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  "body": "\n## Proof\n\nWe argue by strong induction on the length $n$ of the first series.\n\n1. Let $(X_1, \\dots, X_n)$ and $(Y_1, \\dots, Y_m)$ be finite lists of groups. Call them equivalent when $n = m$ and there is a suitable permutation $\\sigma$ of $\\{1, \\dots, n\\}$. It must make $X_i$ [[isomorphic|085PND4KNT]] to $Y_{\\sigma(i)}$ for each $i$.\n2. This relation is symmetric and transitive: invert or compose the permutations and the isomorphisms. A reordering of a list is equivalent to it. If two lists are equivalent, then they stay equivalent after the same group is added at the end of both.\n3. For a [[composition series|0NB8VPB93V]], the factor list is $(G_1/G_0, \\dots, G_n/G_{n-1})$. Let $P(n)$ be the following claim. For every group $G$, a composition series of length $n$ and any composition series of $G$ have equivalent factor lists. This is the theorem.\n4. Each factor $G_i/G_{i-1}$ is [[simple|0C093D1T4M]], so it is not trivial and $G_{i-1} \\neq G_i$. Hence a composition series has length $0$ exactly when $G = \\{e\\}$.\n5. Base case $n = 0$. Then $G = \\{e\\}$, so by step 4 also $m = 0$. Two empty lists are equivalent.\n6. Let $n \\geq 1$ and assume $P(n')$ for all $n' \u003c n$. By step 4, $G \\neq \\{e\\}$ and $m \\geq 1$. Let $A = G_{n-1}$ and $B = H_{m-1}$. Both are [[normal|0W77CVBCHQ]] in $G$, and $G/A$ and $G/B$ are simple.\n7. Case $A = B$. The series $G_0, \\dots, G_{n-1}$ and $H_0, \\dots, H_{m-1}$ are composition series of $A$, the first of length $n - 1$. By $P(n-1)$, their factor lists are equivalent. Adding $G/A = G/B$ to both and using step 2 proves the claim in this case.\n8. Case $A \\neq B$. By the [[second isomorphism theorem|0PPP2J2DJV]], $AB$ is a subgroup of $G$. It is normal in $G$, because $g(ab)g^{-1} = (gag^{-1})(gbg^{-1})$ for $g \\in G$, $a \\in A$ and $b \\in B$.\n9. By the [[correspondence theorem|0S9ZVZ34MD]], $AB/A$ is a normal subgroup of the simple group $G/A$. So $AB/A$ is trivial or equal to $G/A$. By the bijection in that theorem, $AB = A$ or $AB = G$.\n10. Suppose $AB = A$, so $B \\subseteq A$. As in step 9, $A/B$ is normal in the simple group $G/B$, so $A = B$ or $A = G$. But $A \\neq B$, and $A \\neq G$ because $G/A$ is not trivial. So $AB = G$.\n11. Let $D = A \\cap B$. By the second isomorphism theorem with $S = A$ and $N = B$, $D$ is normal in $A$ and $A/D \\cong AB/B = G/B$. In the same way, $D$ is normal in $B$ and $B/D \\cong G/A$. So $A/D$ and $B/D$ are simple.\n12. We build a composition series of $D$. For $0 \\leq i \\leq n-1$, let $D_i = D \\cap G_i$, a subgroup by [[this result|0ZTGMDNVTB]]. Then $D_0 = \\{e\\}$ and $D_{n-1} = D \\cap A = D$.\n13. Fix $1 \\leq i \\leq n-1$. Apply the second isomorphism theorem in the group $G_i$, with $S = D_i$ and the normal subgroup $N = G_{i-1}$. Since $D_i \\cap G_{i-1} = D_{i-1}$, we get $D_{i-1} \\trianglelefteq D_i$ and $D_i/D_{i-1} \\cong D_i G_{i-1}/G_{i-1}$.\n14. The subgroup $D_i$ is normal in $G_i$. Indeed, for $x \\in G_i \\subseteq A$ we have $xDx^{-1} = D$, since $D \\trianglelefteq A$, and $xG_ix^{-1} = G_i$.\n15. As in step 8, $D_i G_{i-1}$ is then a normal subgroup of $G_i$ that contains $G_{i-1}$. By the correspondence theorem, $D_i G_{i-1}/G_{i-1}$ is normal in the simple group $G_i/G_{i-1}$, so it is trivial or all of $G_i/G_{i-1}$.\n16. By steps 13 and 15, each factor $D_i/D_{i-1}$ is trivial or simple. Delete each $D_i$ with $D_i = D_{i-1}$. What remains is a composition series of $D$ of some length $k$, with a factor list $L$.\n17. Adding $A$ on top gives a composition series of $A$ of length $k + 1$, with factor list $(L, A/D)$, by step 11. The series $G_0, \\dots, G_{n-1}$ is a composition series of $A$ of length $n - 1$.\n18. By $P(n-1)$ in the group $A$, the lists $(G_1/G_0, \\dots, G_{n-1}/G_{n-2})$ and $(L, A/D)$ are equivalent. In particular, $k + 1 = n - 1$.\n19. Adding $B$ on top of the series of $D$ gives a composition series of $B$ of length $k + 1 = n - 1$, with factor list $(L, B/D)$. The series $H_0, \\dots, H_{m-1}$ is another composition series of $B$.\n20. By $P(n-1)$ in the group $B$, the lists $(L, B/D)$ and $(H_1/H_0, \\dots, H_{m-1}/H_{m-2})$ are equivalent. In particular, $m - 1 = n - 1$, so $m = n$.\n21. By steps 2, 11 and 18, the factor list of the $G_i$ is equivalent to $(L, A/D, G/A)$, hence to $(L, G/B, G/A)$.\n22. By steps 2, 11 and 20, the factor list of the $H_j$ is equivalent to $(L, B/D, G/B)$, hence to $(L, G/A, G/B)$.\n23. The lists in steps 21 and 22 differ only by the order of the last two entries. By step 2, the two factor lists are equivalent. So $n = m$, and there is a permutation $\\sigma$ with $G_i/G_{i-1} \\cong H_{\\sigma(i)}/H_{\\sigma(i)-1}$ for each $i$.\n\n## Idea\n\nCompare the two maximal normal subgroups $A$ and $B$ just below $G$. If they are equal, the induction hypothesis applies to $A$ directly. If not, their intersection $D$ is a common maximal normal subgroup of both, with $A/D \\cong G/B$ and $B/D \\cong G/A$. One series of $D$ then links the two series of $G$, and the two top factors simply swap places.\n",
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