Let \(p\) be a prime and \(H\) a finite \(p\)-group.
Suppose \(H\) acts on the finite set \(S\text{,}\) and let \(S_0\) be the fixed points of the action, i.e.,
\begin{equation*}
S_0 = \{x \in S | hx = x \text{ for all }h \in H\}.
\end{equation*}
Prove that \(|S| \equiv |S_0|\mod{p}\text{.}\)
Suppose \(G\) is a finite group and \(H \leq G\) (where \(H\) is still a finite \(p\)-group). Prove that \([N_G(H) : H]\equiv[G : H]\mod{p}).\)
Solution.
Let \(p\) be a prime and \(H\) a finite \(p\)-group.
Suppose \(H\) acts on the finite set \(S\text{,}\) and let \(S_0\) be the fixed points of the action. Let \(x\in S_0\text{,}\) and note that \(\Stab_H(x)=H\text{.}\) By [cross-reference to target(s) "cor-orbit-stabilizer" missing or not unique] we see \(|\Orb_H(x)|=1\text{.}\)
The orbits of an action partition the set \(S\text{,}\) and as orbits are either disjoint or equal, we see \(|S|=\sum\limits_{x\in S}|\Orb_H(s)|\text{.}\)
Coming soon!
ActivityC.129.Problem 2 (*).
A normal subgroup \(H\) of \(G\) is called a direct factor of \(G\) if there is some normal subgroup \(K\) of \(G\) such that \(G = H \times K\) (internal direct product).
If \(H\) is a direct factor of \(G\text{,}\)\(L\) is any group, and \(\phi : H \to L\) is any group homomorphism, prove that \(\phi\) extends to a homomorphism \(\widetilde{\phi} : G \to L\) such that \(\widetilde\phi(h) = \phi(h)\) for all \(h \in H\text{.}\)
Provide a counter-example (with justification) to the previous part if \(H\) is assumed to be a normal subgroup of \(G\text{,}\) but not a direct factor. Hint: One such example involves \(G = \Z\text{.}\)
Solution.
Let \(L\) be a group, \(H\) be a direct factor of \(G\text{,}\) and \(\phi : H \to L\) a group homomorphism. As \(H\) is a direct factor of \(G\text{,}\) there exists some \(K\nsg G\) such that \(G=H\times K\text{.}\) Thus there exists some isomorphism \(\psi:G\to H\times K\text{.}\) Define \(\widetilde{\phi} : G \to L\) such that \(\widetilde{\phi}(g)=h\text{,}\) where \(\psi(g)=(h,k)\text{.}\) Let \(x,y\in G\text{,}\) and notice
as \(\psi\) is an isomorphism. Observe \(\widetilde{\phi}(xy)=hh'=\widetilde{\phi}(x)\widetilde{\phi}(y)\text{.}\) Thus \(\widetilde{\phi}\) is a homomorphism, and for any \(h\in H\) we have \(\widetilde{\phi}(h)=h\text{.}\)
Coming soon!
ActivityC.130.Problem 3.
Suppose \(G\) is a group of order \(5 \cdot 7 \cdot 23^2\) and that \(G\) contains an element of order \(35\text{.}\) Prove \(G\) is abelian.
Solution.
By Sylow’s Theorems we know the number of Sylow \(23\)-subgroups of \(G\) must divide \(35\) and be congruent to \(1\mod{23}\text{,}\) the only option of which is \(1\text{.}\) Let \(H\) denote the unique Sylow \(23\)-subgroup and let \(K\) be the cyclic subgroup generated by the element of order \(35\text{.}\)
As \(H\) is unique it is normal in \(G\text{,}\) and it also means we have \(HK\leq G\text{.}\) Notice that as \(H\) and \(K\) are groups of relatively prime order we have \(H\cap K=\{e\}\text{.}\) Thus \(|HK|=|G|\text{,}\) and so \(G=HK\text{,}\) making \(G\cong H\times K\text{,}\) a direct product of abelian groups. Thus \(G\) is abelian.
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ActivityC.131.Problem 4.
Let \(E\) be the splitting field of \(x^4 + 7 \in \Q[x]\text{.}\)
Prove that \(\sqrt[4]{28}\in E\text{.}\)
Find a basis for \(E\) as a \(\Q\)-vector space.
Solution.
Let \(E\) be the splitting field of \(f(x)=x^4 + 7 \in \Q[x].\)
Notice that \(\a_1+\a_4=\sqrt[4]{28}\text{,}\) and thus \(\sqrt[4]{28}\in E\text{.}\)
The polynomial \(g(x)=x^4-28\) is irreducible in \(\Q[x]\) by Eisenstein’s Criterion (\(p=7\)). Thus it is the minimal polynomial of \(\sqrt[4]{28}\text{,}\) and so \(F=\Q(\sqrt[4]{28})\) is a field extension of \(\Q\) of degree \(4\text{.}\) After adjoining \(i\) as well we see that \(E\) is an extension of degree \(8\text{.}\) Thus a basis of \(E\) as a \(\Q\)-vector space is the following:
Let \(F\) be a field and \(G\) a finite subgroup of \(F^*\text{,}\) the multiplicative group of units of \(F\text{.}\) Prove that \(G\) is cyclic.
Solution.
Let \(G\) be a finite subgroup of \(F^*\text{.}\) Let \(|G|=n\text{.}\)
Let \(k\) be the LCM of all orders of elements in \(G\text{.}\) Then \(g^k=1\) and thus \(g\) is a root of the polynomial \(x^k-1\) for all \(g\in G\text{.}\) By Lagrange’s Theorem every element divides \(n\text{,}\) and so we have \(k\leq n\text{.}\) However, by the Factor Theorem CITEX the polynomial \(x^k-1\) can have at most \(k\) roots, and we have \(n\) distinct elements, and thus we have \(k=n\text{.}\) Thus there must exist an element of order \(n\) in \(G\text{,}\) making \(G\) cyclic, as desired.
ActivityC.133.Problem 6 (*).
Let \(L\) be a finite Galois field extension of \(\Q\text{.}\) Let \(E\) and \(F\) be subfields of \(L\) such that \(EF = L, E/\Q\) is normal, and \(E\cap F = \Q\text{.}\) Prove that \([L : Q] = [E : Q][F : Q].\)
Solution.
Coming Soon!
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ActivityC.134.Problem 7.
Let \(V\) be a finite dimensional vector space over a field \(F\) and let \(\t : V \to V\) be an \(F\)-linear operator on \(V\text{.}\) Prove \(\t\) is diagonalizable over \(F\) if and only if its minimum polynomial \(p(x)\) factors into distinct linear terms in \(F[x]\text{.}\)
Solution.
\((\Rightarrow)\) Suppose that \(g\) is diagonalizable. Thus there exists a change of basis matrix \(A\) such that \(A\) is diagonal. As it is diagonal, its diagonal entries are the eigenvalues of \(g\text{,}\) and thus the roots of the minimal polynomial of \(g\text{.}\) Using row and column operations we can rearrange \(A\) so that all repeated linear factors are next to each other in the diagonal, for convenience.
We know that \(\mp_A(x)\) is the smallest monic polynomial that sends \(A\) to \(0\text{.}\) Take all the distinct eigenvalues \(\lambda_i\) and consider \(q(x)=\prod(x-\lambda_i)\text{.}\)
We examine \(q(A)\text{.}\) It will be a product of matrices, one for each \(\lambda_i\text{.}\) First, take the first matrix in this product, \((A-\lambda_1I)\text{,}\) and note that it sends all \(\lambda_1\) in \(A\) to 0. Thus all of the rows and columns that contained a \(\lambda_1\) are now 0, and thus all these rows and columns will be 0 in the final product \(q(A)\text{.}\) As this is we set \(q(x)=\prod(x-\lambda_i)\) for all \(l\lambda_i\in A\text{,}\) we see that for each row and column in \(A\) there will exist a matrix \(A-\lambda_iI\) in the product \(q(A)\) such that the row and column will be 0. Thus the entire matrix will be 0, and \(q(A)=0\text{.}\)
Note that if any \(\lambda_i\) were excluded from \(q(x)\) there would exist a non-zero row and column for every matrix in the product, and thus \(q\) would not send \(A\) to 0. Thus \(q(x)\) is indeed the minimal polynomial of \(A\text{.}\) As \(q(x)=\prod(x-\lambda_i)\text{,}\) we see it does indeed factor into distinct linear terms.
\((\Leftarrow)\) Suppose the minimum polynomial of \(g\) factors completely into distinct linear factors, each of which has the form \(x-\lambda_i\) for some \(\lambda_i\in F\text{.}\) As each \(\lambda_i\) is distinct, each elementary divisor is of the form \((x-\lambda_i)^1\text{.}\)
We construct the Jordan Canonical Form of \(g\text{.}\) As the elementary divisors are linear the Jordan blocks are \(1\times1\) matrices, making the \(JCF\) a diagonal matrix. As the JCF is itself a change of basis matrix, we see that \(g\) is diagonalizable.
ActivityC.135.Problem 8 (*).
Prove the ideal \(I = (2, 1 + \sqrt{-13})\) of the commutative ring \(R = \Z[\sqrt{-13}]\) is not a principal ideal.
Solution.
Define norm function, yada yada
ActivityC.136.Problem 9 (*).
Let \(I\) be an ideal in a commutative ring \(R\text{,}\) let \(M\) and \(N\) be \(R\)-modules and let \(f : M \to N\) be an \(R\)-module homomorphism.
Prove there is a unique \(R\)-module homomorphism \(\ov{f} : M/IM \to N/IN\) such that \(\ov{f} \circ p = q \circ f\text{,}\) where \(p : M\to M/IM\) and \(q : N\to N/IN\) are the canonical quotient maps.
Prove that if \(I^2 = 0\) and \(\ov{f}\) is surjective, then so is \(f\) . (Recall that \(I^2\) is the ideal generated by all elements of the form \(ab\text{,}\) where \(a, b \in I\text{.}\))
