Let \(G\) be a group of order \(150\) and suppose by way of contradiction that \(G\) is simple. Notice that \(150=2\cdot 3\cdot 5^2\text{.}\) By Sylow’s Theorems, we know \(|\Syl_5(G)|\equiv 1\mod{5}\) and divides \(6\text{,}\) the only options are thus \(1\) and \(6\text{.}\) Since \(G\) is simple, there must be exactly \(n_5=6\text{,}\) where \(n_5=\Syl_5(G)\text{.}\) Let \(G\) act on \(\Syl_5(G)\) by conjugation. Thus \(\rho: G\to S_6\) is a group homomorphism. Note that \(|S_6|=720\text{,}\) and that the order of \(G\) does not divide \(720\text{.}\) By Part (2) of Sylow’s Theorems this action is transitive, meaning that the kernel of \(\rho\) cannot be trivial. Thus \(\ker(\rho)\) is a nontrivial normal subgroup of \(G\text{,}\) a contradiction. Thus no group of order \(150\) is simple.
ActivityC.183.Problem 2.
Let \(G\) be a finite group.
If \(N\) is a normal subgroup of \(G\) and \(|N | = 2\text{,}\) prove that \(N\) is contained in the center \(Z(G)\) of \(G\text{.}\)
Suppose that \(|Z(G)|\) is odd and that \(G\) contains a non-trivial simple subgroup \(H\) with \([G : H] = 2\text{.}\) Prove that \(H\) is the only non-trivial proper normal subgroup of \(G\text{.}\)
Solution.
Let \(x\in N\) and let \(g\in G\text{.}\) As \(N\) is normal, we see \(gxg\inv\in N\text{,}\) and thus one of the following must be true:
\(gxg\inv=e\text{,}\) where \(e\) is the identity element of \(G\text{,}\) or
\(gxg\inv=x\text{,}\) the only other element of \(N\text{.}\)
However, in the first case we would have \(x=e\) by multiplying \(g\) and \(g\inv\) over, and thus it must be the case that \(gxg\inv=x\text{,}\) or \(gx=xg\text{.}\) Thus \(x\in Z(G)\text{.}\)
From Lagrange’s Theorem we know that \(|G|/|H|=2\text{,}\) meaning that \(G\) has an even number of elements. Thus \(2\) is the smallest prime dividing the order of \(G\text{,}\) making \(H\) normal in \(G\) CITEX.
Suppose by way of contradiction there exists some non-trivial proper normal subgroup \(N\) of \(G\) that is not \(H\text{.}\) As \(H\) is normal, by the Second Isomorphism Theorem we have \(H\cap N\nsg H\text{.}\) However, as \(H\) is simple, this means that \(H\cap N=\{e\}\text{.}\) The Second Isomorphism Theorem also tells us that \(HN\leq G\text{.}\) As \(|H|=|G|/2\text{,}\) this makes \(|N|=2\) and \(G=HN\text{.}\) From part (1), \(N\leq Z(G)\text{.}\) However, this contradicts Lagrange’s Theorem, as \(2\) does not divide any odd numbers.
ActivityC.184.Problem 3.
Let \(H\) be a normal subgroup of a finite group \(G\text{,}\)\(p\) a prime dividing the order of \(H\text{,}\) and \(P\) a Sylow \(p\)-subgroup of \(H\text{.}\) Prove that \(G = H N_G(P )\)
Hint.
For \(g \in G\text{,}\) consider the subgroup \(gPg\inv\text{.}\)
Solution.
First, note that \(N_G(P)=\{g\in G|gPg\inv=P\}\text{.}\)
Let \(G\) act on \(\Syl_p(G)\) by conjugation, which is a transitive action by part (2) of Sylow’s Theorems. Therefore \(H\) acts transitively on this set as well. Under this action, \(N_G(P)=\Stab_G(P)\text{.}\)
Let \(g\in G\text{,}\) and let \(x\in N_G(P)\text{.}\) Consider \(g\inv x\text{.}\) As the action by \(H\) is transitive there exists some \(h\in H\) such that \(hg\inv\cdot x=x\text{.}\) This means that \(hg\inv\) stabilizes \(x\text{.}\) Then \((hg\inv)\inv=gh\inv\) stabilizes \(x\) as well, so \(gh\inv\cdot x=x\text{.}\) But notice that \(gh\inv h=g\text{,}\) where \(gh\inv\in N_G(P)\) and \(h\in H\text{.}\) Thus \(G = H N_G(P )\text{.}\)
ActivityC.185.Problem 4.
Fix a prime number \(p\text{,}\) and let \(A\) denote the abelian group of all complex roots of unity whose orders are powers of \(p\text{;}\) that is
\begin{equation*}
A = \{z \in \C | z^{p^n} = 1 \text{ for some integer } n \leq 1\} .
\end{equation*}
Prove the following statements.
Every non-trivial subgroup of \(A\) contains the group of \(p^{\text{th}}\) roots of unity.
Every proper subgroup of \(A\) is cyclic.
If \(B\) and \(C\) are subgroups of \(A\text{,}\) then either \(B \subseteq C\) or \(C \subseteq B\text{.}\)
For each \(n \geq 0\) there exists a unique subgroup of \(A\) with \(p^n\) elements.
Solution.
Let \(H\) be a non-trivial subgroup of \(A\text{.}\) Then there exists some \(z\) such that \(z^{p^n}=1\text{.}\) Then \(z^{p^{n-1}}\) yields a primitive \(p\th\) root of unity, which can be used to generate the other roots as well. Thus \(H\) contains the \(p\th\) roots of unity.
Suppose \(H\) is a proper subgroup of \(A\text{,}\) meaning it is missing some \(p^{n{\th}}\) root of unity. But the subgroup of those roots of unity is cyclic and is generated by every element, so that entire subgroup must be missing. But that subgroup can be generated with any primitive root of a higher power of \(p\text{,}\) so \(H\) must be finite and there must be some element of maximum order, which can be used to generate the whole group. Thus \(H\) is cyclic.
Suppose \(B\) and \(C\) are subgroups of \(A\) such that \(B\not\subseteq C\text{.}\) Then \(C\) is a proper subgroup of \(A\text{,}\) making it finite as seen above. If there exists a higher power of \(p\) in \(B\) then it generates \(C\text{.}\)
Let \(n\geq 0\text{.}\) Then the subgroup generated by the \(p^{n^{\th}}\) roots of unity have \(p^n\) elements, and it is unique since it is generated by every such root.
SubsectionSection II: Rings and Fields
ActivityC.186.Problem 5 (*).
Let \(F\) be a field, and let \(f (x) ∈ F [x]\text{.}\) Recall that \(f (x)\) is separable provided, for every extension field \(K/F , f (x)\) has no multiple roots in \(K\text{.}\) (A multiple root is an element \(\alpha \in K\) such that \((x - \alpha)^2 | f (x)\) in \(K[x])\text{.}\)
Prove that \(f (x) ∈ F [x]\) is separable if and only if \(f (x)\) and its derivative \(f'(x)\) are relatively prime in \(F [x]\text{.}\)
Suppose that \(f (x)\) is irreducible and that the degree of \(f (x)\) is not a multiple of the characteristic of \(F\text{.}\) Prove that \(f (x)\) is separable.
Solution.
If two polynomials factor completely into linear factors, then they are relatively prime if and only if they have no linear factors in common, which is equivalent to their having no roots in common. Thus they cannot be generated by the same thing
Coming soon!
ActivityC.187.Problem 6.
Let \(E\) be the splitting field of the polynomial \(x^7 - 12\) over \(\Q\text{.}\) Find \([E : \Q]\text{,}\) and describe the elements of \(\Gal(E/\Q)\) explicitly.
Solution.
First, notice that \(f(x)=x^7-12\) is irreducible in \(\Q\text{,}\) using Eisenstein’s Criterion and \(p=3\text{.}\) By the Factor Theorem there are at most \(7\) distinct roots of \(f\text{.}\) Notice that \(\sqrt[7]{12}\) is one such root. The roots of \(f\) are \(\alpha_i=\sqrt[7]{12}e^{{(2\pi i/7)}^{2i-1}}\) for \(1\leq i\leq 7\text{.}\)
Given that \(f\) is irreducible and monic in \(\Q\text{,}\) this makes \(f\) the minimal polynomial of \(\sqrt[7]{12}\) in \(\Q\) and thus that \([\Q(\sqrt[7]{12}):\Q]=7\text{.}\) Take a seventh root of unity, cyclo time, degree \(6\text{,}\) for a grand total of \(42\text{,}\) which is \(2\) times \(3\) times \(7\text{.}\)
ActivityC.188.Problem 7 (*).
Let \(p\) be a prime number, let \(\F_p\) be the field with \(p\) elements, and let \(n\) be a positive integer. Prove that \(x^{p^n}- x =\prod_{d|n} g_d(x)\text{,}\) where, for a positive divisor \(d\) of \(n\text{,}\)\(g_d(x)\) denotes the product of all monic irreducible polynomials of degree \(d\) in \(\F_p[x]\text{.}\) (You may assume basic results on the structure of finite fields and their subfields.)
Solution.
Coming Soon!
ActivityC.189.Problem 8.
Let \(R\) be a commutative ring with \(1\text{.}\) Recall that \(r \in R\) is called nilpotent if \(r^n = 0\) for some integer \(n > 0\text{.}\) Let \(N\) be the set of nilpotent elements of \(R\text{.}\)
Show that \(N\) is an ideal of \(R\text{.}\)
Show that the ring \(R/N\) has no nonzero nilpotent elements.
For a polynomial \(f(x) \in R[x]\text{,}\) prove that \(f(x)\) is a nilpotent element of \(R[x]\) if and only if every coefficient of \(f(x)\) is nilpotent.
Solution.
Let \(R\) be a commutative ring with \(1\) and Let \(N\) be the set of nilpotent elements of \(R\text{.}\)
Let \(r,s\in N\text{.}\) Then \(r^n=0\) and \(s^m=0\) for \(n,m\in\N\text{.}\) Consider \((r+s)^{mn}\text{,}\) which, by the binomial theorem, has each term raised to either the \(n\)th or \(m\)th power, sending the whole thing to \(0\text{.}\)
Additionally, if \(r\in N\) consider \(r\inv=-r\) when viewed as an additive group. Notice \(-r^{2n}=r(2n)=0\text{.}\) Finally, let \(x\not\in N\) and consider \((rx)^n=r^nx^n=0\text{,}\) so \(rx\in N\text{.}\) Thus we have our ideal.
Let \(rN\) be a nilpotent element in \(R/N\text{.}\) Then \(r^nN=N\text{,}\) meaning \(r^n\in N\text{.}\) This means there exists some \(m\in\N\) such that \(r^{n^m}=r^{nm}=0\text{,}\) placing \(r\in N\text{.}\) Thus all nilpotent elements in \(R/N\) are actually \(0\) (possibly four \(0\)’s in a trench coat).
Let \(f\in R[x]\text{.}\)
First, suppose \(f\) is a nilpotent element of \(R[x]\text{.}\) Thus \(f^n=0\) for some \(n\in\N\text{.}\) By the binomial shenanigans every product of coefficients must to go \(0\text{.}\)
Next, suppose every coefficient of \(f\) is nilpotent. Let \(n\) be equal to the product of all the smallest powers that send each coefficient to \(0\text{.}\) Thus by more binomial shenanigans we have \(f^n=0\text{.}\)
Show that the characteristic and minimal polynomials of \(A\) are, respectively, \((x + 1)^4\) and \((x + 1)^3\text{.}\)
Find the rational canonical form \(R\) of \(A\) and the Jordan canonical form \(J\) of \(A\text{.}\)
Find an invertible matrix \(P\) such that \(P\inv AP = J\text{.}\)
Solution.
Coming soon!
As \(\cp_A(x)=(x + 1)^4\) and \(\mp_A(x)=(x + 1)^3\text{,}\) we know that our characteristic polynomial factors into linear factors, so \(A\) does indeed have a JCF. This also means that \(A\) only has one elementary divisor, \((x + 1)^4\text{,}\) which corresponds to the Jordan Block \(J_4(-1)\text{,}\) a matrix with \(-1\)s along the diagonal, \(1\)s along the subdiagonal, and \(0\) everywhere else. This is the JCF of \(A\text{.}\)
Coming soon!
ActivityC.191.Problem 10.
Let \(A\) and \(B\) be \(n\times n\) matrices with entries in \(F\text{.}\) Recall that \(A\) and \(B\) are said to be similar over \(F\) if there exists an invertible \(n\times n\) matrix, with entries in \(F\text{,}\) such that \(B = P\inv AP\text{.}\) Prove the following statements about matrices \(A\) and \(B\) with entries in \(F\text{:}\)
If \(F \subseteq K\) is a field extension, and \(A\) and \(B\) are similar over \(K\text{,}\) then they are similar over \(F\text{.}\)
\(A\) is similar over \(F\) to its transpose \(A^T\text{.}\)
Solution.
Suppose \(A\) and \(B\) are similar in \(\Mat_{n \times n}(K)\text{.}\) As \(A\) and \(B\) have entries in \(F\text{,}\) then they are both in \(\Mat_{n \times n}(F)\text{.}\) Thus there exist matrices \(C,D\in\Mat_{n \times n}(F)\) in RCF such that \(A\) is similar to \(C\) and that \(B\) is similar to \(D\text{.}\) However, \(A\) is similar to \(C\) and that \(B\) is similar to \(D\) in \(\Mat_{n \times n}(K)\) as well. Notice \(C\) and \(D\) are still in RCF. However, as the RCF is unique, this means that \(C=D\) in \(\Mat_{n \times n}(K)\text{,}\) making them equal in \(\Mat_{n \times n}(F)\) as well. Thus \(A\) is similar to \(B\text{,}\) as similarity is transitive.
Let \(L\) be the algebraic closure of \(F\text{.}\) Thus \(A\) has a Jordan Canonical Form in \(L\text{.}\) For each Jordan block \(J_i\) in the JCF of \(A\text{,}\) let \(B_i\) denote the transpose of the identity matrix, and notice that \(B_iJ_iB_i\inv=J_i^T\text{.}\) As this is the case for every Jordan block, we see that the JCF of \(A\text{,}\)\(J\text{,}\) is similar to its transpose. As the \(A\) is similar to \(J\text{,}\)\(A^T\) is similar to \(J^T\text{,}\) and \(J\) is similar to \(J^T\text{,}\) we see that \(A\sim A^T\) in \(L\) by transitivity. From the first part of this problem, this yields \(A\sim A^T\) in \(F\text{.}\)
ActivityC.192.Problem 11.
Let \(R\) be a commutative ring with \(1\neq 0\text{,}\) and let \(f : R^m\to R^n\) be a surjective homomorphism of free \(R\)-modules. Prove that \(m \leq n\text{.}\)
Solution.
Let \(I\) be a maximal ideal in \(R\text{.}\) Thus \(R/I\) is a field. Lemma 1.58 CITEX tells us that \(M/IM\) and \(N/IN\) are \(R/I\)-vector spaces. Additionally, this gives rise to \(\overline{p}:M/IM\to N/IN\text{,}\) which is a surjective \(R/I\)-module linear transformation.
Note that \(M\) is generated by \(e_i+IM\) for \(i\leq m\text{.}\) Let \(a_i\in R\) and consider \(\sum a_ie_i=0\text{.}\) For this to be \(0\) we need it to be in \(IM\text{,}\) and thus all \(a_i\in I\text{.}\) So the set of \(e_i\) is a basis for \(M/IM\) with \(m\) elements. Likewise \(N/IN\) has a basis with \(n\) elements. As we are surjective, \(\Rank=n\text{,}\)\(\dim=m\text{.}\) So by CITEX Rank-Nullity \(\dim(\ker)=m-n\) which is only positive with \(m\geq n\text{.}\)
