and with the same argument we see that if $p < q$ are primes, then every group of order $pq^n, \ n \geq 0,$ is solvable.

3. a more direct method: again let $P$ be the Sylow 5-subgroup of $G.$ we know that $P$ contains subgroups $P_1 \subset P_2$ such that $|P_1|=5, \ |P_2|=5^2.$ now look at this subnormal series:

$(1) \lhd P_1 \lhd P_2 \lhd P \lhd G.$

4. How do you know that 5^2 is normal in 5^3 ?

5. Code:
How do you know that 5^2 is normal in 5^3 ?
Let $|P_1| = 5^2, |P_2| = 5^3$ then $[P_2:P_1] = 5$ thus $P_2/P_1$ 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

6. Originally Posted by Dinkydoe
Code:
How do you know that 5^2 is normal in 5^3 ?
Let $|P_1| = 5^2, |P_2| = 5^3$ then $[P_2:P_1] = 5$ thus $P_2/P_1$ 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