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Inverse Galois problem
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<h2 id="partial-results">Partial results</h2> <p>Many cases are known. It is known that every finite group is realizable over any <a href="/facts/Function_field_of_an_algebraic_variety/54cc62wP">function field</a> in one variable over the <a href="/facts/Complex_number/2w7mqI7e">complex numbers</a> C {\displaystyle \mathbb {C} } , and more generally over function fields in one variable over any <a href="/facts/Algebraically_closed_field/HuVQClok">algebraically closed field</a> of <a href="/facts/Characteristic_(algebra)/D2VzcQaG">characteristic</a> zero. <a href="/facts/Igor_Shafarevich/n6UINBX1">Igor Shafarevich</a> showed that every finite <a href="/facts/Solvable_group/Na7lKIns">solvable group</a> is realizable over Q {\displaystyle \mathbb {Q} } .<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> It is also known that every simple <a href="/facts/Sporadic_group/hjNHTKVv">sporadic group</a>, except possibly the <a href="/facts/Mathieu_group/ExjDPjWa">Mathieu group</a> <i>M</i>23, is realizable over Q {\displaystyle \mathbb {Q} } .<a class="footnote-ref" id="fnref:3" href="#fn:3"><sup>3</sup></a> </p><p><a href="/facts/David_Hilbert/1vhlHn4b">David Hilbert</a> showed that this question is related to a <a href="/facts/Rationality_question/d1wTEQBR">rationality question</a> for G: </p> If K is any extension of Q {\displaystyle \mathbb {Q} } on which G acts as an <a href="/facts/Automorphism_group/65vImKwe">automorphism group</a>, and the <a href="/facts/Invariant_theory/5fIUh9h3">invariant field</a> <i>KG</i> is rational over Q {\displaystyle \mathbb {Q} } , then G is realizable over Q {\displaystyle \mathbb {Q} } . <p>Here <i>rational</i> means that it is a <a href="/facts/Purely_transcendental/ubxWrbjJ">purely transcendental</a> extension of Q {\displaystyle \mathbb {Q} } , generated by an <a href="/facts/Algebraically_independent/H8wV21eD">algebraically independent</a> set. This criterion can for example be used to show that all the <a href="/facts/Symmetric_group/RUksK5l5">symmetric groups</a> are realizable. </p><p>Much detailed work has been carried out on the question, which is in no sense solved in general. Some of this is based on constructing G geometrically as a <a href="/facts/Galois_covering/bDRGasl7">Galois covering</a> of the <a href="/facts/Projective_line/PhRgf87G">projective line</a>: in algebraic terms, starting with an extension of the field Q ( t ) {\displaystyle \mathbb {Q} (t)} of <a href="/facts/Rational_function/2nJGu0Ko">rational functions</a> in an indeterminate t. After that, one applies <a href="/facts/Hilbert%2527s_irreducibility_theorem/1GCqas32">Hilbert's irreducibility theorem</a> to specialise t, in such a way as to preserve the Galois group. </p><p>All permutation groups of degree 23 or less, except the <a href="/facts/Mathieu_group/ExjDPjWa">Mathieu group</a> <i>M</i>23, are known to be realizable over Q {\displaystyle \mathbb {Q} } .<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a> <a class="footnote-ref" id="fnref:5" href="#fn:5"><sup>5</sup></a> </p><p>All 13 non-<a href="/facts/Abelian_group/FfWwmx4R">abelian</a> <a href="/facts/Simple_group/e5hGF7nT">simple groups</a> smaller than PSL(2,25) (order 7800) are known to be realizable over Q {\displaystyle \mathbb {Q} } .<a class="footnote-ref" id="fnref:6" href="#fn:6"><sup>6</sup></a> </p> <h2 id="a-simple-example-cyclic-groups">A simple example: cyclic groups</h2> <p>It is possible, using classical results, to construct explicitly a <a href="/facts/Polynomial/Lzak8VVx">polynomial</a> whose Galois group over Q {\displaystyle \mathbb {Q} } is the <a href="/facts/Cyclic_group/JL5zfFuL">cyclic group</a> Z/<i>n</i>Z for any positive <a href="/facts/Integer/PUwwrawx">integer</a> n. To do this, choose a <a href="/facts/Prime_number/L20FLkgn">prime</a> p such that <i>p</i> ≡ 1 (mod <i>n</i>); this is possible by <a href="/facts/Dirichlet%2527s_theorem_on_arithmetic_progressions/yFyJaE2d">Dirichlet's theorem</a>. Let Q(<i>μ</i>) be the <a href="/facts/Cyclotomic_field/cGxMMEoR">cyclotomic extension</a> of Q {\displaystyle \mathbb {Q} } generated by μ, where μ is a primitive <i>p</i>-th <a href="/facts/Root_of_unity/EqH0ho1Y">root of unity</a>; the Galois group of Q(<i>μ</i>)/Q is cyclic of order <i>p</i> − 1. </p><p>Since n <a href="/facts/Divisor/DKNa2GvC">divides</a> <i>p</i> − 1, the Galois group has a cyclic <a href="/facts/Subgroup/r91mBgpa">subgroup</a> H of order (<i>p</i> − 1)/<i>n</i>. The <a href="/facts/Fundamental_theorem_of_Galois_theory/srpbXtxW">fundamental theorem of Galois theory</a> implies that the corresponding fixed field, <i>F</i> = Q(<i>μ</i>)<i>H</i>, has Galois group Z/<i>n</i>Z over Q {\displaystyle \mathbb {Q} } . By taking appropriate sums of conjugates of μ, following the construction of <a href="/facts/Gaussian_period/nx3pvsNF">Gaussian periods</a>, one can find an element α of F that generates F over Q {\displaystyle \mathbb {Q} } , and compute its <a href="/facts/Minimal_polynomial_(field_theory)/dJv9Q0Kh">minimal polynomial</a>. </p><p>This method can be extended to cover all finite <a href="/facts/Abelian_group/FfWwmx4R">abelian groups</a>, since every such group appears in fact as a quotient of the Galois group of some cyclotomic extension of Q {\displaystyle \mathbb {Q} } . (This statement should not though be confused with the <a href="/facts/Kronecker%25E2%2580%2593Weber_theorem/U3pq1GAq">Kronecker–Weber theorem</a>, which lies significantly deeper.) </p> <h3>Worked example: the cyclic group of order three</h3> <p>For <i>n</i> = 3, we may take <i>p</i> = 7. Then Gal(Q(<i>μ</i>)/Q) is cyclic of order six. Let us take the generator η of this group which sends μ to <i>μ</i>3. We are interested in the subgroup <i>H</i> = {1, <i>η</i>3} of order two. Consider the element <i>α</i> = <i>μ</i> + <i>η</i>3(<i>μ</i>). By construction, α is fixed by H, and only has three conjugates over Q {\displaystyle \mathbb {Q} } : </p> <i>α</i> = <i>η</i>0(<i>α</i>) = <i>μ</i> + <i>μ</i>6, <i>β</i> = <i>η</i>1(<i>α</i>) = <i>μ</i>3 + <i>μ</i>4, <i>γ</i> = <i>η</i>2(<i>α</i>) = <i>μ</i>2 + <i>μ</i>5. <p>Using the identity: </p> 1 + <i>μ</i> + <i>μ</i>2 + ⋯ + <i>μ</i>6 = 0, <p>one finds that </p> <i>α</i> + <i>β</i> + <i>γ</i> = −1, <i>αβ</i> + <i>βγ</i> + <i>γα</i> = −2, <i>αβγ</i> = 1. <p>Therefore α is a <a href="/facts/Root_of_a_polynomial/mlK3dhoT">root</a> of the polynomial </p> (<i>x</i> − <i>α</i>)(<i>x</i> − <i>β</i>)(<i>x</i> − <i>γ</i>) = <i>x</i>3 + <i>x</i>2 − 2<i>x</i> − 1, <p>which consequently has Galois group Z/3Z over Q {\displaystyle \mathbb {Q} } . </p> <h2 id="symmetric-and-alternating-groups">Symmetric and alternating groups</h2> <p><a href="/facts/David_Hilbert/1vhlHn4b">Hilbert</a> showed that all symmetric and <a href="/facts/Alternating_group/yethKwyo">alternating</a> groups are represented as Galois groups of polynomials with rational <a href="/facts/Coefficient/5Hwq2Wuq">coefficients</a>. </p><p>The polynomial <i>xn</i> + <i>ax</i> + <i>b</i> has discriminant </p> ( − 1 ) n ( n − 1 ) 2 ( n n b n − 1 + ( − 1 ) 1 − n ( n − 1 ) n − 1 a n ) . {\displaystyle (-1)^{\frac {n(n-1)}{2}}\!\left(n^{n}b^{n-1}+(-1)^{1-n}(n-1)^{n-1}a^{n}\right)\!.} <p>We take the special case </p> <i>f</i>(<i>x</i>, <i>s</i>) = <i>xn</i> − <i>sx</i> − <i>s</i>. <p>Substituting a prime integer for s in <i>f</i>(<i>x</i>, <i>s</i>) gives a polynomial (called a specialization of <i>f</i>(<i>x</i>, <i>s</i>)) that by <a href="/facts/Eisenstein%2527s_criterion/ETY6bUAr">Eisenstein's criterion</a> is <a href="/facts/Irreducible_polynomial/PTeuKm4W">irreducible</a>. Then <i>f</i>(<i>x</i>, <i>s</i>) must be irreducible over Q ( s ) {\displaystyle \mathbb {Q} (s)} . Furthermore, <i>f</i>(<i>x</i>, <i>s</i>) can be written </p> x n − x 2 − 1 2 − ( s − 1 2 ) ( x + 1 ) {\displaystyle x^{n}-{\tfrac {x}{2}}-{\tfrac {1}{2}}-\left(s-{\tfrac {1}{2}}\right)\!(x+1)} <p>and <i>f</i>(<i>x</i>, 1/2) can be factored to: </p> 1 2 ( x − 1 ) ( 1 + 2 x + 2 x 2 + ⋯ + 2 x n − 1 ) {\displaystyle {\tfrac {1}{2}}(x-1)\!\left(1+2x+2x^{2}+\cdots +2x^{n-1}\right)} <p>whose second factor is irreducible (but not by Eisenstein's criterion). Only the reciprocal polynomial is irreducible by Eisenstein's criterion. We have now shown that the group Gal(<i>f</i>(<i>x</i>, <i>s</i>)/Q(<i>s</i>)) is <a href="/facts/Doubly_transitive/JAfAegsM">doubly transitive</a>. </p><p>We can then find that this Galois group has a transposition. Use the scaling (1 − <i>n</i>)<i>x</i> = <i>ny</i> to get </p> y n − { s ( 1 − n n ) n − 1 } y − { s ( 1 − n n ) n } {\displaystyle y^{n}-\left\{s\left({\frac {1-n}{n}}\right)^{n-1}\right\}y-\left\{s\left({\frac {1-n}{n}}\right)^{n}\right\}} <p>and with </p> t = s ( 1 − n ) n − 1 n n , {\displaystyle t={\frac {s(1-n)^{n-1}}{n^{n}}},} <p>we arrive at: </p> <i>g</i>(<i>y</i>, <i>t</i>) = <i>yn</i> − <i>nty</i> + (<i>n</i> − 1)<i>t</i> <p>which can be arranged to </p> <i>yn</i> − <i>y</i> − (<i>n</i> − 1)(<i>y</i> − 1) + (<i>t</i> − 1)(−<i>ny</i> + <i>n</i> − 1). <p>Then <i>g</i>(<i>y</i>, 1) has 1 as a <a href="/facts/Double_root/dndUyetA">double zero</a> and its other <i>n</i> − 2 zeros are <a href="/facts/Simple_zero/dndUyetA">simple</a>, and a transposition in Gal(<i>f</i>(<i>x</i>, <i>s</i>)/Q(<i>s</i>)) is implied. Any finite <a href="/facts/Doubly_transitive_permutation_group/JAfAegsM">doubly transitive permutation group</a> containing a transposition is a full symmetric group. </p><p><a href="/facts/Hilbert%2527s_irreducibility_theorem/1GCqas32">Hilbert's irreducibility theorem</a> then implies that an infinite set of rational numbers give specializations of <i>f</i>(<i>x</i>, <i>t</i>) whose Galois groups are <i>Sn</i> over the rational field Q {\displaystyle \mathbb {Q} } . In fact this set of rational numbers is dense in Q {\displaystyle \mathbb {Q} } . </p><p>The discriminant of <i>g</i>(<i>y</i>, <i>t</i>) equals </p> ( − 1 ) n ( n − 1 ) 2 n n ( n − 1 ) n − 1 t n − 1 ( 1 − t ) , {\displaystyle (-1)^{\frac {n(n-1)}{2}}n^{n}(n-1)^{n-1}t^{n-1}(1-t),} <p>and this is not in general a perfect square. </p> <h3>Alternating groups</h3> <p>Solutions for alternating groups must be handled differently for <a href="/facts/Parity_(mathematics)/7zq2UM4V">odd</a> and <a href="/facts/Parity_(mathematics)/7zq2UM4V">even</a> degrees. </p> <h4>Odd degree</h4> <p>Let </p> t = 1 − ( − 1 ) n ( n − 1 ) 2 n u 2 {\displaystyle t=1-(-1)^{\tfrac {n(n-1)}{2}}nu^{2}} <p>Under this substitution the discriminant of <i>g</i>(<i>y</i>, <i>t</i>) equals </p> ( − 1 ) n ( n − 1 ) 2 n n ( n − 1 ) n − 1 t n − 1 ( 1 − t ) = ( − 1 ) n ( n − 1 ) 2 n n ( n − 1 ) n − 1 t n − 1 ( 1 − ( 1 − ( − 1 ) n ( n − 1 ) 2 n u 2 ) ) = ( − 1 ) n ( n − 1 ) 2 n n ( n − 1 ) n − 1 t n − 1 ( ( − 1 ) n ( n − 1 ) 2 n u 2 ) = n n + 1 ( n − 1 ) n − 1 t n − 1 u 2 {\displaystyle {\begin{aligned}(-1)^{\frac {n(n-1)}{2}}n^{n}(n-1)^{n-1}t^{n-1}(1-t)&=(-1)^{\frac {n(n-1)}{2}}n^{n}(n-1)^{n-1}t^{n-1}\left(1-\left(1-(-1)^{\tfrac {n(n-1)}{2}}nu^{2}\right)\right)\\&=(-1)^{\frac {n(n-1)}{2}}n^{n}(n-1)^{n-1}t^{n-1}\left((-1)^{\tfrac {n(n-1)}{2}}nu^{2}\right)\\&=n^{n+1}(n-1)^{n-1}t^{n-1}u^{2}\end{aligned}}} <p>which is a perfect square when n is odd. </p> <h4>Even degree</h4> <p>Let: </p> t = 1 1 + ( − 1 ) n ( n − 1 ) 2 ( n − 1 ) u 2 {\displaystyle t={\frac {1}{1+(-1)^{\tfrac {n(n-1)}{2}}(n-1)u^{2}}}} <p>Under this substitution the discriminant of <i>g</i>(<i>y</i>, <i>t</i>) equals: </p> ( − 1 ) n ( n − 1 ) 2 n n ( n − 1 ) n − 1 t n − 1 ( 1 − t ) = ( − 1 ) n ( n − 1 ) 2 n n ( n − 1 ) n − 1 t n − 1 ( 1 − 1 1 + ( − 1 ) n ( n − 1 ) 2 ( n − 1 ) u 2 ) = ( − 1 ) n ( n − 1 ) 2 n n ( n − 1 ) n − 1 t n − 1 ( ( 1 + ( − 1 ) n ( n − 1 ) 2 ( n − 1 ) u 2 ) − 1 1 + ( − 1 ) n ( n − 1 ) 2 ( n − 1 ) u 2 ) = ( − 1 ) n ( n − 1 ) 2 n n ( n − 1 ) n − 1 t n − 1 ( ( − 1 ) n ( n − 1 ) 2 ( n − 1 ) u 2 1 + ( − 1 ) n ( n − 1 ) 2 ( n − 1 ) u 2 ) = ( − 1 ) n ( n − 1 ) 2 n n ( n − 1 ) n − 1 t n − 1 ( t ( − 1 ) n ( n − 1 ) 2 ( n − 1 ) u 2 ) = n n ( n − 1 ) n t n u 2 {\displaystyle {\begin{aligned}(-1)^{\frac {n(n-1)}{2}}n^{n}(n-1)^{n-1}t^{n-1}(1-t)&=(-1)^{\frac {n(n-1)}{2}}n^{n}(n-1)^{n-1}t^{n-1}\left(1-{\frac {1}{1+(-1)^{\tfrac {n(n-1)}{2}}(n-1)u^{2}}}\right)\\&=(-1)^{\frac {n(n-1)}{2}}n^{n}(n-1)^{n-1}t^{n-1}\left({\frac {\left(1+(-1)^{\tfrac {n(n-1)}{2}}(n-1)u^{2}\right)-1}{1+(-1)^{\tfrac {n(n-1)}{2}}(n-1)u^{2}}}\right)\\&=(-1)^{\frac {n(n-1)}{2}}n^{n}(n-1)^{n-1}t^{n-1}\left({\frac {(-1)^{\tfrac {n(n-1)}{2}}(n-1)u^{2}}{1+(-1)^{\tfrac {n(n-1)}{2}}(n-1)u^{2}}}\right)\\&=(-1)^{\frac {n(n-1)}{2}}n^{n}(n-1)^{n-1}t^{n-1}\left(t(-1)^{\tfrac {n(n-1)}{2}}(n-1)u^{2}\right)\\&=n^{n}(n-1)^{n}t^{n}u^{2}\end{aligned}}} <p>which is a perfect square when n is even. </p><p>Again, Hilbert's irreducibility theorem implies the existence of infinitely many specializations whose Galois groups are alternating groups. </p> <h2 id="rigid-groups">Rigid groups</h2> <p>Suppose that <i>C</i>1, …, <i>Cn</i> are <a href="/facts/Conjugacy_class/AFuPIFNW">conjugacy classes</a> of a finite group G, and A be the set of n-tuples (<i>g</i>1, …, <i>gn</i>) of G such that <i>gi</i> is in <i>Ci</i> and the product <i>g</i>1…<i>gn</i> is trivial. Then A is called rigid if it is <a href="/facts/Empty_set/ffEk3eA6">nonempty</a>, G acts transitively on it by conjugation, and each element of A generates G. </p><p>Thompson (1984) showed that if a finite group G has a rigid set then it can often be realized as a Galois group over a cyclotomic extension of the rationals. (More precisely, over the cyclotomic extension of the rationals generated by the values of the irreducible characters of G on the conjugacy classes <i>Ci</i>.) </p><p>This can be used to show that many finite simple groups, including the <a href="/facts/Monster_group/MlwslQ7H">monster group</a>, are Galois groups of extensions of the rationals. The monster group is generated by a triad of elements of orders 2, 3, and 29. All such triads are conjugate. </p><p>The prototype for rigidity is the symmetric group <i>Sn</i>, which is generated by an n-cycle and a transposition whose product is an (<i>n</i> − 1)-cycle. The construction in the preceding section used these generators to establish a polynomial's Galois group. </p> <h2 id="a-construction-with-an-elliptic-modular-function">A construction with an elliptic modular function</h2> <p>Let <i>n</i> > 1 be any integer. A lattice Λ in the <a href="/facts/Complex_plane/vE0QrxJp">complex plane</a> with period ratio τ has a sublattice Λ′ with period ratio <i>nτ</i>. The latter lattice is one of a finite set of sublattices permuted by the <a href="/facts/Modular_group/wL6l14Hk">modular group</a> PSL(2, Z), which is based on changes of basis for Λ. Let j denote the <a href="/facts/Elliptic_modular_function/H1a38AJF">elliptic modular function</a> of <a href="/facts/Felix_Klein/WmL9guqK">Felix Klein</a>. Define the polynomial <i>φn</i> as the product of the differences (<i>X</i> − <i>j</i>(Λ<i>i</i>)) over the conjugate sublattices. As a polynomial in X, <i>φn</i> has coefficients that are polynomials over Q {\displaystyle \mathbb {Q} } in <i>j</i>(<i>τ</i>). </p><p>On the conjugate lattices, the modular group acts as PGL(2, Z/<i>n</i>Z). It follows that <i>φn</i> has Galois group isomorphic to PGL(2, Z/<i>n</i>Z) over Q ( J ( τ ) ) {\displaystyle \mathbb {Q} (\mathrm {J} (\tau ))} . </p><p>Use of Hilbert's irreducibility theorem gives an infinite (and dense) set of rational numbers specializing <i>φn</i> to polynomials with Galois group PGL(2, Z/<i>n</i>Z) over Q {\displaystyle \mathbb {Q} } . The groups PGL(2, Z/<i>n</i>Z) include infinitely many non-solvable groups. </p> <h2 id="see-also">See also</h2> <ul><li>Semiabelian group (Galois theory)</li></ul> <h2 id="notes">Notes</h2> <ul><li>MacBeath, A. M. (1969). "Extensions of the Rationals with Galois Group PGL(2,Zn)". <i>Bulletin of the London Mathematical Society</i>. 1 (3): 332–338. <a href="/facts/Doi_(identifier)/muM9Etpq">doi</a>:<a href="https://doi.org/10.1112%2FBLMS%2F1.3.332">10.1112/BLMS/1.3.332</a>.</li> <li><a href="/facts/John_G._Thompson/M6tF5WNE">Thompson, John G.</a> (1984), "Some finite groups which appear as Gal L/K, where K ⊆ Q(μ n)", <i>Journal of Algebra</i>, 89 (2): 437–499, <a href="/facts/Doi_(identifier)/muM9Etpq">doi</a>:<a href="https://doi.org/10.1016%2F0021-8693%2884%2990228-X">10.1016/0021-8693(84)90228-X</a>, <a href="/facts/MR_(identifier)/uP137L11">MR</a> <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=0751155">0751155</a></li> <li>Helmut Völklein, <i>Groups as Galois Groups, an Introduction</i>, Cambridge University Press, 1996. ISBN 978-0521065030 .</li> <li><a href="/facts/Jean-Pierre_Serre/1xSr7BCh">Serre, Jean-Pierre</a> (1992). <i>Topics in Galois Theory</i>. Research Notes in Mathematics. Vol. 1. Jones and Bartlett. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 0-86720-210-6. <a href="/facts/Zbl_(identifier)/P6rFxKKx">Zbl</a> <a href="https://zbmath.org/?format=complete&q=an:0746.12001">0746.12001</a>.</li> <li>Gunter Malle, Heinrich Matzat, <i>Inverse Galois Theory</i>, Springer-Verlag, 1999, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 3-540-62890-8.</li> <li>Gunter Malle, Heinrich Matzat, <i>Inverse Galois Theory</i>, 2nd edition, Springer-Verlag, 2018.</li> <li>Alexander Schmidt, Kay Wingberg, <i><a href="https://web.archive.org/web/20050830000818/http://www.math.uiuc.edu/Algebraic-Number-Theory/0136/">Safarevic's Theorem on Solvable Groups as Galois Groups</a></i> (<i>see also</i> <a href="/facts/J%25C3%25BCrgen_Neukirch/nzVc4G3F">Neukirch, Jürgen</a>; Schmidt, Alexander; Wingberg, Kay (2000), <i>Cohomology of Number Fields</i>, <i>Grundlehren der Mathematischen Wissenschaften</i>, vol. 323, Berlin: Springer-Verlag, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-3-540-66671-4, <a href="/facts/MR_(identifier)/uP137L11">MR</a> <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=1737196">1737196</a>, <a href="/facts/Zbl_(identifier)/P6rFxKKx">Zbl</a> <a href="https://zbmath.org/?format=complete&q=an:0948.11001">0948.11001</a>)</li> <li>Christian U. Jensen, Arne Ledet, and <a href="/facts/Noriko_Yui/Ipcy5MRk">Noriko Yui</a>, <i>Generic Polynomials, Constructive Aspects of the Inverse Galois Problem</i>, Cambridge University Press, 2002.</li></ul> <h2 id="external-links">External links</h2> <ul><li><a href="https://web.archive.org/web/20230216095734/https://people.maths.bris.ac.uk/~matyd/InvGal/">"Inverse Galois Problem PCMI 2021 Graduate Summer School Program - Number Theory Informed by Computation - July 26-30, 2021"</a>. Archived from <a href="https://people.maths.bris.ac.uk/~matyd/InvGal/">the original</a> on 2023-02-16.</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>"Mathematical Sciences Research Institute Publications 45" (PDF). MSRI. Archived from the original (PDF) on 2017-08-29. Retrieved 2016-04-17. <a href="https://web.archive.org/web/20170829003635/http://library.msri.org/books/Book45/files/book45.pdf" target="_blank">https://web.archive.org/web/20170829003635/http://library.msri.org/books/Book45/files/book45.pdf</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>Igor R. Shafarevich, The imbedding problem for splitting extensions, Dokl. Akad. Nauk SSSR 120 (1958), 1217-1219. <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> <li id="fn:3"><p>p. 5 of Jensen et al., 2002 <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li> <li id="fn:4"><p>"Home". galoisdb.math.upb.de. <a href="http://galoisdb.math.upb.de/" target="_blank">http://galoisdb.math.upb.de/</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li> <li id="fn:5"><p>"17T7 is a Galois group over the rationals". <a href="https://arxiv.org/abs/2411.07857" target="_blank">https://arxiv.org/abs/2411.07857</a> <a href="#fnref:5" class="footnote-back-ref">↩</a></p></li> <li id="fn:6"><p>Malle and Matzat (1999), pp. 403-424 <a href="#fnref:6" class="footnote-back-ref">↩</a></p></li> </ol>