Let \(G = A_7\) and \(S\) be the set of elements of \(G\) of order \(7\text{.}\) Prove that \(S\) is not a conjugacy class of \(G\text{.}\)
Solution.
Elements of order \(7\) in \(G\) must permute all \(7\) elements, making them of the form \((abcdefg)\text{.}\) There are \(6!\) such permutations, given that we can always reorder them so that \(a\) is first, leaving \(6\) remaining spots without replacement. Let \(S\) denote the set of these \(6!\) elements.
Suppose by way of contradiction there existed some \(s\in G\) such that \(S=C_s(G)\text{.}\) Let \(G\) act on itself through conjugation, making \(S=C_s(G)=\Orb(s)\) by the definition of conjugacy class. The [cross-reference to target(s) "cor-orbit-stabilizer" missing or not unique] tells us that \(|G|=|\Orb(s)|\cdot|\Stab(s)|\text{,}\) meaning that the order of \(S\) must divide the order of \(G\text{.}\) However, \(|G|=\frac{7!}{2}\) and \(|S|=6!\text{,}\) so this is a contradiction. Thus \(S\) is not a conjugacy class of \(G\text{.}\)
ActivityC.47.Problem 2.
Let \(G\) be a group acting transitively on a set \(S\text{.}\) For \(a\in S\text{,}\) let \(G_a = \{g\in G | ga = a\}\) be the stabilizer of \(a\) under the action from \(G\text{.}\)
Let \(a, b\in S\text{.}\) Prove there exists \(g\in G\) such that \(G_b = gG_ag\inv\text{.}\)
Suppose \(S\) has more than one element and that \(G\) is finite. Prove that there exists \(g\in G\) which has no fixed point, that is, for all \(a\in S, ga\neq a\)
Solution.
Let \(G\) be a group acting transitively on a set \(S\text{.}\)
The action is transitive, so there exists some \(g\in G\) such that \(ga=b\text{.}\)
Let \(x\in\Stab(a)\text{,}\) consider \(gxg\inv b\text{.}\) Since \(g\inv b=a\text{,}\) we have \(gxa\text{,}\)\(x\) stabilizes \(a\) so now \(ga\text{,}\) but \(ga=b\text{.}\) So \(g\Stab(a)g\inv\subseteq\Stab(b)\text{.}\)
Let \(x\in\Stab(b)\text{.}\) We know \(xb=b\) and also \(ga=b\text{,}\) and so \(xga=ga\text{.}\) Applying \(g\inv\) to both to see \(g\inv xga=a\text{.}\) This puts \(g\inv xg\in\Stab(a)\text{.}\) Thus \(x\in g\Stab(a)g\inv\text{,}\) completing the proof.
The action is transitive, so there is only one orbit, and so \(\Orb(a)=\Orb(b)=G\text{.}\) Since \(G\) is finite, the [cross-reference to target(s) "cor-orbit-stabilizer" missing or not unique] tells us \(|\Orb(a)|=|G|/|\Stab(a)|\text{.}\) Since every orbit has the same order as \(G\text{,}\) we see \(\Stab(a)=\{e\}\) for all \(a\in S\text{.}\) Thus there cannot exist any fixed points.
ActivityC.48.Problem 3.
Let \(G\) be a group of order \(175 = 3^2\cdot 5^2\) and suppose \(G\) contains an element of order \(25\text{.}\) Prove that \(G\) is abelian.
Solution.
First, note that \(3^2\cdot 5^2=225\text{,}\) not \(175\text{.}\) You hate to see it. Anyway, let \(x\) be an element of order \(25\) and consider \(H=\langle x\rangle\text{,}\) a cyclic subgroup of order \(25\text{.}\) The possible number of Sylow \(5\)-subgroups of \(G\) is exactly \(1\text{,}\) making \(H\) this subgroup and thus normal in \(G\text{.}\) Let \(K\) be a Sylow \(3\) subgroup, it intersects \(H\) trivially and thus \(G\cong H\sdp_{\rho}K\text{,}\) where \(\rho:K\to\Aut(H)\text{.}\) The order of \(\Aut(H)\) is \(20\text{,}\) which is relatively prime to \(|K|\text{,}\) making \(\rho\) trivial and \(G=H\times K\) and thus abelian.
SubsectionSection II: Rings, Modules, and Linear Algebra
ActivityC.49.Problem 4.
Let \(R\) be a commutative ring with identity, and assume \(1\neq0\text{.}\)
Prove that every maximal ideal of \(R\) is a prime ideal.
Assume \(R\) is a finite ring. Prove that every prime ideal is a maximal ideal.
Solution.
See idenitical problem CITEX
ActivityC.50.Problem 5.
Let \(R\) be an integral domain and let \(M\) be an \(R\)-module. Recall that a subset \(S\) of \(M\) is linearly independent if whenever \(∑_i r_is_i = 0\) for ring elements \(r_i\) and elements \(s_i ∈ S\text{,}\) we must have \(r_i = 0\) for all \(i\text{.}\) We say \(S\) is maximally linearly independent if it is linearly independent and it is not properly contained in a linear independent subset of \(M\text{.}\) Finally, recall that we say an \(R\)-module \(P\) is torsion if for all \(p \in P , rp = 0\) for some non-zero \(r\in R\text{.}\)
Suppose \(S\) is a linearly independent subset of \(M\) and let \(N\) be the submodule of \(M\) generated by \(S\text{.}\) Prove it is maximally linearly independent if and only if \(M/N\) is torsion.
Prove that if for every module \(M\) every maximally independent subset of \(M\) generates \(M\) as an \(R\)-module, then \(R\) must be a field.
Solution.
Suppose \(S\) is a linearly independent set \(S\subseteq M\) and let \(N=\text{span}(S)\text{.}\)\((\implies)\) Assume \(S\) is maximal linearly independent. To show \(M/N\) is torsion we need to show for an arbitrary \(m+n\in M/N\) that there exists \(r\neq 0\in R\) such that \(r(m+N)=0+N\text{.}\) We will skip \(0+N\) since this is true for all \(r\in R\text{.}\)
Let \(m+N\in M/N\) such that \(m+N\neq 0+N\text{.}\) This means that \(m\not \in N=\text{span}(S)\) so \(m\not \in S\text{.}\)\(S\) is maximal linearly independent so \(S\cup \{m\}\) is linearly dependent. Therefore there exists
where not all \(r_i=0\text{.}\) Note that \(m\) must be in this sum (and its coefficient can’t be 0), since if it isn’t, it breaks the linear independence of \(S\text{.}\) We will rearrange so that \(m\) is first and we get
so \(-r_1m\in \text{span}(S)=N\text{.}\) Since \(r_1\in R, -r_1\in R\) and \(r_1\neq 0,-r_1\neq 0\text{.}\) For this \(-r_1, -r_1(m+N)=-r_1m+N=0+N\text{.}\) So \(M/N\) is torsion.
\((\impliedby)\) Assume \(M/N\) is torsion. To show \(S\) is maximal linearly independent we just need to show for any \(T\) such that \(S\subsetneq T\subseteq M\) that \(T\) is linearly dependent. Let \(T\) be such a set, thus there exists \(t\in T\setminus S\text{.}\) Consider \((t+N)\text{.}\)\(M/N\) is torsion so there exists \(r\neq 0\in R\) such that \(r(t+N)=(rt+N)=0+n\implies rt\in N=\text{span}(S)\text{.}\) Thus
\begin{equation*}
rs=\sum_{i=1}^nr_is_i,r_i\in R,s_i\in S
\end{equation*}
with not all 0 coefficients so \(T\) is linearly dependent. Note that this works when \(t\) is already in the span of \(S\text{,}\) the \(r\) just equals \(1\text{.}\) Finally we find that \(S\) is maximal.
Assume for every \(R\)-module \(M\) that each maximal linearly independent subset \(S\) generates \(M\text{.}\) By HW2 for \(R\) commutative and \(0\neq 1\) (which we have since \(R\) is an integral domain), if every \(R\)-module is free, then \(R\) is a field. For every \(M\) to be free we need it to have a linearly independent basis, which these assumptions imply for all modules with a maximal linearly independent subset. Since if one such \(S\) exists, then \(M=\text{span}(S)\) and \(S\) is linearly independent, \(M\) is free. So we just need to show every \(R\)-module \(M\) has a maximal linearly independent subset.
To show every module has a maximal linearly independent subset we use Zorn’s lemma.
Let \(P\) be a collection of linearly independent subsets of \(M\text{.}\)\(P\) is nonempty since \(\emptyset\) is linearly independent. Let \(B\) be a totally ordered subset of \(P\text{.}\) If \(P\) is empty then \(B\) is bounded above by \(\emptyset\text{.}\) So assume \(B\) is nonempty. We will show that
Each \(b_i\) is in \(J_i\) for some \(J_i\in B\) and there is a finite amount of them so there is one \(J_i\) that contains all of the others. Since \(M\) is abelian we can arrange these in any order so we will arrange them so that \(J_1\subseteq J_2\subseteq \cdots \subseteq J_n\text{.}\) (These could all be the same, but we don’t care.) Since all of the \(b_i\) must be in \(J_n\) and \(J_n\) is linearly independent, we must have that \(r_i=0\) for each \(i\) and thus \(I\) is linearly independent so \(B\) is bounded above by an element in \(P\text{.}\) Thus we can apply Zorn’s lemma \(P\) has some maximal element.
(just to clarify) Every \(M\) has a maximal linearly independent subset, \(S\text{,}\) as an \(R\)-module. For this \(R\text{,}\) this \(S\) generates \(M\) so \(M\) is free. So under this \(R\) every module is free, and by HW2 since \(R\) commutative with \(1\neq 0\) (because it is an integral domain), \(R\) is a field.
ActivityC.51.Problem 6.
Let \(R\) be a commutative ring with \(1\neq 0\) and let \(f : R^m\to R^n\) be a surjective \(R\)-module homomorphism, with \(m\) and \(n\) integers. Prove \(m \geq n\text{.}\)
Solution.
Let \(R^n=N\) and \(R^m=M\text{.}\)
Let \(I\) be a maximal ideal in \(R\text{.}\) Thus \(R/I\) is a field. Lemma tells us that \(M/IM\) and \(N/IN\) are \(R/I\)-vector spaces. Additionally, this gives rise to \(\overline{p}:M/IM\to N/IN\text{,}\) which is a surjective \(R/I\)-module linear transformation.
Note that \(M\) is generated by \(e_i+IM\) for \(i\leq m\text{.}\) Let \(a_i\in R\) and consider \(\sum a_ie_i=0\text{.}\) For this to be \(0\) we need it to be in \(IM\text{,}\) and thus all \(a_i\in I\text{.}\) So the set of \(e_i\) is a basis for \(M/IM\) with \(m\) elements. Likewise \(N/IN\) has a basis with \(n\) elements. As we are surjective, Rank \(=n\text{,}\)\(\dim=m\text{.}\) So by Rank-Nullity CITEX \(\dim(\ker)=m-n\) which is only positive with \(m\geq n\text{.}\)
and let \(V\) the \(\Q\)-vector space obtained from \(M\) by restriction of scalars along the evident inclusion \(\Q \subseteq \Q[x]\) and let \(t : V \to V\) be the \(\Q\)-linear transformation given as multiplication by \(x\text{.}\) Find, with justification, the rational canonical form of \(t\text{.}\)
and let \(W\) the \(\C\)-vector space obtained from \(N\) by restriction of scalars along \(\C \subseteq \C[x]\) and let \(t : W \to W\) be the \(\C\)-linear transformation given as multiplication by \(x\text{.}\) Find, with justification, the Jordan canonical form of \(t\text{.}\)
Solution.
Coming Soon!
ActivityC.53.Problem 8.
Let \(f (x) = x^p-5 \in \Q[x]\) where \(p\) is an odd prime, and let \(L\) be the splitting field of \(f (x)\) over \(\Q\text{.}\) Find, with justification, \([L : \Q]\text{.}\)
Solution.
Let \(f (x) = x^p-5 \in \Q[x]\) where \(p\) is an odd prime, and let \(L\) be the splitting field of \(f (x)\) over \(\Q\text{.}\) Using Eisenstein’s Criterion with \(p=5\) we see that \(f\) is irreducible in \(\Q[x]\text{.}\) Notice that \(\sqrt[p]5\) is a root of this polynomial. As \(f\) is monic and irreducible it is the minimum polynomial of \(\sqrt[p]5\text{,}\) and thus \([\Q(\sqrt[p]5):\Q]=p\text{.}\) Let \(\z\) be a primitive \(p\th\) root of unity. Notice that \(\z\) is a root of the \((p-1)\th\) cyclotomic polynomial, \(g(x)\text{,}\) which is irreducible in \(\Q[x]\) by the Gospel of Mark.
As \(L\) is the splitting field of \(g\text{,}\) there exists a root \(\a\) of \(g\in L\text{.}\) Consider \(\Q(\sqrt[p]5,\a)\text{.}\) As \(\a\) is algebraic in \(\Q(\sqrt[p]5,\a)\) we know there exists some unique irreducible minimal polynomial \(h\) of degree \(q\text{,}\) and thus that \([\Q(\sqrt[p]5,\a):\Q(\sqrt[p]5)]=q\text{.}\) By the The Degree Formula we see
However, \([\Q(\sqrt[p]5,\a):\Q]=[\Q(\sqrt[p]5,\a):\Q(\a)][\Q(\a):\Q]\) and so \(pq=(p-1)m\) for some \(m\in\Z\text{,}\) so \((p-1)|pq\text{.}\) As \(\gcd((p-1),p)=1\) we must have \((p-1)|q\text{.}\) But \(q\) was defined to be the degree of \(h\text{,}\) which divides \(g\text{.}\) As \((p-1)|q\) and \(q\leq (p-1)\text{,}\) we see that \((p-1)=q\text{,}\) so \(g=kh\) for some \(k\in \Q\text{.}\) As irreducible polynomials multiplied by a constant are still irreducible, we see that \(g\) is indeed irreducible in \(\Q(\sqrt[p]5)\text{.}\)
As \(g\) is monic and irreducible in \(\Q(\sqrt[p]5)\) we see that it is the minimum polynomial of \(\z\text{,}\) and thus \([\Q(\sqrt[p]5,\a):\Q(\sqrt[p]5)]=(p-1)\text{.}\) As \(\Q(\sqrt[p]5,\a)=L\text{,}\) we have \([L:\Q]=p(p-1)\text{.}\)
ActivityC.54.Problem 9.
Let \(F\) be a field, \(V\) an \(F\)-vector space, and \(W\) a subspace of \(V\text{.}\) A subspace \(U\) of \(V\) is called a complement of \(W\) in \(V\) if \(V\) is the internal direct sum of \(W\) and \(U\text{;}\) that is, \(V=W\oplus U\text{.}\)
Prove that for every \(V\) and \(W\) as above, \(W\) has at least one complement in \(V\text{.}\)
Prove that if \(U\) is a complement of \(W\) in \(V\) and \(V\) is finite dimensional, then \(\dim_F (V ) = \dim_F (W ) + \dim_F (U )\) (where \(\dim_F\) denotes the dimension of an \(F\)-vector space).
Solution.
Let \(F\) be a field, \(V\) an \(F\)-vector space, and \(W\) a subspace of \(V\text{.}\)
Let \(A\) denote the set of all subspaces of \(V\) such that \(A\cap W=\{0\}\text{.}\) We can order \(A\) with respect to inclusion. Let \(X\) be a totally ordered subset of \(A\text{,}\) and let \(B\) be the union of all the elements in \(X\text{.}\) Unions of subspaces are subspaces, and by DeMorgan’s Laws we see that \(B\cap W=\{0\}\text{.}\) Thus by Zorn’s Lemma there exists a maximal element of \(A,\) which we denote \(U\text{.}\) So \(U\cap W=\{0\}\) by definition.
Suppose by way of contradiction there exists some \(x\in V\) such that \(x\not\in U+W\text{.}\)
Consider \(U+Vx\text{.}\) As \(x\not\in W\text{,}\)\(vx\not\in W\) for all \(v\in V\text{,}\) as we could just multiply by \(v\inv\text{.}\) Thus \(W\cap U+Vx=\{0\}\) and \(U\subsetneq U+Vx\text{,}\) a contradiction, as \(U\) was maximal. Thus \(U\) is a complement of \(W\text{.}\)
The Second Isomorphism Theorem tells us that \(U+W/W\cong U/(U\cap W)\text{.}\) We also know
