Let \(G\) be a group. A subgroup \(H\) of \(G\) is called a characteristic subgroup of \(G\) if \(\a(H)=H\) for every automorphism \(\a\) of \(G\text{.}\) Show that if \(H\) is a characteristic subgroup of \(N\) and \(N\) is a normal subgroup of \(G\text{,}\) then \(H\) is a normal subgroup of \(G\text{.}\)
Solution.
Let \(G\) be a group, \(N\) is a normal subgroup of \(G\text{,}\) and \(H\) a characteristic subgroup of \(N\text{.}\) Let \(g\in G\) and consider the automorphism \(a_g:G\to G\) given by \(a_g(g')=gg'g^{-1}\text{.}\)
Let \(n\in N\) and notice \(a_g(n)=gng\inv\in N\text{,}\) as \(N\nsg G\text{.}\) Thus \(a_g:N\to N\) is well defined. Let \(n\in N\text{.}\) As \(gNg\inv=N\text{,}\) we can write \(n=gn'g\inv\) for some \(n'\in N\text{.}\) Then \(a_g(n')=gn'g\inv=n\text{,}\) making \(a_g\) surjective. As \(|N|=|N|\) we see that \(a_g\) is a bijection. The homomorphism piece we get for free from \(G\text{,}\) making \(a_g\in\Aut(N)\text{.}\)
Let \(h\in H\text{.}\) As \(H\) is a characteristic subgroup of \(N\text{,}\) we see that \(a_g(h)=ghg\inv=h\) for all \(h\in H\) and for all \(g\in G\text{.}\) Thus \(H\) is normal in \(G\text{.}\)
ActivityC.84.Problem 2.
Let \(G\) be a group acting on a set \(A\text{,}\) and let \(H\) be a subgroup of \(G\) satisfying that the induced action of \(H\) on \(A\) is transitive (that is, for all a\(, b\in A\) there is an \(h\in H\) with \(h\cdot a = b\)). Let \(t\in A\text{,}\) and let \(\Stab_G(t)\) be the stabilizer of \(t\) in \(G\text{.}\) Show that \(G = H\Stab_G(t)\text{.}\)
Solution.
Let \(G\) be a group acting on a set \(A\text{,}\) and let \(H\) be a subgroup of \(G\) satisfying that the induced action of \(H\) on \(A\) is transitive. Let \(t\in A\text{,}\) and let \(\Stab_G(t)\) be the stabilizer of \(t\) in \(G\text{.}\)
Let \(g\in G\text{.}\) Note that \(e\in\Stab_G(t)\) for all \(t\in A\text{.}\) If \(g\in H\) then \(g=ge\text{.}\) Suppose then that \(g\not\in H\text{.}\) Suppose \(gt=a\) for some \(a\in A\text{.}\) As \(H\) acts transitively on \(A\) there exists some \(h\) such that \(ht=a\text{.}\) So \(h\inv gt=a\text{,}\) placing \(h\inv g\in\Stab(t)\text{.}\) Thus \(g=h(h\inv g)\text{,}\) so \(G=H\Stab_G(t)\text{.}\)
ActivityC.85.Problem 3.
Let \(G\) be a simple group of order \(60\text{.}\) Determine the number of elements of \(G\) of order \(5\text{.}\)
Show that there is no simple group of order \(30\text{.}\)
Solution.
Let \(G\) be a simple group of order \(60=2^2\cdot3\cdot 5\text{.}\) By Sylow’s Theorems we know that \(n_5|12\) and that \(n_5\equiv 1\mod{5}\text{.}\) Thus the options for \(n_5\) are \(1\) and \(6\text{.}\) Since \(G\) is simple we see that \(n_5=6\text{.}\) As each Sylow \(5\)-subgroup of \(G\) has \(4\) unique elements of order \(5\) and the identity we see that the number of elements or order \(5\) in \(G\) is \((5-1)\cdot 6=24\text{.}\)
Suppose by way of contradiction that \(G\) is a simple group of order \(30\text{.}\) Similarly to above, \(n_5=6\text{,}\) yielding \(24\) elements of order \(5\text{.}\) Now, \(n_3|10\) and \(n_3\equiv 1\mod{3}\text{,}\) so \(n_3=10\text{,}\) yielding far too many elements to fit in \(G\text{.}\)
SubsectionSection
ActivityC.86.Problem 4.
Let \(d\) be a square-free integer. The ring \(\Q(\sqrt d)\) is the subring of \(\C\) defined by
\begin{equation*}
\Q(\sqrt d) = \{a + b\sqrt d | a, b \in \Q\}.
\end{equation*}
Show that there is a ring isomorphism \(\Q[x]/(x^2-d)\cong\Q(\sqrt d)\text{.}\)
Solution.
Let \(f:\Q[x]\to\Q[\sqrt d]\) such that \(f(x)=\sqrt{d}\text{.}\)
Let \(a+b\sqrt d\in\)\(\Q(\sqrt d)\text{,}\) and notice that \(f(a+bx)=a+b\sqrt d\text{,}\) giving us surjectivity. Also,
Let \(K=\ker(f)\text{,}\) and let \(p=a+bx\in K\text{.}\) As \(\Q[x]\) is a division ring we can write \(p(x)=(x^2-d)q(x)+r(x)\) for some \(p(x),r(x)\) such that \(r(x)=0\) or \(\deg r(x)<\deg (x^2-d)\text{.}\) Thus \(r(x)\) has degree \(1\text{,}\) making \(r(x)=sx+t\text{.}\)
ActivityC.87.Problem 5.
Let \(A\) be a \(\Z\)-module and let \(n\) be any integer. Show that there is a \(\Z\)-module isomorphism
\begin{equation*}
\Hom_\Z(\Z/n\Z, A)\cong A_n, \text{ where }A_n = \{a \in A | na = 0\}
\end{equation*}
is a submodule of \(A\text{.}\) (Note: You may use the fact that \(A_n\) is a \(\Z\)-module without proof.)
Solution.
In \(\mathbb{Z}\)-modules, \(f\) is a module homomorphism iff \(f\) is a group homomorphism. One direction is trivial since it is in the definition. To show that for \(\mathbb{Z}\)-modules a group homomorphism is a module homomorphism let \(f\) be a group homomorphism. Let \(n\in \mathbb{Z}\text{.}\)\(f(ng)=f(\underbrace{g+g+\cdots +g)}_\text{n times}=\underbrace{f(g)+\cdots f(g)}_\text{n times}=nf(g)\) so \(f\) preserves scalar multiplication. It already preserved addition, so therefore it as a \(\mathbb{Z}\)-module homomorphism. Thus for \(\mathbb{Z}\)-modules we can utilize the UMP for cyclic groups when our abelian group is cyclic, as is \(\mathbb{Z}/n\mathbb{Z}\text{.}\) Every \(h\) in \(\text{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},A)\) is uniquely defined by where \(h\) sends 1. Specifically, it has to go to an element in \(A\) such that \(na=0\text{.}\) So let \(f:\text{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},A)\to A_n\) such that \(f(h)=h(1)\text{.}\) First we will show this is well defined. Since by assumption \(nh(1)=0\) and \(h\) maps to elements in \(A\text{,}\)\(h(1)\in A_n\text{.}\) So it is well-defined. Now to show that \(f\) is one-to-one and onto. For one-to-one let \(f(h)=0\) and therefore \(h(1)=0\text{.}\) If the generator is sent to 0, then everything is sent to 0, so \(f\) is the zero map, and thus the kernel is trivial. For onto let \(a\in A_n\text{.}\) Since it is in \(A_n\) we know that \(na=0\) so we can define a group/module homomorphism \(f:\mathbb{Z}/n\mathbb{Z}\to A\) by \(f(1)=a\) due to the UMP. \(h(f)=f(1)=a\) so it is onto. Bijective module homomorphisms are isomorphisms so we have that they are isomorphic.
ActivityC.88.Problem 6.
Determine all similarity classes of matrices with entries in \(\Q\) with characteristic polynomial \((x^4-1)(x^2-1)\text{.}\) Provide an explicit representative for each of these similarity classes.
Solution.
Let \(A\) be a matrix with entries in \(\Q\) with characteristic polynomial \((x^4- 1)(x^2 -1)\text{.}\)
Every matrix is similar to a unique matrix in RCF by Corollary 15.17. Note that RCF is based on the invariant factors if a matrix, and thus if two matrices have the same invariant factors they will have the same RCF, making them both similar to the same (unique) matrix, making them similar to each other. By part (1) of this [[Theorem – Cayley-Hamilton Theorem|Theorem]], the characteristic polynomial of a matrix is equal to the product of the invariant factors of that same matrix.
Recall that the invariant factors must divide all preceding invariant factors in RCF (See: Rational Canonical Form), and observe that \((x^4-1)\) factors as \((x^2-1)(x^2+1)\) and \((x^2-1)\) factors as \((x-1)(x+1)\text{.}\) Given this information, after some fiddling with the factors, we find four possible options for invariant factors of \(A\text{:}\)
\(x^6-x^4-x^2+1=(x^2 -1)(x^4- 1)\text{,}\)
\((x^2-1),(x^4-1)\text{,}\)
\((x+1), (x-1)(x^4-1)\text{,}\) and
\((x-1),(x+1)(x^4-1)\text{.}\)
Let \(f(x)=x^6-x^4-x^2+1, g(x)=(x^4-1),\)\(h(x)=x^2-1\text{,}\)\(j(x)=(x-1),\) and \(k(x)=(x+1)\text{.}\) Observe the companion matrices of each of these polynomials:
Behold: explicit representatives of each similarity class:
\(C(f)\text{,}\)
\(C(g)\oplus C(h)\text{,}\)
\(C(k)\oplus C(gj)\text{,}\) and
\(C(j)\oplus C(gk)\text{.}\)
SubsectionSection
ActivityC.89.Problem 7.
Let \(p\) be a prime integer and let \(\a\) be a root of the polynomial \(x^3-p\text{.}\)
Find, with justification, the degree of the field extension \(\Q(\a, i)\) over \(\Q\text{.}\)
Deduce that the polynomial \(x^3-p\) is irreducible in \(\Q(i)[x]\text{.}\)
Solution.
Let \(p\) be a prime integer and let \(\a\) be a root of the polynomial \(f(x)=x^3-p\text{.}\)
First, notice that by Eisenstein’s Criterion (\(p=p\)) we have \(f\) irreducible in \(\Q[x]\text{.}\)
Let \(\z\) be a primitive \(3^{\text {rd}}\) root of unity. The roots of \(f\) are the following: \(\a_1=\sqrt[3]p\cdot\z\text{,}\)\(\a_2=\sqrt[3]p\cdot\z^2\text{,}\) and \(\a_3=\sqrt[3]p\cdot\z^3=\sqrt[3]p\)
\([\Q(\sqrt[3]p):\Q]=3\text{.}\)\([\Q(\sqrt[3]p,i):\Q\sqrt[3]p]=2\) Total is \(6\) by The Degree Formula.
The minimal polynomial degree being equal to the degree of the field extension implies that it is irreducible.
ActivityC.90.Problem 8.
Show that any finite extension of fields \(K/F\) is algebraic.
Let \(\overline{\Q}\) denote the subfield of \(\C\) consisting of all the complex numbers which are algebraic over \(\Q\text{.}\) (You may use that \(\overline{\Q}\) is a field without proof.) Show that \(\overline{\Q}/\Q\) is an algebraic extension, but not a finite extension.
Solution.
Let \(K/F\) be a finite extension of fields. Let \(\a\in L\text{.}\) By the The Degree Formula we have \([K:F(a)]=[K:F(\a)][F(\a):F]\text{.}\) Thus \([F(\a):F]\) is finite, making \(\a\) algebraic over \(F\text{.}\)
Notice that \(\a_n\sqrt[n]2\in\overline{\Q}\) for all \(n\in\N\) and that \(\a_n\) is a root of the polynomial \(x^n-2\text{,}\) which is irreducible in \(\Q[x]\) by Eisenstein’s Criterion (\(p=2\)). Thus \([\Q(\a_n):\Q]=n\text{.}\) As \(\a_n\) must be added to \(\Q\) for all \(n\text{,}\) we see that this extension is not finite. To show \(\bar{\mathbb{Q}}\) is an algebraic extension, we need to show that for each \(\alpha\in \bar{\mathbb{Q}}\) there exists and \(f(x)\in \mathbb{Q}[x]\) such that \(\alpha\) is a root of \(f\text{.}\) But this is true by the definition of \(\bar{\mathbb{Q}}\text{.}\) To show this let \(n\geq 1\) be an integer. By CITEX Eisenstein’s \(x^n-p\) for some prime \(p\) is irreducible over \(\mathbb{Q}\text{.}\) Therefore \(\sqrt[n]{p}\in \bar{\mathbb{Q}}\) and \([\mathbb{Q}(\sqrt[n]{p}):\mathbb{Q}]=n\implies [E:\mathbb{Q}]\geq n\) for each \(n\geq 1\) therefore \([E:\mathbb{Q}]=\infty\text{.}\)
ActivityC.91.Problem 9.
Let \(K\) be a Galois extension of \(F\) with \(|\Gal(K/F)| = 20\text{.}\)
Prove that there exists a subfield \(E\) of \(K\) containing \(F\) with \([E : F ] = 5\text{.}\)
Determine whether there must also exist a subfield \(L\) of \(K\) containing \(F\) with \([L : F ] = 10\text{.}\)
Solution.
Let \(K\) be a Galois extension of \(F\) with \(|\Gal(K/F )| = 20\text{.}\)
Let \(P\) be a Sylow \(2\)-subgroup of \(\Gal(K/F)\text{.}\) Notice that \(|P|=4\text{.}\) By the Fundamental Theorem of Galois Theory there exists an intermediate field extension \(E\) such that \(|\Gal(K/E)|=4\text{.}\) By the The Degree Formula, we have \([K:F]=[K:E][E:F]\text{,}\) with \([K:F]=20\) and \([K:L]=4\text{.}\) Thus \([L:F]=5\text{.}\)
Notice that as \(2\) is a prime dividing the order of \(\Gal(K/F )\) there must exist an element of order \(2\) by Cauchy’s Theorem. The cyclic subgroup, \(H\text{,}\) generated by this element has order \(2\text{.}\) By the Fundamental Theorem of Galois Theory there exists an intermediate field extension \(L\) such that \(|\Gal(K/L)|=2\text{.}\) By the The Degree Formula, we have \([K:F]=[K:L][L:F]\text{,}\) with \([K:F]=20\) and \([K:L]=2\text{.}\) Thus \([L:F]=2\text{.}\)
