0
$\begingroup$

Dummit and Footes definition of solvable groups is

$G$ has a chain of subgroups: $1=N_0\trianglelefteq N_1 \trianglelefteq N_2 \trianglelefteq \dots \trianglelefteq N_t = G$ such that each $N_i$ is a normal subgroup of $G$ and $N_{i+1}/N$ is abelian, $0\le i\le t-1$

Wikipedia has the same definition and then adds,

For finite groups, an equivalent definition is that a solvable group is a group with a composition series all of whose factors are cyclic groups of prime order. This is equivalent because a finite group has finite composition length, and every simple abelian group is cyclic of prime order.

I don't understand how these are equivalent. Since composition series have simple factors, is this saying every abelian group is simple?

$\endgroup$
  • 2
    $\begingroup$ No. Since the subgroup of an abelian group is always normal, an abelian group is simple whenever they are cyclic prime ordered groups. $\endgroup$ – fantasie Nov 1 '19 at 2:25
6
$\begingroup$

No: the composition series described in the second definition is not necessarily the same as the chain of subgroups described in the first definition. Rather, you obtain the composition series by refining the chain (adding more intermediate groups in between). As a sketch of how to do this, since $N_{i+1}/N_i$ is abelian, it has a composition series where all the factors are cyclic of prime order. We can now lift all the subgroups of $N_{i+1}/N_i$ in that composition series up to subgroups of $N_{i+1}$ which contain $N_i$, and insert them into our chain of subgroups in between $N_i$ and $N_{i+1}$. Doing this for each $i$, we refine the original chain of subgroups into a longer one in which the factors are not just abelian but cyclic of prime order.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.