Let \(G\) be a group, and \(H, K \leq G\) be subgroups of \(G\text{.}\)
Give an example of \(G, H,\) and \(K\) such that \(HK\) is not a subgroup of \(G\text{.}\)
Suppose now that \(H\nsg G\) and \([G : H] = p\) where \(p\) is prime. Prove that either \(K \leq H\) or \(G = KH\text{.}\)
Solution.
Let \(G=S_3\text{,}\)\(H=\langle(12)\rangle\text{,}\) and \(K=\langle(13)\rangle\text{.}\) Notice that \(H\cap K=\{e\}\text{,}\) and thus \(|HK|=4\text{,}\) which does not divide \(6\text{.}\) Thus \(HK\) is not a subgroup of \(G\text{,}\) as this would contradict Lagrange’s Theorem
Suppose there exists some \(k\in K\) such that \(k\not\in H\text{.}\) We examine \(kH\in G/H\text{.}\) Notice that since \(G/H\) has prime order it is cyclic by [provisional cross-reference: cite], and thus generated by any non-identity element, such as \(kH\text{.}\) Thus any element in \(G/H\) can be written in the form \(k^iH\text{,}\) and any element in \(G\) can subsequently be written as \(k^ih\) for some \(h\in H\text{.}\) Thus \(G=KH\text{.}\)
ActivityC.29.Problem 2.
Let \(G\) be a group of order \(2835 = 3^4\cdot 5\cdot 7\text{.}\)
Show that there are at most two options for \(n_3\text{,}\) the number of Sylow \(3\)-subgroups of \(G,\) and list them.
Prove that \(G\) is not simple.
Solution.
Let \(G\) be a group of order \(2835\text{.}\)
By Sylow’s Theorems we know that \(n_3=1\mod{3}\) and \(n_3|35\text{.}\) The possible options are thus \(1\) and \(7\text{.}\)
Suppose by way of contradiction that \(G\) is simple. Thus \(n_3\neq 1\text{,}\) so \(n_3=7\text{.}\) Let \(G\) act on the \(\Syl_3(G)\) by conjugation, yielding the homomorphism
\begin{equation*}
\rho: G\to S_7
\end{equation*}
granted via the Permutation Representation. By (2) in Sylow’s Theorems we see that \(\rho\) is not trivial. As \(|S_7|=7!\) we see that \(|G|\not\big||S_7|\text{,}\) meaning that the \(\ker(\rho)\) is non-trivial, yielding a non-trivial normal subgroup of \(G\text{,}\) a contradiction.
ActivityC.30.Problem 3.
Let \(K\) be a subgroup of a group \(G\text{.}\) Prove that \(K\nsg G\) if and only if there is a group \(H\) and a homomorphism \(\varphi : G\to H\) such that \(K = \ker(\varphi)\)
Let \(G\) be the group of all \(2\times 2\) matrices with entries from \(\Z\) having determinant \(1\text{.}\) Let \(p\) be prime number, and take \(K\) to be the subset of \(G\) consisting of all \(\begin{bmatrix}a & b\\ c & d \end{bmatrix}\) with \(a\equiv d\equiv 1 \mod{p}\) and \(b\equiv c\equiv 0\mod{p}\text{.}\) Prove that \(K\) is a normal subgroup of \(G\text{.}\)
Solution.
First, suppose \(K\nsg G\text{.}\) Let \(H=G/K\) (which is a group since \(K\) is normal in \(G\)) and define \(\varphi: G\to G/K\) such that \(\varphi(g)=gK\text{.}\) Let \(k\in K\text{,}\) and observe that \(\varphi(k)=kK=K\text{,}\) and so \(k\in\ker(\varphi)\text{.}\) Let \(g\in\ker(\varphi)\text{,}\) meaning that \(\varphi(g)=K\text{,}\) placing \(g\in K\text{.}\) As quotient maps are homomorphisms, this direction is complete.
Next, suppose there exists a group \(H\) and and a homomorphism \(\varphi : G\to H\) such that \(K = \ker(\varphi)\text{.}\) Let \(g\in G\) and consider \(gKg\inv\text{.}\) Observe
given that \(\ker(\varphi)=K\text{.}\) Thus \(gKg\inv\subseteq K\) for all \(g\in G\text{,}\) making \(K\) a normal subgroup of \(G\text{.}\)
Define \(\varphi : G \rightarrow SL_2(\mathbb{Z}/p\mathbb{Z})\text{,}\) such that for all \(\begin{bmatrix} a & b\\ c & d \end{bmatrix} \in G\text{,}\)\(\varphi(\begin{bmatrix} a & b\\ c & d \end{bmatrix}) = \begin{bmatrix} a\mod p & b \mod p\\ c \mod p & d \mod p \end{bmatrix}\text{.}\) This is easily verifiable to be a well defined function. To show \(\varphi\) is a homomorphism, let \(\begin{bmatrix} a & b\\ c & d \end{bmatrix}, \begin{bmatrix} e & f\\ g & h \end{bmatrix} \in G\text{.}\) Then
\begin{equation*}
\varphi (\begin{bmatrix} a & b\\ c & d \end{bmatrix} \begin{bmatrix} e & f\\ g & h \end{bmatrix})
= \varphi(\begin{bmatrix}ae + bg & af + bh\\ ce + dg & cf +dh \end{bmatrix})
= \begin{bmatrix} ae + bg \mod p & af + bh \mod p\\ ce + dg \mod p & cf +dh \mod p \end{bmatrix}
\end{equation*}
\begin{equation*}
\varphi(\begin{bmatrix}a & b\\ c & d \end{bmatrix}) \cdot \varphi(\begin{bmatrix}e & f\\ g & h \end{bmatrix})
= \begin{bmatrix} a\mod{p} & b\mod{p}\\ c\mod{p} & d\mod{p} \end{bmatrix} \begin{bmatrix} e\mod{p} & f \mod{p}\\ g \mod{p} & h \mod{p} \end{bmatrix}
\end{equation*}
\begin{equation*}
= \begin{bmatrix} ae + bg \mod p & af + bh \mod p\\ ce + dg \mod p & cf +dh \mod p \end{bmatrix}
\end{equation*}
Thus \(\varphi\) is a homomorphism. Then \(\ker\varphi = \{A \in G | \varphi(A) = I_2\}\text{.}\) So for \(k \in K\text{,}\) we have that \(\varphi(k) = K \mod p = I_2\) by definition of \(K\text{.}\) So \(k \in \ker\Phi\) and \(K \subset \ker\varphi\text{.}\) Also for \(A \in \ker\varphi\) we have \(\varphi(A) = A \mod p = I_2\text{,}\) so \(A\) is congruent to \(I_2\) modulo p, and thus \(A \in K\text{,}\) and \(K = \ker\varphi\text{.}\) So we have defined a homomorphism whose kernal is \(K\text{.}\) By part (a) we have \(K\) is a normal subgroup of \(G\text{.}\)
SubsectionSection II: Rings, Modules, and Linear Algebra
ActivityC.31.Problem 4.
Prove that a finite integral domain \(D\) must be a field.
Prove that if \(R\) is a commutative ring and \(P\subseteq R\) is a prime ideal such that \(P\) has finite index as an additive subgroup of \(R\text{,}\) then \(P\) is a maximal ideal. Give an example to show that this implication may fail if the finite index assumption is dropped.
Solution.
Let \(R\) be a finite integral domain. Then take \(r \in R\) such that \(r \neq 0\text{.}\) We’ll show that \(r\) is a unit which implies it has a multiplicative inverse. If \(r = 1\) then \(r\) is a unit and we are done. Suppose \(r \neq 1\text{.}\) Consider set \(\{r^n | n \in \mathbb{N}\}\text{,}\) which is contained in \(R\text{.}\) Since \(R\) is finite it follows that this set is finite and there must exists \(n < m \in \mathbb{N}\) such that \(r^n = r^m\text{.}\) As an integral domain \(R\) has the cancellation law so cancelling \(r^n\) from the left yields \(1 = r^{m-n}\text{.}\) So we have \(1 = r(r^{m-n-1})\) and \(r\) is a unit. We can conclude all nonzero elements of \(R\) have a multiplicative inverse and \(R\) is a field.
Let \(R\) be a commutative ring and \(P \subset R\) a prime ideal. Suppose \(P\) has finite index as a subgroup of \((R, +)\text{.}\) Since \(P\) is prime it follows by theorem that \(R/P\) is an integral domain where \(R/P = \{r +P | r \in R\}\text{.}\) By assumption since the index of \(P\) in \((R, +)\) is finite there are only finitely many elements in \(R/P\text{.}\) By Part \(a\text{,}\) we have that \(R/P\) is a field and again by theorem \(2.74\) since \(R/P\) is a field, \(P\) is a maximal Ideal of \(R\text{.}\)
ActivityC.32.Problem 5.
Let \(F\) be a field, \(V\) and \(W\) finite dimensional \(F\)-vector spaces, and \(g : V\to W\) an \(F\)-linear transformation.
Prove that there exists bases of \(V\) and \(W\) such that the matrix representing \(g\) with respect to these bases has the form
where \(I_{r\times r}\) is the \(r\times r\) identity matrix and all the \(0\)’s denote zero matrices of the appropriate size.
Prove that the number \(r\) appearing in part (a) is independent of choice of bases; that is, if, for another pair of bases of \(V\) and \(W\text{,}\) the matrix representing \(g\) has the form
Let \(F\) be a field, \(V\) and \(W\) finite dimensional \(F\)-vector spaces, and \(g : V\to W\) an \(F\)-linear transformation.
We form \(B\) and \(C\) in steps.
Start by picking an ordered basis \(C' = \{w_1, \dots, w_r\}\) of the image of \(g\text{.}\) For each \(i\) pick \(v_i \in V\) such that \(g(v_i) = w_i\) and set \(B' = \{v_1, \dots, v_r\}\text{.}\) Then pick a basis \(B''\) of the kernel of \(g\text{.}\) Let us list the elements of \(B''\) as \(\{v_{r+1}, \dots, v_n\}\text{.}\)
I claim that \(B := B' \cup B'' = \{v_1, \dots, v_n\}\) is a basis of \(V\text{.}\) (Note that \(B' \cap B'' = \emptyset\) since \(g(v_i) = w_i \ne 0\) for all \(i\) with \(1 \leq i \leq r\) and \(g(v_i) = 0\) for all \(i > r\text{.}\)) Pick \(v \in V\text{.}\) Then, since \(C'\) spans the image of \(g\text{,}\) we have \(g(v) = \sum_{i=1}^r a_i w_i\) for some scalars \(a_i \in F\text{.}\) It follows that \(v - \sum_{i=1}^r a_i v_i \in \ker(g)\) and hence, since the kernel is spanned by \(B''\text{,}\) we have
for some \(a_{r+1}, \dots, a_n \in F\text{.}\) We rearranging this equation, we conclude \(v \in \Span(B)\) and thus \(B\) spans \(V\text{.}\)
Now say \(\sum_{i=1}^n c_i v_i = 0\) for some \(c_i\)’s in \(F\text{.}\) Since \(g(v_i) = 0\) for all \(i > r\) and \(g(v_i) = w_i\) for all \(i \leq r\text{,}\) this gives that \(0 = g(0) = g(\sum_{i=1}^n c_i v_i) = \sum_{i=1}^n c_i g(v_i) =\sum_{i=1}^r c_i w_i = 0\text{.}\) Since \(C'\) is linearly independent, we have \(c_i = 0\) for \(1 \leq i \leq r\text{.}\) Going back to the original equation \(\sum_{i=1}^n c_i v_i = 0\text{,}\) we see that \(\sum_{i=1}^r c_i v_i = 0\) and hence \(c_i=0\) for all \(i\text{,}\) since \(B''\) is linearly independent.
Finally we extend \(C'\) to an ordered basis \(C = \{w_1, \dots, w_r, w_{r+1}, \dots, w_m\}\) of \(W\) arbitrarily.
By our construction, for any \(v_i \in B\) we have
The existence of such a pair of bases and this matrix tells us that for bases \(B=\{b_1,...,b_l\}\) of \(V\) and \(C=\{c_1,...,c_k\}\) of \(W\text{,}\) that \(g(b_i)=c_i\) for \(1\leq i\leq r'\) and \(g(b_i)=0\) for each \(r'<i\leq l\text{.}\) Rank nullity tells us that \(\text{dim}(V)=\text{rank}(g)+\text{nullity}(g)\text{.}\) We will show that the nullity of \(g\) is \(l-r'\) and therefore \(r'=rank(g)=r\text{.}\)
To accomplish this we will show that \(S=\{b_{r'+1},...,b_l\}\) for a basis for \(ker(g)\) and therefore the nullity of \(g\) is \(l-r'\text{.}\) To show it is a basis, we need to show that it is linearly independent and that it spans.
\(S\) is linearly independent since it is a subset of a linearly independent set. To show it spans let \(x\in ker(g)\text{.}\)\(x\in V\) so \(x\) can be written as a linear combination of elements in \(B\text{.}\) Thus \(0=g(x)=g(\lambda_1b_1+\cdots +\lambda_lb_l)=\lambda_1g(b_1)+\cdots+\lambda_{r'+1}\underbrace{g(b_{r'+1})}_\text{0}+\cdots +\lambda_l\underbrace{g(b_l)}_\text{0}=\lambda_1c_1+\cdots +\lambda_{r'}c_{r'}\text{.}\) Those are linearly independent so \(\lambda_1=\cdots =\lambda_{r'}=0\) and therefore \(x\) was just a linear combination of elements in \(S\) so \(S\) spans the kernel and is linearly independent. Therefore \(S\) is a basis for the kernel, and we have that \(\text{nullity}(g)=l-r'\) so filling in the blanks
Let \(A\) and \(B\) be two \(3\times 3\) matrices with entries in \(F\text{.}\) Prove \(A\) and \(B\) are similar if and only if they have the same characteristic polynomial and the same minimum polynomial.
Show, by way of an example with justification, that the previous part would become false if “\(3\times 3\)” were replaced by “\(4\times 4\)”.
Give an example of a field \(F\) and two \(3\times 3\) matrices with entries in \(F\) having the same minimum polynomial that are not similar.
Solution.
Let \(F\) be any field.
Let \(A\) and \(B\) be two \(3\times 3\) matrices with entries in \(F\text{.}\) First, suppose that \(A\sim B\text{.}\) Matrices are similar if and only if they share the same invariant factors. As minimum polynomial is an invariant factor and the characteristic polynomial is a product of the invariant factors, we see that \(A\) and \(B\) must share the same invariant factors.
Next suppose that \(A\) and \(B\) share the same characteristic polynomial and the same minimal polynomial. As \(A\) and \(B\) are \(3\times 3\) matrices, the characteristic polynomial of both \(A\) and \(B\) must be a degree \(3\) polynomial. We proceed via cases based on the degree of \(\mp_{A,B}(x)\text{.}\)
First, suppose \(\deg\mp_{A,B}(x)=3\text{.}\) Then \(\mp_{A,B}(x)=\cp_{A,B}(x)\text{,}\) making \(\cp_{A,B}(x)\) the only invariant factor of both \(A\) and \(B\text{.}\) Thus \(A\) and \(B\) have the same invariant factors and are therefore similar.
Next, suppose \(\deg\mp_{A,B}(x)=2\text{.}\) As \(\mp|\cp\) and the degrees of all invariant factors must sum to the \(\deg\cp\text{,}\) we know that \(\cp=\mp\cdot k(x)\text{,}\) where \(k(x)\) is a degree \(1\) polynomial, which we denote \(f(x)\) for \(A\) and \(g(x)\) for \(B\text{.}\) Since \(A\) and \(B\) share the same minimum and characteristic polynomials, we see \(\mp\cdot f=\cp\) and \(\mp\cdot g=\cp\text{,}\) and thus that \(f=g\text{.}\) Hence \(A\) and \(B\) share the same invariant factors, making \(A\sim B\text{.}\)
Finally, suppose \(\deg\mp=1\text{.}\) The minimum polynomial is the largest invariant factor, and thus the invariant factors of \(A\) and \(B\) must be \(\{\mp,\mp\mp\}\text{,}\) making them similar.
If we replaced \(3\times 3\) with \(4\times 4\) then this would allow for \(\cp_{A,B}(x)=x^4\) and \(\mp_{A,B}(x)=x^2\text{,}\) allowing two sets of invariant factors: \(\{x^2,x^2\}\)\(\{x,x,x^2\}\text{.}\) Notice that \(C(x)=\begin{bmatrix}0\end{bmatrix}\) and \(C(x^2)=\begin{bmatrix}0&0\\1&0\end{bmatrix}\text{.}\) Set \(A=C(x^2)\oplus C(x^2)\) and \(B=C(x)\oplus C(x)\oplus C(x^2)\text{,}\) so
These matrices have the same \(\cp\) and \(\mp\) but are not similar.
Let \(F=\Q\text{.}\) We define \(A=C(x-1)\oplus C(x^2-1)\) and \(B=C(x+1)\oplus C(x^2-1)\text{.}\) Notice that these matrices are in rational canonical form. However, the invariant factors of \(A\) are \(\{(x+1), (x^2-1)\}\) and the invariant factors of \(B\) are \(\{(x-1)(x^2-1)\}\text{.}\) Thus \(A\) is not similar to \(B\text{,}\) but the minimal polynomial of both is \(x^2-1\text{.}\)
SubsectionSection III: Fields and Galois Theory
ActivityC.34.Problem 7.
Let \(E\) be the splitting field of \(x^4 + 5\) over \(\Q\text{.}\)
Prove, by adding two appropriate roots of \(x^4 + 5\) or otherwise, that there exists \(\Q\subseteq F\subseteq E\) such that \(F \sse \R\) and \([F : \Q] = 4\text{.}\)
Determine, with justification, \([E : \Q]\text{.}\)
Solution.
Let \(E\) be the splitting field of \(f(x)=x^4 + 5\) over \(\Q\text{.}\)
First, we note that the roots of \(f\) are the following:
First we examine \(F=\Q(\sqrt[4]{20})\text{.}\) Notice that \(\sqrt[4]{20}\) is a root of the polynomial \(x^4-20\text{,}\) which is irreducible in \(\Q[x]\) using Eisenstein’s Criterion with \(p=5\text{.}\) Thus \(x^4-20\) is monic and irreducible, making it the minimal polynomial of \(\sqrt[4]{20}\text{.}\) Thus \([F:\Q]=4\text{.}\) Notice that \(\a_1+\a_4=\frac{\sqrt[4]{20}}{2}(1+i+1-i)=\sqrt[4]{20}\text{,}\) so \(\sqrt[4]{20}\in E\text{.}\) Thus there exists \(\Q\subseteq F\subseteq E\) such that \(F \sse \R\) and \([F : \Q] = 4\text{.}\)
Now let \(K=F(i)\text{.}\) Notice that \(i\) is a root of the polynomial \(x^2+1\text{,}\) which is irreducible in \(F[x]\) as \(F\subseteq\R\) and \(i\not\in\R\text{.}\) Thus \(x^2+1\) is monic and irreducible, making it the minimal polynomial of \(i\text{.}\) Thus \([K:F]=2\) and \([K:\Q]=8\) by the The Degree Formula.
Notice that \(\a_1+\a_4=\frac{\sqrt[4]{20}}{2}(1+i+1-i)=\sqrt[4]{20}\text{,}\) so \(\sqrt[4]{20}\in E\text{.}\) Thus so is \(\frac{1}{\sqrt[4]{20}}\text{,}\) and so we see that \(i=\a_1\cdot\frac{1}{\sqrt[4]{20}}-1\text{,}\) and thus that \(i\in E\) as well, making \(K\subseteq E\text{.}\) As seen in Part (a), each root can be expressed in terms of \(i\) and \(\sqrt[4]{20}\text{,}\) and so \(E\subseteq K\) as well. Hence \(K=E\) and we have \([E : \Q]=[K:\Q]=8\text{.}\)
ActivityC.35.Problem 8.
Let \(F\) be the splitting field over \(\Q\) of the polynomial \(x^3-2\text{.}\) Prove that the Galois group \(\Gal(F : \Q)\) is isomorphic to \(S_3\text{.}\)
Solution.
Let \(F\) be the splitting field over \(\Q\) of the polynomial \(f(x)=x^3-2\text{,}\) the roots of which are:
\(\displaystyle \sqrt[3]2\)
\(\sqrt[3]2e^{2\pi i/3}\text{,}\) and
\(\sqrt[3]2e^{4\pi i/3}\text{.}\)
Using Eisenstein’s Criterion with \(p=2\) we see that \(f\) is irreducible in \(\Q[x]\text{.}\) As \(f\) is monic and irreducible it is the minimum polynomial of \(\sqrt[3]2\in\Q[x]\) and \([\Q(\sqrt[3]2):\Q]=3\text{.}\) Thus, by the Fundamental Theorem of Galois Theory we know there exists an element of order \(3\) in \(\Gal(F:\Q)\text{.}\)
Notice now that \(f\) has exactly two complex roots, making complex conjugation correspond to a transposition in \(S_3\text{.}\) Thus we have an element of order \(2\) and an element of order \(3\text{,}\) so the order of \(\Gal(F:\Q)\) must be at least \(6\) by Lagrange’s Theorem.
As \(\deg f=3\) we know \(\Gal(F:\Q)\) is isomorphic to a subgroup of \(S_3\) by Theorem 17.6, which has order \(6\text{.}\) Thus \(\Gal(F : \Q)\) is isomorphic to \(S_3\text{.}\)
ActivityC.36.Problem 9.
Let \(F\) be a field and \(f\subseteq F [x]\) be irreducible. Prove the \(F [x]/(f )\) is a field.
Give an explicit construction (with justification) of a field of size \(16\text{.}\) (You may use without proof that the unique irreducible quadratic in \(\F_2[x]\) is \(x^2 + x + 1.)\)
Solution.
Let \(F\) be a field and \(f\subseteq F [x]\) be irreducible. Thus \(f\) is prime, making \((f)\) a prime ideal in \(F[x]\) and \(F[x]/(f)\) a domain. Suppose that there existed a proper ideal \(I\) such that \((f)\subseteq I\text{.}\) However, as \(F\) is a field we know that \(F[x]\) is a PID, meaning \(I=(g)\) for some \(g\in F[x]\text{.}\) If \(f\in(g)\) then \(g|f\text{,}\) meaning that \(f=g\text{.}\) Thus \((f)\) is a maximal ideal and we have \(F[x]/(f)\) a field.
Let \(F=\F_2\text{.}\) Let \(f=x^4+x+1\text{.}\) Suppose \(f=gh\) with \(g,h\) irreducible. Thus either both \(g\) and \(h\) have degree \(2\) or one of them has degree \(1\text{.}\) As \(f\neq (x^2+x+1)^2\text{,}\) we see that without loss of generality \(\deg g=1\text{.}\) Then \(g=x\) or \(x+1\text{,}\) but neither divide \(f\text{.}\) Thus \(f\) is irreducible, making \(F[x]/(f)\) a field, the elements of which are: 1. \(0\) 2. \(1\) 3. \(x\) 4. \(x+1\) 5. \(x^2\) 6. \(x^2+1\) 7. \(x^2+x\) 8. \(x^2+x+1\) 9. \(x^3\) 10. \(x^3+1\) 11. \(x^3+x\) 12. \(x^3+x^2\) 13. \(x^3+x+1\) 14. \(x^3+x^2+1\) 15. \(x^3+x^2+x\) 16. \(x^3+x^2+x+1\text{,}\) making it a field of order \(16\text{.}\)
