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
$\displaystyle P,$ the Sylow 5-subgroup of $\displaystyle G,$ is normal because $\displaystyle [G:P]=2$ and it's solvable because it's nilpotent. also $\displaystyle G/P$ is abelian and hence solvable. thus $\displaystyle G$ is solvable.
more generally and with the same argument we see that if $\displaystyle p < q$ are primes, then every group of order $\displaystyle pq^n, \ n \geq 0,$ is solvable.
a more direct method: again let $\displaystyle P$ be the Sylow 5-subgroup of $\displaystyle G.$ we know that $\displaystyle P$ contains subgroups $\displaystyle P_1 \subset P_2$ such that $\displaystyle |P_1|=5, \ |P_2|=5^2.$ now look at this subnormal series:
$\displaystyle (1) \lhd P_1 \lhd P_2 \lhd P \lhd G.$
Let $\displaystyle |P_1| = 5^2, |P_2| = 5^3 $ then $\displaystyle [P_2:P_1] = 5$ thus $\displaystyle P_2/P_1$ is abelian. A group of p-order is cyclicCode:How do you know that 5^2 is normal in 5^3 ?
There's a little theorem that states that any sub-group with index p is normal