Modern Algebra: Groups, Rings, Modules, Fields

First, notice that \(e^2=e\in H\text{.}\) Let \(x\in H\text{,}\) and consider \(x\inv\text{.}\) Notice that \((x{\inv})^2=x^{-2}\in H\text{.}\) As \(H\) is multiplicatively closed, we see that \(xx^{-2}=x\inv\in H\text{.}\) Thus \(H\) is a subgroup of \(G\) by the Subgroup Tests.

Let \(g\in G\text{,}\) \(h\in H\text{,}\) and consider \(gh\text{.}\) Notice that \((gh)^2=ghgh\in H\text{.}\) Multiplying by \(h\inv\) on the right we see \(ghg\in H\text{,}\) as it is multiplicatively closed and \(h\inv\in H\text{.}\) We rewrite \(ghg=ghggg\inv=ghg\inv g^2\text{,}\) given that elements always commute with their inverses. As \(g^2\in H\text{,}\) we see that \(ghg\inv\in H\) as well. Thus \(H\) is normal in \(G\text{.}\)

Let \(x,y\in G/H\text{.}\) As \(g^2\in H\) for every \(g\in G\text{,}\) every element \(G/H\) has order \(2\text{.}\) Thus \(xyxy=e\) and so \(xy=(xy)\inv=yx\text{,}\) making the group abelian.

Let \(G\) act on \(\Syl_p(G)\) by conjugation, inducing the homomorphism \(\rho: G\to S_{n_p}\) via the Permutation Representation. Notice that the order of \(S_{n_p}\) is conspicuously \(n_p!\text{.}\) The kernel of this map is a normal subgroup of \(G\) by Theorem 3.40. Note that since \(G\) is simple the only normal subgroups of \(G\) are the trivial subgroup and \(G\) itself. However, the kernel cannot be all of \(G\) as this would make \(\rho\) trivial, which cannot be the case given that our action is transitive by Part (2) of Sylow’s Theorems. Thus \(\ker(\rho)=\{e\}\text{,}\) making \(\rho\) injective by Theorem 1.68. Thus \(|\im(\rho)|=|G|\text{.}\) CITEX As the image is a subgroup of \(S_{n_p}\text{,}\) the result follows from Lagrange’s Theorem.

Let \(G\) be a group of order \(1,000,000\text{.}\) Suppose by way of contradiction that \(G\) is simple. \(1,000,000=10^6=2^6\cdot 5^6\text{.}\) Thus the number of Sylow-\(5\) subgroups is congruent to \(1\mod 5\) and divides \(2^6=64\text{,}\) the options of which are 1 and 16 (See: Sylow’s Theorems). As \(|G|\) does not divide \(16!,\) this contradicts part (a). Thus there are no simple groups of order \(1,000,000\text{.}\)

Let \(A\) be an integral domain that contains a field \(F\) as a subring, and that moreover \(A\) is finite dimensional as an \(F\)-vector space.

Say \(A\) has dimension \(n\) over \(K\text{.}\) Let \(x\in A,x\neq0\text{.}\) Consider elements \(1,x,x^2,x^{3,\dots,}\) which cannot be independent because of the finite dimension. So we may choose \(m\) so that \(1,x,\dots,x^m\) is linearly dependent over \(K\) and is as small as possible. This means that we may find \(c_0,c_1,\dots,c^{m}\in K\text{,}\) not all equal to \(0\text{,}\) such that \(c_0+c_1x+\dots+c_mx^m=0\text{.}\) Note that \(c_0\neq0\text{,}\) as this would contradict the minimality of \(m\text{.}\) Then \(x(c_1+\dots+c)_mx^{m-1})\) is invertible, so \(x\) is invertible as well, making \(A\) a field.

Consider \(\Q[x]\text{,}\) which contains the subring \(\Q\text{.}\) However, this is not a finite dimensional vector space over \(F\text{,}\) and \(\Q[x]\) is not a field, as \(x\) has no inverse.

For any integer \(n \geq 1\text{,}\) consider the \(\C\)-vector space \(V = \{p \in \C[x] : \deg(p) \leq n - 1\}\text{.}\)

This time around the change of basis matrix (denoted \(B\) and using the same basis as above) for this matrix has \(0\)s along the diagonal, and increasing natural numbers (starting at \(0\text{,}\) sorry) along the upper diagonal. Thus \(\cp_B(x)=x^n\text{.}\)

Recall that the minimal polynomial corresponding to \(B\) will be the the smallest monic polynomial such that \(B\) is sent to 0. Note that as \(D\) is the operator sending \(p\) to its derivative, and that \(D^{(2)}=D(D(p))=p''=p^{(2)}\text{.}\) Thus \(B^2\) can be viewed as a change of basis matrix for taking the second derivative of the basis, and so on.

As the basis extends to \(x^{n-1}\text{,}\) it requires \(n\) derivatives to make this polynomial become \(0\text{.}\) Thus the minimal polynomial of \(B\) must be \(x^n\text{,}\) as it it monic and \(B^n=0\text{.}\) As the degree of the invariant factors must sum to \(n\) and \(\mp_B(x)=x^n\text{,}\) which is itself an invariant factor, we see that it must in fact be the only one.

As \(x^n\) is already a power of a prime, it is the only elementary divisor as well. Thus the Jordan Canonical Form for \(B\) is an \(n\times n\) Jordan Block with \(0\)s along the diagonal and \(1\)s along the sub-diagonal.