If G is a finite solvable group, show that there exist subgroups of G{e} = H, C H, C H, C · · CH, = G2.such that H/H, has prime order.i+1'

We need to prove that there exist subgroups of G, e=H0⊂H1⊂H2⊂...⊂Hn=G such that Hi+1Hi has prime…

Answered: If G is a finite solvable group, show…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

If G is a finite solvable group, show that there exist subgroups of G
{e} = H, C H, C H, C · · CH, = G
2.
such that H/H, has prime order.
i+1'

Expert Solution

Step 1

We need to prove that there exist subgroups of G,

$\{e\} = H_{0} \subset H_{1} \subset H_{2} \subset . . . \subset H_{n} = G$

such that $\frac{H_{i + 1}}{H_{i}}$ has prime order.

We will prove this by induction on the length of the derived series of G, denoted as D(G).

Step 2

If D(G) = 0, then G is abelian, and we can take $H_{i} = g^{p} : g \in G$ , where p is any prime, and $\frac{H_{i} + 1}{H_{i}} ≅ \frac{Z}{p Z}$ , which has prime order.

Now, suppose that D(G) = n > 0 and the statement holds for all groups with a derived length less than n. Since G is solvable, its derived subgroup G' is a proper subgroup of G, and $\frac{G}{G'}$ is abelian. Therefore, the quotient group $\frac{G}{G'}$ has a composition series:

$e = \frac{G_{0}}{G'} \subset \frac{G_{1}}{G'} \subset . . . \subset \frac{G_{m}}{G'} = \frac{G}{G'}$

such that each factor $\frac{G_{i + 1}}{G_{i}}$ is simple. Moreover, since G is solvable, so is $\frac{G}{G'}$ , and therefore the length of its derived series is less than n. By the induction hypothesis, we can find subgroups of $\frac{G}{G'}$ such that: