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    Series, solvability, and nilpotence

    A composition series of a finite group \(G\) is a chain \(1=G_0\trianglelefteq G_1\trianglelefteq\cdots\trianglelefteq G_r=G\) with every factor \(G_{i+1}/G_i\) simple; by the Jordan–Hölder theorem, the factors, with multiplicity and up to isomorphism, do not depend on the series. A group is solvable if its derived series \(G\trianglerighteq[G,G]\trianglerighteq[[G,G],[G,G]]\trianglerighteq\cdots\) reaches \(1\); a finite group is solvable if and only if its composition factors are cyclic of prime order. A group is nilpotent if its lower central series \(G\trianglerighteq[G,G]\trianglerighteq[G,[G,G]]\trianglerighteq\cdots\) reaches \(1\). Nilpotent groups are solvable, and \(S_3\) is solvable and not nilpotent. A finite group is nilpotent if and only if every Sylow subgroup is normal, if and only if it is the direct product of its Sylow subgroups. The symmetric group \(S_n\) is solvable if and only if \(n\le4\).

    Normal series

    Let \(G\) be a group. A normal series of \(G\) is a chain of subgroups \[ G = G^0 \supseteq G^1 \supseteq G^2 \supseteq \cdots \] such that \(G^{i+1}\) is a normal subgroup of \(G^i\) for every \(i\).

    Composition series

    Let \(G\) be a group with identity \(e\). A composition series of \(G\) is a finite normal series \[ G = G^0 \geq G^1 \geq \cdots \geq G^n = \theset{e}, \qquad G^{i+1}\normal G^i, \] such that for each \(0\leq i<n\), \(G^{i+1}\) is a maximal proper normal subgroup of \(G^i\).

    Composition series of a group

    Let \(G\) be a group. A composition series of \(G\) is a finite sequence of subgroups \[ 1=N_{0} \leq N_{1} \leq \cdots \leq N_{k-1} \leq N_{k}=G \] such that \(N_i \normal N_{i+1}\) and \(N_{i+1}/N_i\) is simple for \(0\leq i<k\). The quotients \(N_{i+1}/N_i\) are the composition factors of the series.

    What is a composition series of a group $G$?

    Jordan--Hölder theorem

    Composition factors are unique up to isomorphism and reordering.Suggest a slogan

    Let \(G\) be a group, and let \[ G=G_0 \trianglerighteq G_1 \trianglerighteq \cdots \trianglerighteq G_r=\theset{e} \quad\text{and}\quad G=H_0 \trianglerighteq H_1 \trianglerighteq \cdots \trianglerighteq H_s=\theset{e} \] be two composition series of \(G\). Then \(r=s\), and there is a permutation \(\pi\) of \(\theset{1,\ldots,r}\) such that \(G_{i-1}/G_i\cong H_{\pi(i)-1}/H_{\pi(i)}\) for every \(i\).

    Commutator subgroup

    Let \(G\) be a group. For \(a, b\in G\), the commutator of \(a\) and \(b\) is \([a,b] \coloneqq aba^{-1}b^{-1}\). The commutator subgroup of \(G\) is the subgroup generated by all commutators: \[ [G, G] \coloneqq \gens{ aba^{-1} b^{-1} \st a, b\in G }. \]

    Let \(G\) be a group. Then \([G,G]\) is a normal subgroup of \(G\) and \(G/[G,G]\) is abelian. If \(N\normal G\) and \(G/N\) is abelian, then \([G, G]\leq N\).

    An element of \([G, G]\) is a product of commutators, and need not itself be a commutator.

    Derived series

    Let \(G\) be a group. Set \(G^{(0)} \coloneqq G\) and \(G^{(i+1)} \coloneqq [G^{(i)}, G^{(i)}]\), the commutator subgroup of \(G^{(i)}\), for \(i\geq 0\). The derived series of \(G\) is the chain \(G^{(0)} \geq G^{(1)} \geq G^{(2)} \geq \cdots\).

    Solvable group

    Let \(G\) be a group with identity \(e\). The group \(G\) is solvable if there is a finite normal series \[ \theset{e} = G_0 \normal G_1 \normal \cdots \normal G_n = G \] such that \(G_{i}/G_{i-1}\) is abelian for every \(1\leq i\leq n\).

    Solvable finite group

    A finite group \(G\) is solvable if it has a composition series whose composition factors are all abelian.

    When is a group solvable, in terms of its composition factors?

    Solvability via the derived series

    A group is solvable exactly when repeated commutator subgroups eventually become trivial.Suggest a slogan

    A group \(G\) is solvable if and only if its derived series reaches the trivial group, that is, \(G^{(n)}=\theset{e}\) for some \(n\geq 0\).

    Groups of order $pqr$ are solvable

    A group whose order is the product of three distinct primes is solvable.Suggest a slogan

    Let \(p < q < r\) be primes and let \(G\) be a group with \(\abs{G} = pqr\). Then \(G\) is solvable.

    Central series

    Let \(G\) be a group with identity \(e\). A central series of \(G\) is a finite chain of subgroups \[ G = G^0 \geq G^1 \geq \cdots \geq G^n = \theset{e} \] with each \(G^i \normal G\) and \([G, G^i] \leq G^{i+1}\) for \(0\leq i < n\), where \([G, G^i]\) is the subgroup generated by the commutators \(ghg^{-1}h^{-1}\) with \(g\in G\) and \(h\in G^i\).

    For subgroups \(G^{i+1}\leq G^i\) normal in \(G\), the condition \([G, G^i] \leq G^{i+1}\) holds if and only if \(G^i/G^{i+1}\) is contained in the center of \(G/G^{i+1}\).

    Lower central series

    Let \(G\) be a group. Set \(G^0 \coloneqq G\) and \(G^{i+1} \coloneqq [G, G^i]\) for \(i\geq 0\), where \([G, G^i]\) is the subgroup generated by the commutators \(ghg^{-1}h^{-1}\) with \(g\in G\) and \(h\in G^i\). The lower central series of \(G\) is the chain \(G^0 \geq G^1 \geq \cdots\).

    The series is descending, while the upper central series is ascending. A group \(G\) with identity \(e\) is nilpotent if and only if \(G^n = \theset{e}\) for some \(n\geq 0\).

    Lower central series

    Let \(G\) be a group. Set \(G^0 \coloneqq G\) and \(G^{i+1} \coloneqq [G, G^i]\) for \(i\geq 0\), where \([G, G^i]\) is the subgroup generated by the commutators \(ghg^{-1}h^{-1}\) with \(g\in G\) and \(h\in G^i\). The lower central series of \(G\) is the chain \(G^0 \geq G^1 \geq \cdots\).

    Upper central series

    Let \(G\) be a group. Define normal subgroups \(Z_i\normal G\) inductively by \(Z_0\coloneqq 1\) and, for \(i\geq 0\), \[ Z_{i+1}\coloneqq \pi_i^{-1}\bigl(Z(G/Z_i)\bigr), \] where \(\pi_i\colon G\to G/Z_i\) is the quotient map and \(Z(G/Z_i)\) is the center of \(G/Z_i\). The upper central series of \(G\) is the chain \(Z_0\leq Z_1\leq Z_2\leq\cdots\).

    The induction is well defined: if \(Z_i\normal G\), then \(Z(G/Z_i)\) is normal in \(G/Z_i\), so its preimage \(Z_{i+1}\) is normal in \(G\). \(Z_{i+1}\) is the unique subgroup of \(G\) containing \(Z_i\) with \(Z_{i+1}/Z_i=Z(G/Z_i)\); in particular \(Z_1=Z(G)\).

    Nilpotent group

    A group \(G\) is nilpotent if its upper central series \(Z_0\leq Z_1\leq\cdots\) satisfies \(Z_n = G\) for some \(n\geq 0\).

    Nilpotent groups via the lower central series

    A group is nilpotent exactly when its lower central series reaches the identity.Suggest a slogan

    Let \(G\) be a group with identity \(e\) and lower central series \(G^0\geq G^1\geq\cdots\). Then \(G\) is nilpotent if and only if \(G^n=\theset{e}\) for some \(n\geq0\).

    Nilpotent groups via the upper central series

    A group is nilpotent exactly when its upper central series reaches the whole group.Suggest a slogan

    Let \(G\) be a group with upper central series \(Z_0\leq Z_1\leq\cdots\). Then \(G\) is nilpotent if and only if \(Z_n=G\) for some \(n\geq 0\).

    A finite group is nilpotent if and only if its Sylow subgroups are normal

    A finite group is nilpotent exactly when every Sylow subgroup is normal.Suggest a slogan

    Let \(G\) be a finite group. Then \(G\) is nilpotent if and only if, for every prime \(p\) dividing \(\abs{G}\), the Sylow $p$-subgroup of \(G\) is normal.

    A finite group is nilpotent if and only if its maximal subgroups are normal

    A finite group is nilpotent exactly when every maximal subgroup is normal.Suggest a slogan

    A finite group \(G\) is nilpotent if and only if every maximal subgroup of \(G\) is normal in \(G\).

    Normalizers grow

    Let \(G\) be a group. Normalizers grow in \(G\) if for every proper subgroup \(H < G\), the subgroup \(H\) is a proper subgroup of its normalizer \(N_G(H)\).