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 Homework Statement:
 Let ##G## and ##H## be finite groups. If there are normal series of ##G## and ##H## having the same set of factor groups, then ##G## and ##H## have the same composition factors.
 Relevant Equations:

A normal series of a group ##G## is
$$G = G_0 \ge G_1 \ge G_2 \ge \dots \ge G_n = 1$$
where ##G_i \trianglelefteq G_{i+1}## for all ##i##. The factor groups of this series are ##G_{i}/G
_{i+1}## for all ##i##. The length of this series is the number of distinct factors.
A composition series is a normal series whose factors are simple or trivial.
Attempt so far: We're given that ##G## and ##H## have equivalent normal series
$$G = G_0 \ge G_1 \ge \dots \ge G_n = 1$$
and
$$H = H_0 \ge H_1 \ge \dots \ge H_n = 1$$
We can assume they have the same length because they are equivalent. I think from here I need to construct two composition series; one for ##G## and one for ##H##, and then show they are equivalent. By Jordan Holder theorem, any two composition series of a group will have the same factors. So this would be enough, I think.
But i'm having trouble constructing a composition series, starting with ##G##... I think I need to write
##G = G_0 \ge G_{0, 1} \ge G_{0, 2} \ge \dots \ge G_{0, k} \ge G_1 \ge G_{1, 1} \ge \dots##, but i'm not sure how to define ##G_{i, j}## to such that the series is a composition series. Can I have a hint, please?
Edit: For each ##G_i/G_{i+1}## there is an ##H_j/H_{j+1}## such that ##G_i/G_{i+1} \cong H_j/H_{j+1}##.
By correspondence theorem, there is a bijection between the set of subgroups of ##G_i## that contain ##G_{i+1}## and the set of subgroups of ##G_{i}/G_{i+1}##. Similarly, there is a bijection between the set of subgroups of ##H_i## that contain ##H_{i+1}## and the set of subgroups of ##H_i/H_{i+1}##. Hence, there is a bijection between the set of subgroups of ##G_i## that contain ##G_{i+1}## and the set of subgroups of ##H_{i}## that contain ##H_{i+1}##.
$$G = G_0 \ge G_1 \ge \dots \ge G_n = 1$$
and
$$H = H_0 \ge H_1 \ge \dots \ge H_n = 1$$
We can assume they have the same length because they are equivalent. I think from here I need to construct two composition series; one for ##G## and one for ##H##, and then show they are equivalent. By Jordan Holder theorem, any two composition series of a group will have the same factors. So this would be enough, I think.
But i'm having trouble constructing a composition series, starting with ##G##... I think I need to write
##G = G_0 \ge G_{0, 1} \ge G_{0, 2} \ge \dots \ge G_{0, k} \ge G_1 \ge G_{1, 1} \ge \dots##, but i'm not sure how to define ##G_{i, j}## to such that the series is a composition series. Can I have a hint, please?
Edit: For each ##G_i/G_{i+1}## there is an ##H_j/H_{j+1}## such that ##G_i/G_{i+1} \cong H_j/H_{j+1}##.
By correspondence theorem, there is a bijection between the set of subgroups of ##G_i## that contain ##G_{i+1}## and the set of subgroups of ##G_{i}/G_{i+1}##. Similarly, there is a bijection between the set of subgroups of ##H_i## that contain ##H_{i+1}## and the set of subgroups of ##H_i/H_{i+1}##. Hence, there is a bijection between the set of subgroups of ##G_i## that contain ##G_{i+1}## and the set of subgroups of ##H_{i}## that contain ##H_{i+1}##.
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