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ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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Two composition series of C12 have the same factors in different orders

Example

Let C12=x. The chains C12>x2>x4>1 and C12>x3>x6>1 are composition series. Their factor orders are respectively (2,2,3) and (3,2,2), so they display the same composition factors in different orders.

Facts & Assumptions

Given: The cyclic group C12=x.

[L1]

Every subgroup of a cyclic group G=g is cyclic; a nontrivial subgroup is generated by the least positive power of g that it contains (Every subgroup of a cyclic group is cyclic; the least positive exponent in a nontrivial subgroup supplies a generator). A quotient of G=g is generated by the image of g and so is cyclic as well.

[L2]

Composition factors are invariant up to isomorphism and permutation (The Jordan–Hölder theorem for groups).

[L3]

The order of a finite group is the product of the orders of the factors in a composition series (The order of a finite group is the product of the orders of its composition factors).

Verification

technique · direct
1.1

Listing the powers of x shows that x2, x4, x3, and x6 have orders 6,3,4,2, respectively; [L1] confirms that all displayed terms are cyclic subgroups.

givenL1algebra
2.1

Each adjacent quotient therefore has prime order: the first list is 2,2,3, and the second is 3,2,2. A group of prime order is simple, so both chains are composition series.

step 1.1algebra
3.1

Each quotient is cyclic by [L1], hence the two factor lists are (C2,C2,C3) and (C3,C2,C2). Their products both equal 12 as [L3] requires, and their agreement up to permutation illustrates [L2].

step 1.1step 2.1L1L2L3

Depends on

Used by

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