Let \(G\) be a finite group and let \(H\) be a subgroup of \(G\) of index \(p\text{,}\) where \(p\) is the smallest prime dividing the order of \(G\text{.}\) Prove that \(H\) is a normal subgroup of \(G\text{.}\)
Hint.
Use an action of \(G\) on the set of cosets of \(H\text{.}\)
Solution.
Let \(G\) act on left cosets of \(G/H\) by left multiplication. This induces a homomorphism \(f:G\rightarrow Perm(G/H)\text{.}\) Let \(K\) be \(ker(f)\text{.}\) Since \(kH=H\) for all \(k\in K\) we know that \(K\subseteq H\text{.}\) Let \([H:K]=n\text{.}\) By the First Isomorphism Theorem, \(G/K\cong P\leq Perm(G/H)\) so \([G:K]|p!\text{.}\) We have \([G:K]=[G:H][H:K]=pn\text{,}\) so \(pn\divides p!\implies n\divides (p-1)!\text{.}\) All of \(n\)’s prime factors must be greater than or equal to \(p\) since \(p\) is the smallest prime that divides \(|G|\text{.}\) However since the only factors of \((p-1)!\) are less than \(p\) it must be that \(n=1\text{,}\) so \([H:K]=1\) and \(K\subseteq H\) together imply \(H=K\text{.}\) So \(H\) is the kernel of a homomorphism and thus is normal in \(G\text{.}\)
ActivityC.11.Problem 2.
Give two groups of order \(21\) which are not isomorphic, with justification.
Prove that there are at most three groups of order \(21\) up to isomorphism.
Solution.
First, notice \(21=pq\) where \(p=3,q=7\) and because \(3|7-1=6\) there are exactly \(2\) groups of order \(21\) up to isomorphism, \(C_{21}\) and the nonabelian group \(C_7\rtimes_{\rho} C_3\) for a nontrivial homomorphism \(\rho:C_3\rightarrow \Aut(C_7)\text{.}\) These two groups are not isomorphic since abelianness is an isomorphism invariant.
Let \(G\) be a group of size \(21\text{.}\) There is only \(1\) Sylow \(7\)-subgroup since the number of them has to be congruent to \(1 mod 7\) and divide \(3\text{.}\) Thus the unique Sylow \(7\) subgroup, call it \(N\text{,}\) is normal. Letting \(H\) be any Sylow \(3\)-subgroup we know\(|NH|=\frac{|N||H|}{|N\cap H|}=\frac{7\cdot 3}{1}=21\text{,}\) since the intersection has to be trivial and thus \(NH=G\text{.}\) By the Recognition Theorem for Internal Semidirect Products\(G\cong N\rtimes_{\rho} H\) for some \(\rho:H\rightarrow \Aut(N)\text{.}\)\(N\cong \mathbb{Z}_7\) and all automorphisms will be determined by where a generator is sent to. We need autmorphisms that, when applied 3 times, will return to the identity. There is the identity automorphism, but then there are only two more. Those are the automorphisms defined by sending 1 to 2 and by sending 1 to 4. This is because \(2^3=8=1mod7\) and \(4^3=64=1mod7\text{.}\) The others don’t work because \(3^3=5^3=6^3=6mod7\) so there are at most 3 groups of size 21. (Technically there are just the two from part a, but this question is actually very generous because it doesn’t make you prove the conjugacy of the two nontrivial homomorphisms)
ActivityC.12.Problem 3.
Let \(G\) be a finite abelian group and call the exponent of \(G\) the smallest positive integer \(n\) which satisfies \(g^n=e_G\) for all \(g\in G\text{,}\) where \(e_G\) is the identity element of \(G\text{.}\)
Give, with justification, a formula for the exponent of \(G\) in terms of the invariant factors of \(G\text{.}\) (As a reminder for those who confuse “invariant factors" and “elementary divisors": the invariant factors form a list in which each one divides the next.)
For \(p\) prime, determine the exponent of the abelian group \(\mathbb{Z}_p^{\times}\) and use it to show that this group is cyclic.
Solution.
For invariant factors, we can write \(G\cong \mathbb{Z}_{n_1}\times \cdots \times \mathbb{Z}{n_t}\) for \(t\geq 0\text{,}\)\(n_i\geq 2\) where each \(n_{i+1}|n_i\text{.}\) The exponent will be \(n_1\text{.}\) This can be seen because each \(n_i\) divides it, and thus for an arbitrary \((g_1,g_2,...,g_t)\)\(g_1^{n_1}=g_2^{n_1}=g_2^{n_2\cdot k_2}=g_3^{n_1}=g_3^{n_3\cdot k_3}=\cdots =g_t^{n_1}=g_t^{n_t\cdot k_t}=0\) for integers \(k_j\text{.}\) So \((g_1,\dots,g_t)^{n_1}=(0,\dots,0)\text{.}\) So the exponent is less than or equal to \(n_1\) since it is the smallest such integer that works. No integer less than \(n_1\) will work though since \((1,0,\dots,0)\) has order \(n_1\) so the exponent is both greater than or equal to \(n_1\) and less than or equal to \(n_1\) so it is therefore \(n_1\text{.}\)
Let \(p\) be prime. \(\mathbb{Z}_p^{\times}\) is a finite abelian group so by Fundamental Theorem of Finitely Generated Abelian Groups (FTFGAG) we can write in its invariant factor form: \(\mathbb{Z}_p^{\times}\cong \mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times \cdots \mathbb{Z}_{n_t}\) where \(t\geq 0\text{,}\)\(n_i\geq 2\forall i\) and \(n_{i+1}|n_i\forall i\) and this is unique. We have that in \(\mathbb{Z}_p\) for all \(m>0\) there are at most \(m\) solutions to \(x^m=1\text{.}\) From the previous part, we know that the exponent is \(n_1\) so there are at most \(n_1\) solutions to \(x^{n_1}=1\text{.}\) But the definition of exponent tells us that all elements of \(\mathbb{Z}_p^{\times}\) satisfy this. So \(p-1\leq n_1\text{.}\) Therefore \(t\) must be 1 because we have \(p-1=n_1\cdot n_2\cdots n_t\) where each \(n_i\geq 2\) which would be a contradiction. Thus \(\mathbb{Z}_p^{\times}\cong \mathbb{Z}_{n_1}\) which is cyclic.
SubsectionSection II: Rings, Modules, and Linear Algebra
ActivityC.13.Problem 4.
Prove that in a principal ideal domain a nonzero ideal is prime if and only if it is maximal.
Solution.
\((\implies)\) Let \(R\) be a PID and \(I\) a prime nonzero ideal. Let \(J\) be an ideal such that \(I\subseteq J\subseteq R\text{.}\) We need to show that \(I=J\) or \(J=R\text{.}\) Since \(R\) is a PID, \(I=(a),J=(b)\) for some \(a,b\in R\text{.}\) There is some \(r\in R\) such that \(a=rb\) since \(I\subseteq J\text{.}\)\(I\) is a prime ideal so \(a=rb\in I\implies b\in I\) or \(r\in I\text{.}\)
If \(b\in I\) then \(J\subseteq I\) because the generator of \(b\) is in \(I\) and thus all of \(J\) is in \(I\) and thus they are equal. If \(r\in I\) then \(r=sa\) for some \(s\in R\text{.}\) Thus \(a=rb=sab=asb\) so \(a-asb=0\) so \(a(1-sb)=0\text{.}\) But this shows that \(1-sb=0\) since \(R\) is an ID and \(I\) is a nonzero ideal, so \(1=sb\) and \(b\) is a unit and \((b)=R\text{.}\) So \(I\) is maximal.
\((\impliedby)\) Let \(R\) be a PID and let \(I\) be a maximal ideal. \(R\) is a PID so it is commutative. Thus \(R/I\)is a field since \(I\) is maximal, and thus \(R/I\) is an ID. Now we want to show that \(I\) is prime and thus take \(xy\in I\text{.}\) We need to show that either \(x\in I\) or \(y\in I\text{.}\)\(xy\in I\implies (x+I)(y+I)=0+I\) and since the quotient ring is an ID this implies that \(x+I=0+I\) or \(y+I=0+I\) and thus one of them is in \(I\text{.}\) So \(I\) is prime.
ActivityC.14.Problem 5.
Let \(I\) be a nonzero ideal of the ring of Guassian integers \(\mathbb{Z}[i]\text{.}\) Prove that the quotient ring \(\mathbb{Z}[i]/I\) is finite.
Solution.
Recall that \(\mathbb{Z}[i]\) is a ED with norm \(N(c+di) = c^2+d^2\text{.}\) Thus, \(\mathbb{Z}[i]\) is a PID by Theorem 10.12. Then there exists \(a = x+yi \in \mathbb{Z}[i]\) such that \(I = (a)\text{.}\) Let \(b \in \mathbb{Z}[i]\text{.}\) Then there exists \(q,r \in \mathbb{Z}[i]\) such that \(b = qa + r\) such that either \(r = 0\) or \(N(r) < N(a)\text{.}\) Thus, as \(qa \in (a) = I\text{,}\)\(b + I = r + I\text{.}\) As \(0\) is unique, there is one \(r\) such that \(r = 0\text{.}\) Note as well that there are at most \(1000N(a)^2\) (this is quite an overestimate but it works just fine) elements \(r \in \mathbb{Z}[i]\) such that \(N(r) < N(a)\text{.}\) Hence, there are at most \(1000N(a)^2+ 1\) elements in \(\mathbb{Z}[i]/I\text{.}\)
ActivityC.15.Problem 6.
Determine, with justification, if the following two matrices with complex entries are similar.
They are not similar. \(A\) is already in RCF and we can see that we get \(C((x-2)^2)\oplus C((x-2)^2)\text{.}\) This means that the invariant factors are \((x-2)^2\) and \((x-2)^2\text{.}\)
\(B\) is already in JCF and \(B=J_1(2)\oplus J_1(2)\oplus J_2(2)\) which tells us that the elementary divisors are \((x-2),(x-2)\text{,}\) and \((x-2)^2\text{.}\) We can see that the invariant factors of \(B\) are \((x-2),(x-2)\text{,}\) and \((x-2)^2\text{.}\) Thus \(A\) and \(B\) have different smith normal forms and therefore aren’t similar.
SubsectionSection III: Fields and Galois Theory
ActivityC.16.Problem 7.
Let \(K\subseteq L\) be a finite extension of fields and assume \(f(x)\) is a polynomial with coefficients in \(K\) that is irreducible in the ring \(K[x]\text{.}\) Prove \(f(x)\) remains irreducible when regarded as an element of the ring \(L[x]\) provided \([L:K]\) is relatively prime to the degree of \(f(x)\text{.}\)
Solution.
Let \(d\) be the degree of \(f\text{.}\)Assume \(\gcd(d,[L:K])=1\text{.}\) By way of contradiction assume \(f(x)\) is reducible in \(L[x]\text{.}\) We will divide out the leading term so that \(f(x)\) is monic. This will not change the irreducibility of \(f\) in \(L[x]\) since if the result was reducible, then the original must have been reducible. We just do this because all of our theorems deal with monic polynomials specifically.
We are assuming \(f(x)\) is reducible in \(L[x]\text{,}\) so \(f(x)=cp_1(x)\cdots p_n(x)\) where \(c\) is a unit in \(L\text{,}\) and each \(p_i\) has degree \(0<r_i<d\) and is itself irreducible in \(L[x]\) and monic. Now let \(\alpha\) be a root of \(p_1(x)\) and consider \(L(\alpha)\text{,}\) a field extension of \(L\text{.}\)\(L\) contains \(K\) and \(\alpha\) so it is a field extension of \(K(\alpha)\) which is itself a field extension of \(K\text{.}\) So utilizing the degree formula, we get the following equality:
However since \(\alpha\) is the root of \(p_1(x)\) an irreducible monic polynomial in \(L[x]\text{,}\) the expression on the left is just \(r_1\cdot [L:K]\) since \(p_1=m_{\alpha,L}(x)\) and has degree \(r_1\text{.}\) The expression on the left is \([L(\alpha):K(\alpha)]\cdot d\) since \(f\) has degree \(d\) and in our assumption it is irreducible over \(K[x]\text{.}\) So we actually have
\begin{equation*}
r_1\cdot [L:K]=[L(\alpha):K(\alpha)]\cdot d
\end{equation*}
and because \(\gcd(d,[L:K])=1\) it must be that \(d\big\vert r_1\) a contradiction since \(0<r_1<d\text{.}\)
ActivityC.17.Problem 8 (*).
Let \(L\) be the splitting field of \(x^3-2\) over \(\mathbb{Q}\text{.}\)
Prove there is a unique intermediate field \(\mathbb{Q}\subseteq K\subseteq L\) such that \([K:\mathbb{Q}]=2\text{.}\)
Find, with justification, a primitive generator for the field \(K\) you found in part (a); that is, find an explicit element \(\alpha\in K\) such that \(K=\mathbb{Q}(\alpha)\text{.}\)
Solution.
Coming soon!
ActivityC.18.Problem 9 (*).
Let \(F\) be a field of characteristic \(p>0\) and let \(F\subseteq L\) be an extension of fields, and assume there is an element \(a\in L\) such that \(a^p\in F\) but \(a\not \in F\text{.}\) Prove \(\#Aut(F(a)/F)<[F(a):F]\text{.}\)
