Let \(G\) be a group of order \(p^mq\) where \(p\) and \(q\) are distinct primes and \(m \geq 1\text{.}\) Assume that \(G\) has at least one element of order \(p^m\text{.}\) Let \(S\) be the set of all elements of \(G\) of order \(p^m\text{,}\) that is \(S = \{x \in G : |x| = p^m\}\text{.}\)
Prove that \(G\) acts on \(S\) by conjugation.
Prove that if \(Z(G) = \{e\}\) then \(q\) divides \(|S|\text{.}\)
Solution.
Let \(G\) be a group of order \(p^mq\) where \(p\) and \(q\) are distinct primes and \(m \geq 1\text{.}\) Assume that \(G\) has at least one element of order \(p^m\text{.}\) Let \(S\) be the set of all elements of \(G\) of order \(p^m\text{.}\)
First, notice that \(e\cdot x=exe=x\) for all \(x\in S\text{.}\) Next, let \(g,h\in G\) and consider \(gh\cdot x=ghx(gh)\inv=ghxh\inv g\inv=g\cdot(h\cdot x)\text{.}\) Thus \(G\) acts on \(S\) by conjugation.
Suppose \(Z(G)\) is trivial, and let \(G\) act on \(S\) by conjugation. Thus the orbits of this action correspond the normalizers of elements and the stabilizers correspond to the centralizers. Recall that the orbits of this action partition \(S\text{,}\) so we have
for distinct orbits. However, by [cross-reference to target(s) "cor-orbit-stabilizer" missing or not unique] we also have \(|G|=|\Orb(x)|\cdot |\Stab(x)|\) for all \(x\in S\text{.}\) Now recall the Class Equation:
With all that out of the way, let \(x,y\in S\text{.}\) Notice that \(x^{p^{m}}=e=y^{p^{m}}\)
\(xyx\inv\)
ActivityC.102.Problem 2.
Prove there are exactly three groups, up to isomorphism, of order \(75\text{.}\)
Solution.
Let \(G\) be a group of order \(75=3\cdot 5^2\text{.}\)
By Sylow’s Theorems we know the following: - \(n_3|25\) and \(n_{3}\equiv1\mod{3}\text{,}\) so our options are \(1\) and \(25\text{.}\) - \(n_5|3\) and \(n_{3}\equiv1\mod{5}\text{,}\) so there can only be one. Let \(P\) denote the unique Sylow \(5\)-subgroup, which has order \(25\) and is normal in \(G\text{.}\) As \(P\) is a group of order \(p^2\) we know that it is abelian. Thus there are only two options for \(P\) up to isomorphism: \(\Z_5\times\Z_5\) and \(\Z_{25}\text{.}\)
Let \(Q\) denote a Sylow \(3\)-subgroup. All Sylow \(3\)-subgroups are isomorphic to each other and there is only one group of order \(3\text{,}\) which is \(\Z_3\text{.}\)
Suppose \(P=\Z_{25}\text{.}\) As \(G = PQ\) and \(P\cap Q = \{e\}\text{,}\) we have that \(G = P\sdp_\phi Q\text{,}\) where \(\phi : Q \to\Aut(P)\text{.}\) Note that \(\Aut(P)\cong\Z^\times_{25}\text{,}\) which is cyclic of order \(20\text{.}\) Since \(|Q| = 3\) and \(\gcd(3,20) = 1\text{,}\) we conclude that \(\phi\) must be the trivial map. Hence, \(G = P\times Q\text{.}\)
Now suppose \(P=\Z_5\times\Z_5\text{.}\)
ActivityC.103.Problem 3.
Prove that no group of order \(2^m \cdot 7\) with \(m \geq 1\) is simple.
Solution.
Let \(G\) be a group of order \(2m\cdot 7\) with \(m\geq1\) and suppose by way of contradiction that \(G\) is simple.
By Sylow’s Theorems we know the following: - \(n_{7}|2^m\) and \(n_7\equiv1\mod{7}\text{,}\) and - \(n_{2}|7\) and \(n_{2}\equiv1\mod{2}\text{,}\) so our options are \(3,5,7\text{.}\)
Let \(G\) act on \(\Syl_2(P)\) by conjugation, yielding the permutation representation homomorphism \(\rho:G\to S_{n_2}\text{.}\) The kernel of this homomorphism cannot be trivial as the conjugation action on Sylow subgroups is transitive by Part (2) of Sylow’s Theorems. Notice that if \(n_2\) is \(3\) or \(5\) then \(|G|\) does not divide the order of \(S_{n_2}\text{,}\) so our kernel cannot be all of \(G\text{,}\) making \(\ker(\rho)\) a nontrivial normal subgroup of \(G\text{.}\)
If \(n_2=7\text{,}\) then \(G\) must divide \(7!\text{,}\) meaning that \(m=1\text{.}\) However, by the cyclic subgroup generated by an element of order \(7\) has index \(2\) in \(G\text{,}\) making it normal.
Thus \(G\) cannot be simple.
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ActivityC.104.Problem 4.
Prove that in a UFD an element \(p\) is irreducible if and only if the ideal \((p)\) is prime.
Prove that \(\Z[\sqrt{-5}]\) is not a UFD.
Solution.
Let \(R\) be a UFD.
First, suppose \(p\) is irreducible, and consider the ideal \((p)\text{.}\) Let \(a,b\in R\) such that \(ab\in(p)\text{.}\) Thus \(p|ab\text{,}\) so there exists some \(k\in R\) such that \(pk=ab\text{.}\) As we are in a UFD the elements \(a,b,\) and \(k\) all have unique (up to associates) factorizations of irreducible elements. Thus \(p\) must be an associate of one of the irreducible elements in the factorization of \(a\) or \(b\text{,}\) and so \(a\in(p)\) or \(b\in(p)\text{,}\) making \((p)\) prime.
Now suppose that \((p)\) is prime and that \(p=ab\) for some \(ab\in R\text{.}\) Thus \(p|ab\) and \(ab\in(p)\text{,}\) so either \(a\in(p)\) or \(b\in(p)\text{,}\) as \((p)\) is a prime ideal. Assume without loss of generality that \(a|p\text{.}\) Thus \(ak=p\) for some \(k\in R\text{.}\) Thus \(ab=pkb\) and \(p=pkb\) As UFDs are integral domains we see \(kb=1\text{,}\) making \(b\) a unit. Thus \(p\) is irreducible.
First, notice that \(2\cdot 3=6=(1+\sqrt{-5})(1-\sqrt{-5})\text{.}\) Define a function
Thus \((a^2+5b^2)=\pm1,\pm2,\) 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 \(\Z[\sqrt{-5}]\text{.}\)
Suppose by way of contradiction that \(2\) is prime in \(\Z[\sqrt{-5}]\text{.}\) Note that \(2|6=(1+\sqrt{-5})(1-\sqrt{-5})\text{.}\) Thus \(2\) divides one of these factors.
First, suppose there exists some \(a+b\sqrt{-5}\) such that \(2(a+b\sqrt{-5})=(1\pm\sqrt{-5})\text{.}\) Thus \(2a+2b\sqrt{-5}=1\pm\sqrt{-5}\text{,}\) and so \(2a=\pm1\text{.}\) However, \(\pm\frac12\) is not an integer, and thus \(2\) cannot divide either of these factors. Thus \(2\) is not prime in \(\Z[\sqrt{-5}]\text{.}\) By Part (a), this is not a UFD.
ActivityC.105.Problem 5.
Let \(F\) be a field and \(f(x) \in F[x]\) a monic polynomial of degree \(n\text{.}\) Prove: all matrices in \(M_{n\times n}(F)\) having characteristic polynomial \(f (x)\) are similar if and only if the irreducible factorization of \(f(x)\) has no repeated factors.
Solution.
Let \(F\) be a field, \(f(x) \in F[x]\) a monic polynomial of degree \(n\text{,}\) and \(A,B\) matrices in \(M_{n\times n}(F)\) having characteristic polynomial \(f(x)\text{.}\)
ActivityC.106.Problem 6 (*).
Make \(\R^3\) into a \(\R[x]\)-module by letting \(f (x)v= f (A)v\) for any vector \(v\) in \(\R^3\text{,}\) where \(A\) is the matrix
Solution.
Coming Soon!
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ActivityC.107.Problem 7.
Let \(F \sse L\) be a finite extension of fields and let \(f(x) \in F[x]\) be an irreducible polynomial of degree \(d\text{.}\) Prove that, if \(\gcd(d,[L : F ]) = 1\text{,}\) then \(f (x)\) is also irreducible in \(L[x]\text{.}\)
Solution.
Let \(F \sse L\) be a finite extension of fields and let \(f(x) \in F[x]\) be an irreducible polynomial of degree \(d\) such that \(\gcd(d,[L : F ]) = 1\text{.}\) Let \([L:F]=n\)
First, note that if \(d=1\) then \(f\) is always irreducible, so we may reduce to the case where \(d>1\text{.}\)
Let \(K\) denote the algebraic closure of \(F\text{,}\) let \(\a\) be a root of \(f\in K\text{,}\) and consider \(F(\a)\) and \(L(\a)\text{.}\) Let \(g(x)\) denote the minimal polynomial of \(\a\in L(\a)\text{,}\) and let \(\deg(g)=q\text{.}\) Thus \([L(\a):L]=q\text{.}\) By the The Degree Formula we have - \([L(\a):F]=[L(\a):L][L:F]=qn\) - \([L(\a):F]=[L(\a):F(a)][F(a):F]=md\) for some \(m\in \N\text{.}\) Thus \(qn=md\) and \(md|qn\text{.}\) As \(n\) and \(d\) are relatively prime we have \(d|q\text{.}\) However, \(q\) is the minimal polynomial of \(\a\text{,}\) and thus \(q\leq d\text{,}\) yielding \(q=d\text{.}\) Thus \(f=kg\) for some \(k\in F\text{.}\) As irreducible polynomials multiplied by a constant are still irreducible, we see that \(f\) is indeed irreducible in \(K[x]\text{.}\)
ActivityC.108.Problem 8.
Let \(F \sse L\) be a field extension with \(L\) algebraically closed. Consider the set
\begin{equation*}
K = \{\a\in L : f (\alpha) = 0 \text{ for some monic polynomial }f (x) \in F [x]\}.
\end{equation*}
Show that \(K\) is a field.
Show that \(K\) is algebraically closed.
Solution.
Let \(a,b\in K\text{.}\) Notice that \(a\inv, ab,\) and \(a+b\) are contained in \(F(a,b)\text{.}\) As \(b\in K\text{,}\) there exists some polynomial with coefficients in \(F\) such that \(b\) is a root. However, this polynomial also lives in \(F(a)\text{,}\) so we have \([F(a,b):F(a)]\) algebraic. As \([F(a):F]\) is also algebraic, we have \(F[a,b]\) algebraic as well, as it is a finite extension by the The Degree Formula. Thus \(a+b, a\inv,\) and \(ab\) are algebraic over \(F\) as well, making \(K\) a field.
Let \(\a\in K\text{.}\) Thus \(\a\) is the root of a polynomial \(f(x)=a_nx^n+\dots+a_0\text{,}\) where \(a_{i}\in F\text{.}\) Notice that \(f\) is a polynomial in \(F(a_n,\dots,a_0)\) as well, and thus \(\a\) is algebraic over this extension as well. Consider the chain of extensions
As \(a_i\) is algebraic over \(F\) for all \(i\) and \(\a\) is algebraic over \(F(a_0,\dots,a_n)[x]\) we see that each step in this chain has finite degree by. By the The Degree Formula, \([F(a_0, \dots, a_{n}, \a): F]\) is finite and thus so is \([F(\a):F]\text{.}\) Thus \(\a\) is algebraic over \(F\text{,}\) hence \(\a\in K\text{,}\) making \(K\) algebraically closed.
ActivityC.109.Problem 9.
Let \(f (x) \in \Q[x]\) be an irreducible cubic (degree \(3\)) polynomial having exactly one real root. Let \(L\) be the splitting field of \(f(x)\) over \(\Q\text{.}\) Show that \(\Gal(L/\Q) \cong S_3\text{.}\)
Solution.
Let \(f(x) \in \Q[x]\) be an irreducible cubic (degree \(3\)) polynomial having exactly one real root, and let \(L\) be the splitting field of \(f(x)\) over \(\Q\text{.}\)
As \(L\) is the splitting field of \(f\text{,}\) it is a normal extension. As \(\Q\) has characteristic \(0\text{,}\)\(L\) is separable (because \(L\) is algebraic extension, as its the extension caused by adjoining each root of \(f\text{,}\) and algebraic extensions of algebraic extensions are algebraic). Thus by we see \(|\Aut(L/\Q)|=[L:F]\text{.}\)
By Proposition 2.83 CITEX we see that \(\Aut(L/F)\) is isomorphic to some subgroup of \(S_m\text{,}\) where \(m\) is the number of distinct roots of \(f\text{.}\) As \(f\) cubic and irreducible we know that the real root must be irrational, which we will denote \(\alpha\text{.}\) Consider the extension \(\Q(\alpha)\text{.}\) As \(\alpha\not\in\Q\text{,}\) we see that \([\Q(\alpha):\Q]\geq 2\text{.}\) However, neither of our complex roots are in this extension, and so another extension is needed to reach \(L\text{.}\) But this extension would also have a degree larger than \(1\text{,}\) so \([L:\Q]\geq4\text{.}\) As \(|S_2|=2\text{,}\) there exists no subgroup of it that \(\Aut(L/F)\) can be isomorphic to, given that \(|\Aut(L/\Q)|=[L:F]\text{.}\) Thus we see that \(m=3\text{,}\) meaning that our complex roots are distinct.
Then \(|\Aut(L/F)|=1,2,3,\) or \(6\text{,}\) the only possible sizes of subgroups of \(S_3\text{.}\) However, by the previous argument we see that as \([L:\Q]\geq4\text{,}\) the only viable subgroup of \(S_3\) is \(S_3\) itself. Thus \(\Aut(L/\Q) \cong S_3\text{.}\)
