Show that the direct product group \(\Aut(H)\times \Aut(K)\) is isomorphic to a subgroup of \(\Aut(H\times K)\text{.}\)
Give an example, with justification, of groups \(H\) and \(K\) for which \(\Aut(H) \times\Aut(K)\) is not isomorphic to \(\Aut(H\times K)\text{.}\)
Solution.
Let \(H\) and \(K\) be groups.
First, let \(\psi\in\Aut(H)\) and \(\phi\in\Aut(K)\text{.}\) Let \(\varphi:H\times K\to H\times K\) be defined by \(\varphi(h,k)=(\psi(h),\phi(k))\text{.}\) Let \((h,k)\in H\times K\text{.}\) As \(\psi,\phi\) automorphisms, there exist \(h',k'\) such that \(\psi(h')=h\) and \(\phi(k')=k\text{.}\) Thus we have surjectivity, and hence bijectivity, given the equal cardinalities. The homomorphism element is verifiable, I’ll do it later. Thus \(\varphi\) is indeed an automorphism of \(H\times K\text{.}\)
Define \(f:\Aut(H)\times \Aut(K)\to\Aut(H\times K)\) by \(f(\psi,\phi)=\varphi\text{.}\) Suppose two elements map to the same place. Then they were the same! Thus, by the First Isomorphism CITEX, we have the result.
Let \(H = K = \mathbb{Z}_2\text{,}\) the cyclic group of order \(2\text{.}\) Then \(\Aut(H) \cong \mathbb{Z}_2\) since there is only one non-identity automorphism (namely, the map that sends the generator to its inverse), and \(\Aut(K) \cong \mathbb{Z}_2\) for the same reason. Therefore, \(\Aut(H) \times \Aut(K) \cong \mathbb{Z}_2 \times \mathbb{Z}_2\text{,}\) which is abelian.
On the other hand, \(H\times K \cong \mathbb{Z}_2\times\mathbb{Z}_2\) is also abelian. Let \(\varphi\) be the nontrivial automorphism of \(\mathbb{Z}_2\text{,}\) which sends \(0\) to \(1\) and \(1\) to \(0\text{.}\) Then the automorphism \((g,h) \mapsto (\varphi(g), \varphi(h))\) is nontrivial and sends \((0,0)\) to itself, so it is an automorphism of \(H\times K\text{.}\) Therefore, \(\Aut(H\times K)\) contains a nonabelian subgroup, and in particular \(\Aut(H) \times \Aut(K)\) is not isomorphic to \(\Aut(H\times K)\text{.}\)
ActivityC.93.Problem 2.
Let \(K\) be a normal subgroup of a group \(H\text{.}\) Recall that a group \(G\) is solvable if there is a sequence of subgroups
\begin{equation*}
1 = N_0\nsg N_1\nsg\dots\nsg N_j = G
\end{equation*}
for some \(j \leq 0\) such that for all \(i \in \{0,\dots, j-1\}\) the quotient group \(N_{i+1}/N_i\) is abelian. Show that if both \(K\) and \(H/K\) are solvable groups, then \(H\) is solvable.
Solution.
Suppose that \(G\) is solvable, and \(H\) is a subgroup of \(G\text{.}\) Since \(G\) is solvable, it has a filtration:
\begin{equation*}
1 = N_0\nsg N_1\nsg\dots\nsg N_j = G
\end{equation*}
where \(N_{i+1}/N_i\) is abelian. Let \(H_{i} = H\cap N_i\text{.}\) Since \(N_i\) is normal in \(N_{i+1}\text{,}\) it is obvious that \(H_i\) is normal in \(H_{i+1}\text{.}\)
ActivityC.94.Problem 3.
Determine all of the groups of order \(45\text{,}\) up to isomorphism
\(n_3|5\) and \(n_3\equiv 1\mod{3}\text{,}\) so \(n_3=1\)
\(n_5|\) and \(n_5\equiv1\mod{5}\text{,}\) so \(n_5=1\) as well.
Thus there is exactly one Sylow \(5\)-subgroup, \(P\text{,}\) and exactly one Sylow \(3\)-subgroup, \(Q\text{.}\) Both are normal in \(G\text{.}\) Notice that \(Q\) has order \(9\text{,}\) a prime squared. Thus \(Q\) is abelian. By the [cross-reference to target(s) "thm-ftfgab" missing or not unique], \(Q\) is either isomorphic to \(\Z_3\times\Z_3\) or \(\Z_9\text{.}\) Thus \(G\cong\Z_5\times\Z_9\) or \(G\cong\Z_5\times\Z_3\times\Z_3\text{.}\)
SubsectionSection
ActivityC.95.Problem 4.
In the commutative ring \(R = \Z[\sqrt{-3}]\text{,}\) show that the element \(2\) is irreducible but not prime.
Solution.
First, notice that \(2\cdot 2=6=(1+\sqrt{-3})(1-\sqrt{-3})\text{.}\) Define a function
Thus \((a^2+3b^2)=\pm1,\pm2,\) 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 \(\Z[\sqrt{-3}]\text{.}\)
Suppose by way of contradiction that \(2\) is prime in \(\Z[\sqrt{-3}]\text{.}\) Note that \(2|6=(1+\sqrt{-3})(1-\sqrt{-3})\text{.}\) Thus \(2\) divides one of these factors.
First, suppose there exists some \(a+b\sqrt{-3}\) such that \(2(a+b\sqrt{-3})=(1\pm\sqrt{-3})\text{.}\) Thus \(2a+2b\sqrt{-5}=1\pm\sqrt{-3}\text{,}\) 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 \(\Z[\sqrt{-3}]\text{.}\)
ActivityC.96.Problem 5.
Let \(R\) be a commutative ring (with \(1\)) and let \(M\) be an \(R\)-module. Show that if \(M\neq 0\) and the only submodules of \(M\) are \(0\) and \(M\text{,}\) then there is a maximal ideal \(I\) of \(R\) such that \(M\) is isomorphic to \(R/I\text{.}\)
Solution.
Let \(R\) be a commutative ring (with \(1\)) and let \(M\) be an \(R\)-module such that \(M\neq 0\) and the only submodules of \(M\) are \(0\) and \(M\text{.}\)
As \(M\neq 0\text{,}\) there exists a non-zero element \(m\in M\text{.}\) Let \(f:R\to M\) be defined by \(f(r)=rm\text{.}\) Since \(1\in R\) and \(1m=m\in Rm\text{,}\) we have \(Rm=M\text{.}\) By the [cross-reference to target(s) "thm-fit-module" missing or not unique] we see that \(R/\ker(f)\cong M\text{,}\) making \(\ker(f)\) an ideal in \(R\text{,}\) which we shall conspicuously denote as \(I\) henceforth. By the Theorem 9.26 the only two ideals of \(R/I\) are \(0\) and \(R/I\text{,}\) making \(R/I\) a field, and \(I\) a maximal ideal in \(R\text{.}\)
Not so bad! Now, the invariant factors all divide the characteristic polynomial and must divide the following factor, so our options for sets of invariant factors are the following: - \(\{-x(x+1)^2\}\) - \(\{(x+1),-x(x+1)\}\) However, the largest invariant factor is also the minimal polynomial. So we check to see if \(-A(A+I)=0\text{.}\) Luckily, the very first calculation shows that this is not the case. Thus the minimum polynomial is the characteristic polynomial is the only invariant factor of \(A\text{.}\) Thus the RCF of \(A\) is
Luckily for us, \(\cp_A(x)\) factors completely into linear terms! So our elementary divisors are \(-x\) and \((x+1)^2\text{.}\) We see \(J_1(0)=\begin{bmatrix} 0 \end{bmatrix}\) and \(J_2(-1)=\begin{bmatrix} -1 & 0 \\ 1 & -1\end{bmatrix}\text{,}\) so the Jordan Canonical form of \(A\) is
Suppose that \(K/F\) is a finite extension of fields such that the degree \([K : F ]\) is odd. Show that if \(b \in K\text{,}\) then \(F (b) = F (b^2)\text{.}\)
Solution.
Let \(K/F\) be a finite extension of fields such that the degree \([K : F ]\) is odd.
Notice that \(F(b^2)\subseteq F(b)\) as everything in \(b^2\) can be written in terms of \(b\text{.}\) Suppose by way of contradiction that there exists some \(\a\in F(b)\) such that \(\a\not\in F(b^2)\text{.}\) Then \(b\not\in F(b^2)\text{.}\) Notice that \(b\) is a root of the polynomial \(f(x)=x^2-b^2\in F(b^2)\text{,}\) which is irreducible in \(F(b^2)\) as it is a degree \(2\) polynomial that has no roots in \(F(b^2)\text{.}\) Thus \(f\) is the minimum polynomial of \(b\) and \([F(b):F(b^2)]=2.\)
However, by the The Degree Formula\([K:F]=[K:F(b)][F(b):F(b^2)][F(b^2):F]\text{,}\) which is a problem, given \(2\) now divides \([K:F]\text{,}\) an odd number. Thus \(F (b) = F (b^2)\text{.}\)
ActivityC.99.Problem 8.
Let \(L/F\) be an extension of fields and let \(a, b \in L\text{.}\) Show that
Let \(\s\in\Aut(L/F(a,b))\text{.}\) Then \(\s\) is an automorphism of \(L\) that restricts to the identity on \(F(a,b)\text{.}\) As \(F(a),F(b)\subseteq F(a,b)\) we see that \(\s\) restricts both of these to the identity as well. Thus \(\s\in\Aut(L/F (a)) \cap \Aut(L/F (b))\text{.}\)
Next, suppose \(\s\in\Aut(L/F (a)) \cap \Aut(L/F (b))\text{.}\) Thus \(\s\) restricts to the identity on \(F(a)\) and \(F(b)\text{.}\) Let \(x\in F(a,b)\text{.}\)
ActivityC.100.Problem 9.
Let \(K\) be the splitting field of \(x^4-3\) over \(\Q\)
Find a basis for \(K\) as a vector space over \(\Q\text{.}\)
Show that \(\Aut(K/Q)\) is not abelian.
Solution.
First, note that \(f\) is irreducible in \(\Q[x]\) by Eisenstein’s Criterion (\(p=3\)). The roots of \(f\) are - \(\sqrt[4]3\text{,}\) - \(\sqrt[4]3i\text{,}\) - \(-\sqrt[4]3\text{,}\) and - \(-\sqrt[4]3i\text{.}\) Let \(F=\Q(\sqrt[4]3)\) and notice \([F:\Q]=4\text{.}\) As \(F\subseteq\R\) we see that \([F(i):F]=2\text{,}\) and thus \([K:\Q]=8\text{,}\) meaning our basis will have eight elements: 1. 1 2. \(\sqrt[4]3\) 3. \((\sqrt[4]3)^2\) 4. \((\sqrt[4]3)^3\) 5. \(i\) 6. \(\sqrt[4]3i\) 7. \((\sqrt[4]3)^2i\) 8. \((\sqrt[4]3)^3i\)
Now that \(K\) has been verified to be a splitting field, we see that \(\Aut(K/Q)\) is isomorphic to a subgroup of \(S_4\) of degree \(8\text{.}\)
Any subgroup of \(S_4\) of order \(8\) is a Sylow \(2\)-subgroup. By Sylow’s Theoremss, the number of Sylow \(2\)-subgroups is either \(1\) or \(3\text{.}\)
If there are three they are all conjugate, and conjugation induces an isomorphism on the group, we see that all three subgroups are isomorphic.
Let \(X\) be the set of left cosets of the subgroup \(\{e,s\}\) of \(D_8\text{.}\) Note that \(|X| = 4\text{.}\) Let \(G\) act on \(X\) by left multiplication. This action induces a homomorphism \(\varphi : G \to P(X)\) where \(P(X)\cong S_4\) is the permutation group on \(X\text{.}\) As shown in class, the kernel of this homomorphism is the largest normal subgroup contained in \(\{e,s\}\text{,}\) which is \(\{e\}\text{.}\) Thus, \(\varphi\) is injective and the image of \(\varphi\) is a subgroup of \(S_4\) isomorphic to \(D_8\text{.}\)
