Modern Algebra: Groups, Rings, Modules, Fields

By Sylow’s Theorems we know \(n_{11}|21\) and is congruent to \(1\mod{11}\text{.}\) The only possibility for such is \(1\text{,}\) meaning that \(G\) has exactly \(1\) Sylow \(11\)-subgroup.

Let \(G\) act on \(P\) by conjugation. This gives rise to the Permutation Representation homomorphism \(\rho: G\to S_{11}\text{.}\) However, \(P\) has prime order, making it a cyclic group isomorphic to \(\Z_{11}\text{.}\)

By the First Isomorphism Theorem we know that \(G/\ker(\rho)\cong\im(\rho)=\Aut(P)\text{.}\) However, \(\Aut(P)\cong\Aut(\Z_{11})\cong \Z^x\text{,}\) which has order \(10\text{.}\) As this is a homomorphism, we see the order of \(G/\ker(\rho)\) must divide both \(10\) and \(231\text{,}\) two numbers that are relatively prime. Thus \(\ker(\rho)=1\text{,}\) meaning that conjugation is equivalent to the identity map, or that \(gxg\inv=x\) for all \(g\in G\text{.}\) Thus \(P\in Z(G)\text{.}\)

Let \(A\) be the set of all linearly independent subsets of \(M\) that contain \(T\text{.}\) We can order \(A\) with respect to inclusion. Let \(X\) be a totally ordered subset of \(A\text{,}\) and let \(U\) be the union of all elements in \(X\text{.}\) Let \(\{u_i\}\) be a set of elements in \(U\) such that \(\sum_{i=0}^nr_iu_i=0\) for some \(r_i\in R\text{,}\) where \(i\leq n\) for some \(n\in\N\text{.}\) As \(U\) is the union of all elements in \(X\text{,}\) there exists some \(B_1\) such that \(u_1\in B_1\text{.}\) However, as \(X\) is totally ordered, there exists some \(B_2\) such that \(B_2\) contains \(B_1\) and \(u_2\in B_2\text{.}\) Continuing in this way, we see that there exists some \(B_n\in X\) such that \(\{u_i\}\subseteq B_n\text{.}\) As \(B_n\) is linearly independent, we know that \(\sum_{i=0}^nr_iu_i=0\) means that \(r_i=0\) for all \(i\text{.}\) Thus \(U\) is indeed linearly independent, making it an upper bound for \(X\text{.}\) Thus by Zorn’s Lemma there exists a maximal element of \(A\text{,}\) which we denote \(S\text{.}\) Thus \(S\) is linearly independent, contains \(T\text{,}\) and is maximal.

\((\Rightarrow)\) Suppose \(T\) is maximal linearly independent, and suppose by way of contradiction that \(M/N\) is not torsion. Thus there exists some \(xN\in M/N\) such that for all \(r\in R\text{,}\) we see that \(rx\neq0\text{.}\) However, as \(x\neq 0\) and \(x\not\in N\text{,}\) this means that \(x\not\in T\text{.}\) Consider \(T\cup\{x\}\text{.}\) This set is linearly independent, contradicting the assumption that \(T\) was maximal.

\((\Leftarrow)\) Suppose \(M/N\) is torsion. Let \(x\neq0\in M\) and consider \(T\cup\{x\}\text{.}\) Consider \(x+N\in M/N\text{.}\) As \(M/N\) is torsion, there exists an \(r\neq0\in R\) such that \(rxN=0\text{.}\) Thus \(rx\in N\text{.}\) (Note, if \(x\in N\text{,}\) then \(r=1\)). As \(rx\in N\) and \(N\) is generated by \(T\) (\(N=RT\)), \(rx=\sum r_it_i\text{.}\) Subtracting over we see that \(rx-\sum r_it_i=0\text{.}\) But as \(r\neq 0\text{,}\) we see that each \(t_i\) and \(x\) are in \(T\cup\{x\}\text{,}\) but the sum is 0. Thus \(T\cup\{x\}\) is linearly dependent.