Modern Algebra: Groups, Rings, Modules, Fields

Let \(f:G/H\to gHg\inv\) be defined by \(f(gH)=gHg\inv\text{.}\) Suppose \(xH=yH\text{.}\) Notice that \(f(xH)=xHx\inv\) and \(f(H)=yHy\inv\text{,}\) but as \(x=y\) we have equality, and thus \(f\) is well defined. Let \(gHg\inv\) be a conjugate set, and let \(gH\in G/H\text{.}\) Then \(f(gH)=gHg\inv\text{,}\) and so we have surjectivity. As \(|G/H|=|[G:H]|=h\text{,}\) there can be at most \(h\) distinct conjugate sets for \(g\in G\text{.}\)

Let \(G\) act on \(H\) by conjugation. We know that the orbits of this action partition \(H\) by [provisional cross-reference: cite]. However, under this action the orbits are exactly the conjugacy classes of \(H\) by [provisional cross-reference: cite]. There are at most \(h\) conjugacy classes. Each conjugacy class has at most \(|H|\) elements in it, and each one has the identity. As there are at most \(h\) of them then when we add all of their orders we get at most \(h(|H|-1)\text{,}\) which is less than the order of \(G\text{.}\)

First, by Sylow’s Theorems we see that \(n_p=p+1\text{.}\) So there are \((p-1)(p+1)=p^2-1\) elements of order \(p\text{,}\) and \(|G|=p^2+p\text{.}\) So there are \(p+1\) elements left.

Also from Sylow’s Theorems we know \(G\) acts on \(\Syl_p(G)\) by conjugation. The normalizer of \(P\) is the set \(N_G(P)=\{g\in G|gPg\inv=P\}\text{,}\) which is exactly the stabilizer of \(P\) under this action by [provisional cross-reference: cite].

Let \(x\in P\text{,}\) and observe \(xPx\inv=P\text{,}\) so \(x\in N_G(P)\text{,}\) and \(P\subseteq N_G(P)\text{.}\)

The conjugation action on Sylow subgroups is transitive by Part (2) of Sylow’s Theorems, and thus there is exactly one orbit under this action by [provisional cross-reference: cite]. By [cross-reference to target(s) "cor-orbit-stabilizer" missing or not unique] we have \(|G|=|\Orb(x)|\cdot|\Stab(x)|\text{,}\) and thus \(|\Stab(x)|=p+1\) for every \(x\in P\text{.}\) However, \(p\) of these elements are accounted for by \(P\) and the last element is the identity, thus the only elements of \(N_G(P)\) are those in \(P\text{.}\)

Let \(P=\langle t\rangle\text{,}\) \(a\) be an element of order \(2\) in \(G\text{,}\) and \(n<,m\leq p-1\text{.}\) Suppose by way of contradiction that \(t^nat^{-n}=t^mat^{-m}\text{.}\) By strategic multiplication we see \(a=t^kat^{-k}\text{,}\) where \(k=m-n\text{.}\) Thus we can reduce to the case where As \(a=t^kat^{-k}\) for some\(k\leq p-1\text{.}\)

As \(a\) has order two, it is contained in some Sylow \(2\)-subgroup of \(G\text{.}\) Since \(P=\langle t\rangle\) we have \(t^k=x\) for some \(x\in P\text{.}\) Notice then that \(a=xax\inv\) means that \(x=axa\inv\text{,}\) placing \(a\in N_G(x)\text{,}\) contradicting the fact that \(a\not\in P\text{,}\) as it does not have order \(3\text{.}\) Thus \(a\) is distinct from \(t^kat^{-k}\) for all \(k\leq p-1\text{.}\) If any two powers of

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.

Let \(n\in\N\) and suppose that \(F\) has characteristic \(p>0\) and is algebraically closed.

First, suppose by way of contradiction that \(F^*\) has a subgroup of order \(n\text{,}\) \(G\text{,}\) and \(p|n\text{.}\) From Part (a) \(G\) is cyclic and generated by some \(x\) such that \(|x|=n\text{.}\) However, as \(p|n\) we see that \(x^p=0\text{,}\) given that we are in an additive group. This contradicts the fact that \(n\) is the smallest number such that \(x^n=0\text{.}\)

Now we proceed via the contrapositive. Suppose \(F^*\) does not have a subgroup of order \(n\text{.}\) Then there cannot exist a unit \(x\) such that \(|x|=n\text{.}\) Consider the polynomial \(x^n-1\text{.}\) As \(F\) is algebraically closed there exists some root \(r\in F\text{.}\) (how do we know this isn’t 1???) Notice that this means \(r^n=1\) and so \(r\) is a unit in \(F\text{.}\) As \(r\) cannot have order \(n\text{,}\) it must have an order that divides \(n\text{.}\) We also know that \(r^p=1\) as we are in a field of characteristic \(p\text{.}\) Thus \(|r|\) either divides \(p\) or is \(p\text{.}\) Either there is a non-identity root of \(x^n-1=(x-1)^n\text{.}\) Note that in this case \(n>p\) as we are