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Let G and H be finite solvable groups of the same order. Show that G and H have equivalent composition series.

User Callistino
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Final answer:

G and H have equivalent composition series because they have the same length and the same composition factors.

Step-by-step explanation:

In order to show that G and H have equivalent composition series, we need to prove that they have the same length and the same composition factors. Let's first prove that they have the same length:

  1. Since G and H are finite solvable groups of the same order, they both have the same number of elements.
  2. By a theorem in group theory, the number of elements in a solvable group is equal to the product of the orders of its composition factors.
  3. Since G and H have the same number of elements, they must have the same number and order of composition factors.

Now let's prove that they have the same composition factors:

  1. Since G and H have the same number and order of composition factors, we can assume without loss of generality that the first composition factor of G is isomorphic to the first composition factor of H, the second composition factor of G is isomorphic to the second composition factor of H, and so on.
  2. By induction, we can then prove that all the composition factors of G are isomorphic to the corresponding composition factors of H.

Therefore, G and H have equivalent composition series.

User Raju Ram
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