(i.e., \(N\) is the intersection of all the conjugates of K\()\)
Prove \(N\) is a the largest normal subgroup of \(G\) that is contained in \(K\text{.}\)
Prove \([G : N ]\) divides \(n!\text{.}\)
Solution.
Let \(G\) act on the left cosets of \(K\) in \(G\) by left multiplication, yielding the permutation representation homomorphism \(\rho:G\to S_n\text{.}\) Let \(N\) conspicuously denote the kernel of this function.
Let \(x\in N\text{.}\) Then \(\rho(n)\text{,}\) the automorphism \(\s:G/K\to G/K\) defined by \(\s(gK)=xgK\) is precisely the identity permutation. Thus \(xgK=gK\) for all \(g\in G\) and \(g\inv xgK=K\) for all \(g\in G\text{,}\) so \(x\in\text{.}\) This means the elements of \(N\) are precisely those that are in the conjugacy class of \(gKg\inv\) for all \(g\in G\text{.}\) There cannot exist a larger normal subgroup of \(G\) contained in \(K\text{,}\) as it would contain an element that was not in some conjugacy class for \(G\text{,}\) negating the definition of a normal subgroup.
Recall the permutation representation homomorphism \(\rho:G\to S_n\text{.}\) The First Isomorphism Theorem tells us \(G/N\cong\im(\rho)\leq S_n\text{,}\) which has order \(n!\text{.}\) Thus \([G:N]|n!\text{.}\)
ActivityC.120.Problem 2.
Prove that if \(G\) is a finite group of odd order, then for any non-identity element \(x\in G\text{,}\)\(x\) is not conjugate to \(x\inv\text{.}\)
Solution.
Let \(G\) be a finite group of odd order, and let \(G\) act on itself via conjugation. Under this action, the orbit of an element is exactly its conjugacy class. By the [cross-reference to target(s) "cor-orbit-stabilizer" missing or not unique] we have \(|G|=|\Orb(x)|\cdot|\Stab(x)|\text{.}\) Thus the order of an element divides the order of the group, and so every orbit must have odd order.
Suppose there exists an element \(x\) such that \(gxg\inv=x\inv\) for some \(g\in G\text{.}\) Thus \(x\inv\in\Orb(x)\text{.}\) Let \(y\in\Orb(x)\text{.}\) Thus \(y=hxh\inv\) for some \(h\in G\text{.}\) Then \(y\inv=h\inv x\inv h\text{.}\) However, as \(x\inv=gxg\inv\text{,}\) we have \(y\inv=h\inv gxg\inv h\text{.}\) As \(h\inv g\) and \(g\inv h\) are both in \(G\text{,}\) we see that \(y\inv\in\Orb(x)\) as well. As \(e\in\Stab(x)\text{,}\) we see that this means that the orbit of \(x\) would have even order, which is not possible.
ActivityC.121.Problem 3 (*).
Groups of order 75.
Prove that if \(G\) is a group of order \(75\) that contains an element of order \(25\text{,}\) then \(G\) is cyclic.
Prove there exists a non-abelian group of order \(75\text{.}\)
Solution.
Let \(G\) be a group of order \(75\text{.}\)
Suppose \(G\) has an element of order \(25\text{.}\) Let \(H\) be the cyclic subgroup generated by this element, and notice that \([G:H]=3\text{,}\) the smallest prime dividing the order of \(G\text{.}\) Thus \(H\nsg G\text{.}\) Let \(K\) be a Sylow \(3\)-subgroup of \(G\text{.}\) Notice that this group has order \(3\) and is thus cyclic as well. Thus we have \(G\cong H\sdp_{\rho}K\text{,}\) where \(\rho:K\to\Aut(H)\) is a homomorphism. Note that \(\Aut(H)\cong\Z^\times_{25}\text{,}\) which is cyclic of order \(20\text{.}\) Since \(|K| = 3\) and \(\gcd(3,20) = 1\text{,}\) we conclude that \(\rho\) must be the trivial map. Hence, \(G = H\times K\text{,}\) which is a direct product of cyclic groups of relatively prime order. Hence, \(G\) is cyclic.
Coming soon!
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ActivityC.122.Problem 4 (*).
Let \(L\) be the splitting field over \(\Q\) of the polynomial \(x^3-7\text{.}\)
Find all intermediate fields \(E\) with \(\Q \sse E \sse L\) (including possibly \(L\) and \(\Q\)) such that \(E\) is Galois over \(\Q\text{.}\)
For each field \(E\) you found in (a), find with justification a primitive generator (i.e., find \(\a \in E\) so that \(E = \Q(\a))\text{.}\)
Solution.
Let \(L\) be the splitting field over \(\Q\) of the polynomial \(f(x)=x^3-7\)
First, notice that \(f\) is irreducible in \(\Q[x]\) by Eisenstein’s Criterion\((p=7)\text{.}\) Let \(\z\) denote a primitive third root of unity. The roots of \(f\) are the following: 1. \(\a_1=\sqrt[3]{7}\z\text{,}\) 2. \(\a_2=\sqrt[3]{7}\z^2\text{,}\) and 3. \(\a_3=\sqrt[3]{7}\z^3=\sqrt[3]{7}\text{.}\) As \(f\) is irreducible and monic we see that it is the minimal polynomial of \(\a_3\) over \(\Q\text{.}\) Let \(E_1=\Q(\a_3)\) and notice \([E_1:\Q]=3\text{.}\)
Recall that \(G=\Gal(L/\Q)\) is isomorphic to a subgroup of \(S_3\text{.}\) As \(E_1\sse \R\) and \(\a_1\not\in\R\text{,}\) we see another extension is needed, and that extension will have at least degree \(2\text{.}\) Thus, due to size constraints, we see \(G\cong S_3\text{.}\)
By the Fundamental Theorem of Galois Theory each Galois intermediate extension between \(\Q\) and \(L\) corresponds to a normal subgroup of \(G\text{,}\) which are the normal subgroups of \(S_3\text{.}\)
The elements of \(S_3\) are the following: 1. \((1)\) 2. \((12)\) 3. \((13)\) 4. \((23)\) 5. \((123)\) 6. \((132)\) The subgroup \(F=\langle(123)\rangle=\{(123),(132)\}\) has index \(2\) in \(G\) and is thus normal. None of the order \(2\) subgroups are normal in \(S_3\text{,}\) so \(F\) is the only strictly intermediate extension.
Recall \([E_1:\Q]=3\text{,}\) meaning \(|\Gal(L/E_1)|=3\text{,}\) so \(E\) corresponds to a subgroup of order \(2\) in \(G\text{,}\) so its not Galois unfortunately. However, \(E_2=\Q(\z)\) is a degree \(2\) extension that is an intermediate field, as \(\z\) is a root of the irreducible polynomial \(x^2+x+1\text{.}\)
With all this in mind, notice: - \(\Q\) is a splitting field of \(\Q\text{,}\) and has the primitive generator by \(1\text{.}\) - \(E_2\) is our only strictly intermediate field, and has the primitive generator \(\z\) - Finally, \(L\) is Galois over \(\Q\text{,}\) and has the primitive generator \(\z+\sqrt[3]{7}\text{.}\)
Coming soon!
ActivityC.123.Problem 5.
Let \(p\) be any positive prime integer.
Prove that if \(p = k^2 + 1\) for some integer \(k\text{,}\) then \(p\) is not an irreducible element of \(\Z[i]\text{.}\)
Prove \(x^4-p\) is irreducible in \(\Q(i)[x]\text{.}\)
Solution.
Let \(p\) be any positive prime integer.
Suppose \(p = k^2 + 1\) for some integer \(k\text{.}\) As \((k-i)(k+i)=p\text{,}\)\(p\) is not an irreducible element of \(\Z[i]\text{.}\)
As \(x^4-p\) is irreducible in \(Q\) (Eisenstein’s Criterion, \(p=p\)) and \(x^4-p\) is monic, we see that \([\Q(\sqrt[4]{p}):\Q]=4\text{.}\) As \(x^2+1\) is irreducible and monic, we see that \([\Q(\sqrt[4]{p},i):\Q(\sqrt[4]{p})]=2\text{.}\) By the The Degree Formula we have \([\Q(\sqrt[4]{p},i):\Q]=8\text{.}\) Note that the four roots of \((x^4 - p)\) are the following: - \(\sqrt[4]{p}\text{,}\) - \(-\sqrt[4]{p}\text{,}\) - \(\sqrt[4]{p}i,\) and - \(-\sqrt[4]{p}i\text{.}\) As \(\frac{\sqrt[4]{p}i}{\sqrt[4]{p}}=i\text{,}\) we see that \(\Q(\sqrt[4]{p})\) is the splitting field of \(F\text{.}\) Note then that as \([\Q(i):\Q]=2\) and the degree of the splitting field is 8, by the The Degree Formula we see that the degree of the splitting field of \(x^4-p\) over \(\Q(i)=4\text{.}\) As \(x^4-p\) is monic and degree \(4\text{,}\) it must be the minimal polynomial of the second extension and thus irreducible in \(\Q(i)[x]\text{.}\)
ActivityC.124.Problem 6.
For fields \(E\) and \(F\text{,}\) we say \(E\) is a finite splitting field over \(F\) if \(E\) is the splitting field over \(F\) of some polynomial \(f(x)\in F [x]\text{.}\) Assume \(E\) and \(K\) are both finite splitting fields over \(\Q\) and prove \(E \cap K\) is also a finite splitting field over \(\Q\text{.}\)
Solution.
Let \(E\) and \(K\) be finite splitting fields over \(\Q\text{.}\) So there exist polynomials \(f,g\in F[x]\) such that \(f\) splits in \(F\) and \(g\) splits in \(E\text{.}\) Note that both \(E\) and \(F\) are finite extensions, as each can be reached by adjoining each of the roots of \(g\) and \(f\text{,}\) respectively.
If \(E\cap F=\Q\) then we have our splitting field, as \(\Q\) is a splitting field over itself.
Let \(h\) be an irreducible polynomial in \(F[x]\) with a root in \(E\cap K\text{.}\) Then \(h\) has a root in \(E\) and a root in \(F\text{.}\) If a splitting field has one root of a polynomial it has them all, and so we see that \(h\) splits in both of these fields. Thus it splits in \(E\cap K\text{.}\)
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ActivityC.125.Problem 7.
Let \(F\) be a field and recall that a square matrix \(A\) with entries in \(F\) is called nilpotent if \(A^m = 0\) for some positive integer \(m\text{.}\)
Prove that if \(A\) is an \(n \times n\) nilpotent matrix, then \(A^n = 0\text{.}\)
Assume \(n \leq 3\) and prove that two \(n \times n\) nilpotent matrices are similar if and only if they have the same rank. (Recall the rank of a matrix is the dimension of the vector space spanned by its columns.)
Provide an example, with justification, of two \(4 \times 4\) nilpotent matrices that have the same rank but are not similar.
Solution.
Let \(F\) be a field.
Let \(A\) be an \(n \times n\) nilpotent matrix. Let \(\lambda\) be some eigenvalue of \(A\text{.}\) Thus there exists some vector \(v\) such that \(A\lambda=\lambda v\text{.}\) Let’s consider this the base-case of some rather banal induction. Now assume that for \(n\) we have \(A^nv=\lambda^nv\text{.}\) Consider
Recall that as \(A\) is nilpotent, there exists some \(m\in\N\) such that \(A^m=0\text{.}\) As \(\lambda\) is an eigenvalue of \(A\text{,}\) by the above induction we see that \(\lambda^m\) is an eigenvalue for \(A^m=0\text{.}\) As \(F\) is a field and thus an integral domain, we see that \(\lambda^n=0\) implies that \(\lambda=0\) is as well. As this holds in the algebraic closure of \(F\) as well, we see that when factored into linear terms all the \(\lambda=0\text{.}\) Thus \(\cp_A(x)=x^n\text{.}\)
By the Cayley Hamilton Theorem CITEX we know \(\mp_A(x)|\cp_A(x)=x^n\text{,}\) and thus \(A^n=0\text{.}\)
Assume \(n \leq 3\) and let \(A,B\) be nilpotent matrices with entries in \(F\text{.}\)
\((\Rightarrow)\) Suppose \(A\sim B\text{.}\) Thus there exists some invertible matrix \(P\) such that \(PAP\inv=B\) by the definition of similar matrices. Let \(U\in\ker(PA)\text{.}\) Thus \(PA(U)=0\) and \(P(AU)=0\text{.}\) We multiply both sides by \(P\inv\) to see that \(AU=0\text{.}\) Therefore the \(\ker(PA)=\ker(A)\text{,}\) and hence the ranks of \(PA\) and \(A\) are equal by Rank-Nullity Theorem.
as \((P^{-1})^T\) is an invertible matrix. Thus \(\rank(B)=\rank(PAP\inv)=\rank(A)\text{.}\)
\((\Leftarrow)\) Suppose that \(\rank(A)=\rank(B)\text{.}\) From Part (a) we know \(\cp_A(x)=\cp_B(x)=x^n\text{.}\)
We consider the case where \(\cp_A(x)=\cp_B(x)=x^3\text{.}\)
The only possible invariant factors involving \(x^3\) are 1. \(x,x,x\text{;}\) 2. \(x,x^2\text{;}\) and 3. \(x^3\) itself. However, if \(x,x,x\) are the invariant factors then the rank of \(A\) would be 3, making it invertible, contradicting the fact that 0 is an eigenvalue of \(A\text{.}\) Thus we need only consider the latter two cases.
As \(C(x)\oplus C(x^2)\) has rank 1 and \(C(x^3)\) has rank 2, since \(\rank(A)=\rank(B)\) we see that they must have the same invariant factors, making them similar.
If \(n=2\) then the only possible invariant factor is \(x^2\text{,}\) as having two \(x\)’s would make \(A\) and \(B\) invertible again. If \(n=1\) then \(A=B=0\) and we’re done.
Consider \(A=C(x^2)\oplus C(x^2)\) and \(B=C(x)\oplus C(x^3)\text{.}\)
both of which have rank \(2\) and are in RCF. Thus they are not similar.
ActivityC.126.Problem 8.
Let \(R\) be a commutative ring and \(x\) an indeterminant. Prove that \(R[x]\) is a PID if and only if \(R\) is a field.
Solution.
Let \(R\) be a commutative ring and \(x\) an indeterminant.
First, suppose \(R[x]\) is a PID. Let \(I\) be a nonzero ideal in \(R\text{,}\) and consider the ideal \((I,x)\in R[x]\text{.}\) As \(R[x]\) is a PID we see \((I,x)=(z)\) for some \(z\in R[x]\text{.}\) Then \(z|x\text{,}\) but \(x\) is an indeterminant and thus irreducible, so \(z\) must be a unit or \(x\) itself. But \(z\) also generates \(I\text{,}\) so \(z\) can’t be \(x\text{,}\) so \(z\) is a unit, meaning \(z\in R\) and so \(I=(R)\text{.}\) Thus the only ideals of \(R\) are \((0)\) and \(R\text{,}\) making \(R\) a field.
Next, suppose \(R\) is a field. Let \(J\) be an ideal in \(R[x]\text{.}\) If \(J\in R\) then \(J=R\) or \(J=(0)\text{,}\) both of which are principal. Thus \(x\in J\text{,}\) and so \(J=(x)\text{,}\) making \(J\) principal as well. Hence \(R[x]\) is a PID.
ActivityC.127.Problem 9 (*).
Find, with justification, a complete and non-redundant list of conjugacy class representatives for the group \(\GL_2(\F_3)\text{,}\) where \(\F_3\) is the field with three elements.
Solution.
Let \(\F_3\) denote the field with three elements, and consider the group \(\GL_2(\F_3)\text{.}\)
Recall that matrices are in the same conjugacy class if and only if they are similar, and that two matrices are similar if and only if they share the same invariant factors.
Let \(A\in\GL_2(\F_3)\text{.}\) All characteristic polynomials are monic, and as \(A\) is invertible we see the \(a_0\) term in the \(\cp_A(x)\neq0\text{,}\) and thus \(a_0\) is either \(1\) or \(2\text{,}\) the only other elements in \(\F_3\text{.}\) There are only so many monic polynomials of degree \(2\) with coefficients in \(\F_3\text{;}\) hence the only possible characteristic polynomials of \(A\) are the following: 1. \(f(x)=x^2+x+1\text{,}\) 2. \(g(x)=x^2+x+2\text{,}\) 3. \(h(x)=x^2+1\text{,}\) 4. \(j(x)=x^2+2\text{.}\) 5. \(k(x)=x^2+2x+1\text{.}\) 6. \(\ell(x)=x^2+2x+2\) Note that since \(a_0\neq0\text{,}\) 0 cannot be a root of any of these polynomials. Thus all that remains is to check \(1\) and \(2\text{.}\)
