Let \(G\) be a (not necessarily finite) group and \(H\) and \(K\) normal subgroups such that \(G = HK\text{.}\) Prove that \(G/(H\cap K)\cong G/H \times G/K\text{.}\)
Solution.
Let \(f:G\to G/H\times G/K\) defined such that \(f(g)=(gH, gK)\text{.}\) Let \(x,y\in G\text{.}\) Then
Let \((xH,yK)\in G/H\times G/K\text{.}\) As \(x,y\in G=HK\text{,}\) we have \(x=hk\) and \(y=h'k'\) with \(h,h'\in H\) and \(k,k'\in K\text{.}\) As \(k'\in K\) and \(h\in H\) we see \((xH,yK)=(kH,h'K)\text{.}\)
Let \(g\in H\cap K\text{.}\) Then \(f(g)=(gH,gK)=(H,K)\text{,}\) and so \(g\in\ker(f)\text{.}\) Let \(g\in\ker(f)\text{.}\) Then \(f(g)=(H,K)\text{,}\) so \(gH=H\) and \(gK=K\text{,}\) placing \(g\in H\cap K\text{.}\) Thus, by the First Isomorphism Theorem, we have \(G/(H\cap K)\cong G/H \times G/K.\)
ActivityC.147.Problem 2.
Suppose \(G\) is a finite group which has precisely one subgroup of order \(d\) for each divisor \(d\) of \(|G|\text{.}\) Prove that \(G\) is cyclic.
Solution.
First, suppose that \(G\) is a \(p\)-group. Let \(g\in G\) have biggest order. Let \(h\in G\text{.}\) So \(|h|\bigg||g|\text{.}\) Since \((g)\leq G\text{,}\) it also has exactly one subgroup for each divisor. But (h) has the same order as one of those subgroups, so they must be the same group. So \(h\in(g)\text{.}\) Since \(h\) was arbitrary, then \(G\leq (g)\text{.}\) So when \(G\) is a \(p\)-group then it is cyclic.
If its not a \(p\)-group then we can decompose \(|G|\) into relatively prime powers of primes, all of which are \(p\)-groups and maintain this property. Thus \(G\) is the product of relatively prime cyclic groups, making it cyclic itself.
ActivityC.148.Problem 3.
Let \(G\) be a group of order \(90 = 2 · 3^2 · 5\) and let \(\Syl_3(G)\) denote the set of Sylow \(3\)-subgroups of \(G\text{.}\)
Suppose for any \(Q, Q' \in \Syl_3(G)\) either \(Q = Q'\) or \(Q \cap Q' = {1}.\) Prove that \(G\) is not simple.
Suppose there exists \(Q, Q' ∈ \Syl_3(G)\) such that \(|Q \cap Q'| = 3\text{.}\) Prove that \(G\) is not simple. (Hint: Consider the normalizer of \(Q \cap Q'\text{.}\))
Solution.
Let \(G\) be a group of order \(90 = 2 · 3^2 · 5\) and let \(\Syl_3(G)\) denote the set of Sylow \(3\)-subgroups of \(G\text{.}\)
Suppose by way of contradiction that \(G\) is simple. By Sylow’s Theorems we know the following:
\(n_3|10\) and is congruent to \(1\mod{3}\text{.}\) As \(G\) is simple there must be ten of them.
\(n_5|18\) and is congruent to \(1\mod 5\text{.}\) As \(G\) is simple there must be six of them.
Since each Sylow \(3\)-subgroup will have \(3^2-1=8\) non-identity elements and they are all distinct that accounts for \(80\) elements of order \(3\text{.}\) However, there are also \(24\) elements or order \(5\) to account for, which is a problem. Thus \(G\) cannot be simple.
Suppose now that there exists \(Q, Q' ∈ \Syl_3(G)\) such that \(|Q \cap Q'| = 3\text{.}\) Given this intersection, we know that \(|QQ'|=9\cdot 9/3=27\text{.}\) Additionally, note that as \(Q \cap Q'\) is a subgroup of both \(Q\) and \(Q'\) with index 3 in both, the smallest prime dividing the order or both, that \(Q \cap Q'\) is normal in both.
We now consider the normalizer of \(Q \cap Q'\text{,}\) which we denote \(N\) for simplicity. As \(N\) is a subgroup of \(G\) its order must divide \(90\) by Lagrange’s Theorem. However, \(QQ'\leq N\text{,}\) and so \(N\) must have at least \(27\) elements, leaving the options of \(45\) and \(90\text{.}\) If \(|N|=90\) then \(N=G\text{,}\) making \(Q\cap Q'\) normal in \(G\text{,}\) a problem.
If \(|N|=45\) then \([G:N]=2\text{,}\) the smallest prime dividing \(G\text{.}\) Thus \(N\) is still normal, which is still a problem. Thus \(G\) cannot be simple.
SubsectionSection
ActivityC.149.Problem 4.
Let \(F\) be a field of characteristic \(p > 0\text{,}\)\(a \in F\) , and consider the polynomial \(f(x)=x^p -x-a \in F[x]\text{.}\)
Prove that \(f(x)\) is either irreducible over \(F\) or it splits into distinct linear factors over \(F\text{.}\) (Hint: If \(\alpha\) is a root of \(f(x)\text{,}\) consider \(\alpha + j\) for \(j \in \Z/p\text{.}\))
Suppose \(f(x)\) is irreducible over \(F\) and let \(L\) be a splitting field of \(f\) over \(F\text{.}\) Prove that the Galois group of \(L\) over \(F\) is cyclic.
Solution.
Let \(F\) be a field of characteristic \(p > 0\text{,}\)\(a \in F\text{,}\) and consider the polynomial \(f(x)=x^p -x-a \in F[x]\text{.}\)
Suppose \(f\) has a root, \(\alpha\text{,}\) in \(F\text{.}\) Then \(\alpha^p-\alpha-a=0\text{.}\) Consider \(\alpha+j\) for some \(j\in\Z/p\text{,}\) and observe \((\alpha+j)^p-(\alpha+j)-a\text{.}\) By The Freshman’s Dream, we have \(\alpha^p+j^p-\alpha-j-a\text{,}\) but as \(\alpha^p-\alpha-a=0\text{,}\) we really have \(j^p-j\text{.}\) By Fermat’s Little Theorem, \(p|j^p-j\text{,}\) and thus \(f(\alpha+j)=0\text{.}\) Thus we have found \(p\) roots of \(f\text{,}\) and thus \(f\) splits into linear factors.
Suppose then that no root of \(f\) exists in \(F\text{.}\) Let \(L\) be a splitting field of \(f\) over \(F\text{,}\) and note that from the above paragraph we have \(L=F(\alpha)\text{.}\) As \(f'(x)=1\text{,}\) we see \(\gcd(f,f')=1\text{,}\) and thus \(f\) is separable. Hence \(F(\alpha)\) is a Galois extension. Thus there exists a \(\sigma\in\Gal(L/F)\) such that \(\sigma(\alpha)\neq\alpha\text{.}\) So \(\sigma(\alpha)=\alpha+j\) for some \(j\in\Z/p\text{.}\) Notice \(\sigma(\alpha+j)=\sigma(\alpha)+j=\alpha+2j\text{.}\) As \(p\) is prime, we see that \(|j|=p\text{,}\) and thus we need to apply \(\sigma\) to \(\alpha\)\(p\) times in order to get back to \(\alpha\text{.}\) Thus \(\sigma(\alpha)^p=\alpha\text{,}\) so \(|\sigma|=p\text{.}\) Thus \(|\Gal(L/F)|=p\text{.}\) Thus the minimum polynomial of \(\alpha\) must have degree \(p\text{.}\) As \(\alpha\) is a root of \(f\) and \(f\) is monic, it must be the minimal polynomial and is thus irreducible.
Suppose \(f(x)\) is irreducible over \(F\) and let \(L\) be a splitting field of \(f\) over \(F\text{.}\) Let \(\alpha\) be a root of \(f.\) Consider \(F(\alpha)\text{.}\) By part (a), \(F(\alpha)\) contains all the roots of \(f\text{,}\) hence \(L=F(\alpha)\text{.}\) As \(f\) is monic and irreducible, it is the minimum polynomial of \(f\text{,}\) and thus \([L:F]=p\text{.}\) Hence \(|\Gal(L/F)|=p\text{.}\) All groups of prime order are cyclic, completing the proof.
ActivityC.150.Problem 5.
Let \(\a = \sqrt2 + \sqrt5 \in\R\text{.}\)
Find the minimal polynomial \(f(x)\) of \(\a\) over \(\Q\text{.}\)
Let \(E\) be the splitting field of \(f(x)\) over \(\Q\text{.}\) Find the Galois group \(G\) of \(E/\Q\text{.}\)
Find all subgroups of \(G\) and generators for the corresponding intermediate fields of \(E/\Q\text{.}\)
which factors as two irreducible polynomials and has no roots in \(\Q\text{,}\) making it irreducible. Thus \(f\) is the minimal polynomial of \(\a\text{.}\)
Notice that \(E=\Q(\sqrt{ 2 },\sqrt{ 5 })\) which also has degree \(4\text{.}\) Let \(G\) denote the Galois group of \(E/\Q\text{.}\) Thus \(G\) is a group of order \(4\text{,}\) making it isomorphic to \(\Z_2\times\Z_2\) or \(\Z_4\text{.}\) Notice that the elements of \(G\) are the following:
All of these automorphisms have degree \(2\text{,}\) making \(G\cong\Z_2\times\Z_2\text{.}\)
Thus are only two subgroups of \(G\text{,}\)\(H\cong\Z_2\) and \(K\cong\Z_2\text{.}\) The first, \(H\text{,}\) corresponds to \(\Q(\sqrt{ 2 })\) and is generated by \(\sqrt{ 2 }\text{,}\) where \(K\) corresponds to \(\Q(\sqrt{ 5 })\) and is generated by \(\sqrt{ 5 }\text{.}\)
ActivityC.151.Problem 6 (*).
Let \(F\) be any field, and let \(x, y\text{,}\) and \(t\) be indeterminates. Prove in detail that \(F[x,y]/(y^2-x^3)\) and \(F[t^2,t^3]\) are isomorphic rings.
Solution.
Coming Soon!
SubsectionSection III: Linear Algebra and Modules
ActivityC.152.Problem 7 (*).
Consider the following matrix over the complex numbers:
Show that the characteristic polynomial of \(A\) is \((t-2)^{4}\text{.}\)
Find the Jordan canonical form \(J\) of \(A\text{.}\)
Find an invertible matrix \(P\) such that \(P^{-1} A P=J\)
Solution.
Coming Soon!
ActivityC.153.Problem 8 (*).
Let \(N\) be a submodule of an \(R\)-module \(M\text{.}\) Using Zorn’s Lemma, prove that there is a submodule \(N'\) such that
\(N\cap N' = (0)\text{,}\) and
\(N''\cap(N + N')\neq(0)\) for every non-zero submodule \(N''\) of \(M\text{.}\)
Solution.
Coming Soon!
ActivityC.154.Problem 9 (*).
Let \(A\) be a square matrix over the field \(\C\) of complex numbers.
Suppose \(A\) is invertible. Prove that there is a square matrix \(B\) over \(\C\) such that \(B^2=A\text{.}\) (Hint: Reduce to the case that \(A = I + N\) where \(I\) is the identity matrix and \(N\) is a nilpotent matrix.)
Show by example that (1) can fail if \(A\) is not assumed to be invertible.
