Let \(G\) be a group. A subgroup \(H\) of \(G\) is called maximal if \(H\neq G\) (that is, \(H\) is a proper subgroup of \(G\)) and whenever \(K\) is another subgroup of \(G\) containing \(H\text{,}\) either \(K = H\) or \(K = G\text{.}\) Show that every nontrivial finitely generated group \(G\) possesses maximal subgroups.
Solution.
Let \(G\) be a group.
Let \(S\) be the poset of all proper subgroups of \(G\) ordered in terms of inclusion. Consider a string of these. Consider the union of them all. Luckily, unions of subgroups are subgroups if and only if there is containment, which there is, since everything is in the union. Thus its a subgroup. Since union in \(S\) and yields an upper bound, by Zorn’s Lemma CITEX we have a maximal element. Thus \(G\) possesses maximal subgroups.
ActivityC.74.Problem 2.
Let \(G\) be a group and \(H\) a subgroup of \(G\text{.}\) Recall that the centralizer of \(H\) in \(G\) is
\begin{equation*}
C_H(G)=\{g\in G|gh=hg\text{ for all }h\in H\}
\end{equation*}
Prove that if \(H\) is normal in \(G\text{,}\) then so is \(C_G(H)\) and that \(G/C_G(H)\) is isomorphic to a subgroup of the automorphism group of \(H\text{.}\)
Solution.
See identical problem CITEX
ActivityC.75.Problem 3.
Let \(G\) be a group of order \(p^2q\) where \(p\) and \(q\) are distinct primes.
Prove that \(G\) contains a normal Sylow subgroup.
Suppose \(p < q\) and the Sylow \(p\)-subgroup is cyclic and normal. Prove that \(G\) is abelian.
Solution.
Let \(G\) be a group of order \(p^2q\) where \(p\) and \(q\) are distinct primes. Suppose by way of contradiction that \(G\) has no normal Sylow \(p\)-subgroup.
First, suppose \(p<q\text{.}\) By Sylow’s Theorems we know the following:
\(n_p|q\) and \(n_p\equiv 1\mod{p}\text{,}\) so \(n_p=1\) or \(q\text{,}\) so \(q\)
\(n_q|p^2\) and \(n_q\equiv 1\mod{q}\text{,}\) so \(n_q=1\) or \(p^2\text{,}\) so \(p^2\text{.}\)
We know there must be \((q-1)\cdot p^2=p^2q-p^2\) elements of order \(p\text{.}\) Luckily, there is more than one Sylow \(p\)-subgroup with \(p^2\) elements, so there isn’t room for all of them.
Suppose then that \(q>p\text{.}\) By Sylow’s Theorems we know \(n_p|q\) and \(n_p\equiv 1\mod{p}\text{,}\) so \(n_p=1\text{,}\) so we’re definitely good there.
Suppose \(p < q\) and the Sylow \(p\)-subgroup, \(P\text{,}\) is cyclic and normal. We know from Part (a) that there are either \(1\) or \(p^2\) Sylow \(q\)-subgroups, but since there are already \(p^2\) elements of order \(p\) there is only room for one, \(Q\text{,}\) which is also cyclic, given its prime power. As \(P\) and \(Q\) are thus normal in \(G\) and only intersect trivially, we see that \(PQ=G\text{,}\) meaning that \(G=P\times Q\text{.}\) Thus \(G\) is the product of two cyclic groups of relatively prime order, making \(G\) cyclic as well. Cyclic groups are abelian, so we are done.
ActivityC.76.Problem 3?
Let \(G\) be a finite group of order \(p^2q\) with \(p < q\) prime numbers. Show that \(G\) is not a simple group.
Solution.
Let \(G\) be a finite group of order \(p^2q\) with \(p < q\) prime numbers, and suppose by way of contradiction that \(G\) is simple. By Sylow’s Theorems we know \(n_p|q\) and \(n_p\cong 1\mod{p}\text{,}\) and thus \(n_p=q\text{.}\) We also know \(n_q|p^2\) and \(n_q\cong 1\mod{q}\text{,}\) and thus \(n_q=p^2\text{.}\) From this information we see that there are \(p^2(q-1)\) elements of order \(q\) and \((p^2-1)q\) elements of order \(p\text{,}\) for a lovely total of \(p^2q-p^2+p^2q-q\) elements, which is too many.
SubsectionSection
ActivityC.77.Problem 4.
Consider the subring \(\Z[\sqrt{10}] = \{a + b\sqrt{10} | a, b \in \Z\}\) of \(\R\text{.}\) Show that \(2\) is irreducible but not prime in \(\Z[\sqrt{10}].\text{.}\)
Hint.
Consider the function \(N : \Z[\sqrt{10}]\to \Z, N (a + b\sqrt{10}) = a^2-10b^2\)
Solution.
Consider \(2\in\Z[\sqrt{10}]\) and the function \(N : \Z[\sqrt{10}]\to \Z, N (a + b\sqrt{10}) = a^2-10b^2\text{.}\)
Thus \((a^2-10b^2)=\pm1,\pm2,\) or \(\pm4\text{,}\) as these are the only integer divisors of \(4\text{.}\) However, there do not exist integers \(a,b\) such that this is true. Thus \(2\) is irreducible in \(\Z[\sqrt{10}]\text{.}\)
Suppose by way of contradiction that \(2\) is prime in \(\Z[\sqrt{10}]\text{.}\) Note that \(2|6=(4+\sqrt{10})(4-\sqrt{10})\text{.}\) Thus \(2\) divides one of these factors.
First, suppose there exists some \(a+b\sqrt{10}\) such that \(2(a+b\sqrt{10})=(4\pm\sqrt{10})\text{.}\) Thus \(2a+2b\sqrt{10}=4\pm\sqrt{10}\text{,}\) and so \(2b=\pm1\text{.}\) However, \(\pm\frac12\) is not an integer, and thus \(2\) cannot divide either of these factors. Thus \(2\) is not prime in \(\Z[\sqrt{10}]\text{.}\)
ActivityC.78.Problem 5.
Recall that a \(\Z\)-module \(M\) is called torsion-free if its torsion submodule is \(\Tor(M ) =\{0\},\) where \(\Tor(M ) = \{m \in M | rm = 0 \text{ for some }r \in \Z\setminus\{0\}\}.\) Consider the \(\Z\)-module
\begin{equation*}
M= \frac{\Z \oplus \Z}{\langle(7, 11)\rangle }\text{ where }\langle(7, 11)\rangle = \{(7k, 11k) | k \in \Z\}.
\end{equation*}
Show that \(M\) is torsion free.
Solution.
Let \(m=(a,b)\in\Tor(M)\text{.}\) Thus \(rm=0\) for some nonzero \(r\in\Z\text{.}\) Then \((ra,rb)=0\text{,}\) and so \(ra=0\) and \(rb=0\text{.}\) Then there exists \(p,q\in\Z\) such that \(7p=ra\) and \(7q=rb\) or \(11p=ra\) and and \(11q=rb\text{.}\) Notice that in \(M\) we have \(7,11=0\text{,}\) and thus neither \(7\) nor \(11\) can divide \(r\text{.}\) Suppose \(7p=ra\) and \(7q=rb\text{.}\) As \(7\) cannot divide \(r\) we see that \(7\) divides both \(a\) and \(b\text{,}\) placing them both \(\langle(7, 11)\rangle\text{.}\) Thus \(m=0\text{.}\) The same holds true if we use \(11\text{.}\) Thus \(m=0\) and \(\Tor(M)=\{0\}\text{.}\)
(ALTERNATE SOLUTION) To show \(M\) is torsion-free we need to show that \(\text{Tor}(M)=\{0\}\text{.}\) The fact that \(\{0\}\subseteq \text{Tor}(M)\) is beyond trivial, so we just need to show that \(\text{Tor}(M)\subseteq \{0\}\text{.}\) So let \((x,y)+\langle(7,11)\rangle\in \text{Tor}(M)\text{.}\) Thus there exists an \(r\neq 0\in \mathbb{Z}\) such that \(r((x,y)+\langle (7,11)\rangle )=(rx,ry)+\langle (7,11)\rangle=(0,0)+\langle (7,11)\rangle\text{.}\) This implies that \((rx,ry)-(0,0)=(rx,ry)\in \langle (7,11)\rangle\text{.}\) By definition, this means that \((rx,ry)=(7k,11k)\) for some \(k\in \mathbb{Z}\text{.}\) This gives us two equations:
and therefore \(\frac{rx}{7}=\frac{ry}{11}\iff 11rx=7ry\iff 11x=7y\text{.}\) Since 7 and 11 are each prime, it must be that \(7|x\) and \(11|y\text{.}\) So \(x=7m\) and \(y=11n\) for some \(m,n\in \mathbb{Z}\text{.}\) Therefore we get
So we get exactly that \(x=7m,y=11m\implies (x,y)\in \langle(7,11)\rangle\) and therefore \((x,y)+\langle (7,11)\rangle =(0,0)+\langle (7,11)\rangle\text{.}\) Finally we find that \(\text{Tor}(M)\subseteq \{0\}\implies \text{Tor}(M)=\{0\}\text{.}\)
ActivityC.79.Problem 6.
Let \(n\) be a positive integer. Consider the real vector space \(V = \{p \in R[x] | \deg(p) \leq n\}\) and the linear transformation \(T : V \to V, T (p) = p'\text{,}\) where \(p'(x)\) is the derivative of \(p(x)\text{.}\)
Find the characteristic polynomial and the minimal polynomial for \(T\text{.}\)
Find the invariant factors and the elementary divisors for \(T\text{.}\)
Find the rational canonical form and the Jordan canonical form for \(T\text{.}\)
Solution.
Let \(T:V \to V\) be the linear operator given by \(T(p) = xp'\) (where \(p'\) denotes the derivative of \(p).\)
Note that the change of basis matrix for this operator is given by
which is diagonal. Hence \(\cp_AT(x)=\displaystyle\prod_{i=0}^{n-1}(x-i)\text{.}\)
Our \(\cp_AT(x)\) factors into distinct linear polynomials, each of which is in the form \((x-i)^1\) for \(0\leq i\leq n-1\text{.}\) Thus each linear term is an elementary divisor. However, as none of these elementary divisors divide any of the others, we see that the only invariant factor is \(\cp_AT(x)\) itself.
As each linear term is an elementary divisor, each Jordan block a \(1\times1\) matrix with \(i\) as the only entry. Thus the Jordan Canonical form is
As the only invariant factor is \(\cp_AT(x)\text{,}\) we see that the Rational Canonical Form of \(T\) is \(C(\cp_AT(x))\text{.}\)
SubsectionSection
ActivityC.80.Problem 7.
Let \(E, K,\) and \(L\) be subfields of a field \(F\) with \(E \subseteq K\) and \(E \subseteq L\text{.}\) Let \(k = [K : E]\) and \(\ell = [L : E]\text{.}\) Recall that \(KL\) denotes the smallest (with respect to containment) subfield of \(F\) which satisfies \(K\subseteq KL\) and \(L\subseteq KL\text{.}\)
Show that \([KL : E] \leq k\ell\text{.}\)
Show that if \(\gcd(k,\ell) = 1\) then \([KL : E] = k\ell.\)
Give an example satisfying \([KL : E] < k\ell\)
Solution.
Let \(E, K,\) and \(L\) be subfields of a field \(F\) with \(E \subseteq K\) and \(E \subseteq L\text{.}\) Let \(k = [K : E]\) and \(\ell = [L : E]\text{.}\)
We’re going to go slightly out of order. By Part (b), we see \([KL : E] = k\ell\) when \(\gcd(k,\ell) = 1\text{.}\) This is the largest possible size of \(KL\text{,}\) as everything is irreducible where it can be. Thus \([KL : E] \leq k\ell\)
Suppose \(\gcd(k,\ell) = 1\text{.}\) As \(K\) and \(L\) are finite extensions of \(E\) they are also algebraic. Thus there exists \(\a\in K\) and \(\b\in L\) such that \(\mp_\a(x)\) has degree \(k\) and \(\mp_\b(x)\) has degree \(\ell\text{.}\)
Thus \(\gcd(\deg\mp_\a(x),\ell) = 1\text{,}\) making \(\mp_\a(x)\) irreducible in \(L[x]\text{.}\) Thus \([L(\a):L]=k\text{.}\) Notice that \(E(\a,\b)=KL\text{,}\) as it is the smallest (with respect to containment) subfield of \(F\) which satisfies \(K\subseteq KL\) and \(L\subseteq KL\text{.}\) Thus we have \([KL:E]=k\ell\text{,}\) as desired.
Let \(E=\Q\text{,}\)\(L=K=\Q(\sqrt2)\text{,}\) and \(F=\Q(\sqrt2,i)\text{.}\) Notice that \(KL=K=L\text{,}\) and thus \([KL:E]=[L:E]=\ell<\ell^2=k\ell\text{.}\)
ActivityC.81.Problem 8.
Let \(F\subseteq L\) be fields and let \(x, y \in L\) be algebraic elements over \(F\text{.}\) Prove that \(x + y\) and \(xy\) are also algebraic elements of \(L\) over \(F\text{.}\)
Solution.
Let \(F\subseteq L\) be fields and let \(x, y \in L\) be algebraic elements over \(F\text{.}\) First, notice that \(xy\) and \(x+y\) are contained in \(F(x,y)\text{.}\) As \(y\) is algebraic over \(F\text{,}\) it is the root of some polynomial \(f\) with coefficients in \(f\text{.}\) But \(f\) is also contained in \(F(x)\text{,}\) and thus the extension \([F(x,y):F(x)]\) is algebraic. As \([F(x):F]\) is algebraic as well, we see that \([F(x,y):F]\) is an algebraic extension of fields. Thus \(xy\) and \(x+y\) are algebraic over \(F\text{.}\)
ActivityC.82.Problem 9.
Consider \(f (x) = x^5 -20x - 2 \in \Q[x]\text{.}\) This polynomial has exactly three real roots, a fact that you may use without proof.
Show that \(f\) is irreducible in \(\Q[x]\text{.}\)
Let \(L\) be a splitting field of \(f\) over \(\Q\text{.}\) Show that \(L/\Q\) is a Galois extension and find the isomorphism class of the Galois group \(\Gal(L/\Q)\text{.}\)
Solution.
Notice that \(p=2\) is prime in \(\Q[x]\text{,}\) and thus \(f(x)\) is irreducible in \(\Q[x]\) by Eisenstein’s Criterion.
Let \(\alpha\) be a real root of \(f(x)\text{.}\) As \(f\) is irreducible this root is not in \(\Q\text{.}\) As \(\alpha\) is the root of a monic irreducible polynomial of degree \(5\text{,}\) we see that \([\Q(\alpha):\Q]=5\text{.}\) By the FTGT CITEX there exists a subgroup of \(\Gal(L/\Q)\) with order \(5\text{,}\) making it a cyclic subgroup generated by some element of order \(5\text{.}\)
However, as \(f\) only has \(2\) complex roots we see that complex conjugation corresponds to an element of order \(2\) in \(\Gal(L/\Q)\cong H\leq S_5\text{.}\) Thus we have a transposition and a \(5\)-cycle, meaning we can generate all of \(S_5\text{.}\)\(\Gal(L/\Q)\cong S_5\text{.}\)
