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1.\(S_3\) is the Only Non-Abelian Group of Order 6.
Let \(G\) be a non-Abelian group of order 6. Use a group action of \(G\) on the cosets of a carefully chosen subgroup of \(G\) to prove that \(G\cong S_3\text{.}\)
By Cauchy’s Theorem, \(G\) has an element \(a\) of order 3 and an element \(b\) of order 2. The subgroup \(H=\langle a\rangle\) has index 2 in \(G\) and so is a normal subgroup. On the other hand, the subgroup \(K=\langle b\rangle\) must not be normal, or else we would have \(G=H\times K\) and \(G\) would be Abelian since \(H\) and \(K\) are cyclic. Now consider the action of \(G\) on the set of left cosets of \(K\) by left multiplication. Since there are three such cosets (which must be \(K=\{e,b\}, aK=\{a,ab\}, a^2K=\{a^2,a^2b\}\) since if \(aK=a^2K\) we would have \(a\in K\text{,}\) this gives a homomorphism \(G\to S_3\text{.}\) But the left multiplication action of \(G\) has no elements which act trivially on all cosets besides the identity, since \(g(g'K)=g'K\) if and only if \(gg'\in g'K\) if and only if \(g\in g'K(g')^{-1}\) and since \(K\) is not normal we have \(K\leq N(K) \lt G\) so we have \(N(K)=K\) and thus \(g'K(g')^{-1}\neq K\) for all \(g'\in G\setminus K\) so \(g\) can act trivially on at most one coset. So the kernel of this homomorphism is just the identity and therefore the map is 1-to-1 and (since the groups are finite) also onto. Hence \(G\cong S_3\text{.}\)
Problem Specs/Notes: This problem must be done with a group action for a Success, though there are many proofs of the result that do not use group actions.
Let \(G\) be a group of order \(pq\) where \(p\) and \(q\) are distinct primes with \(p\lt q\text{.}\) Prove that if \(p\) does not divide \(q-1\text{,}\) then \(G\) is cyclic.
Since \(n_q\equiv 1 \pmod q\) and \(q\gt p\text{,}\) the condition \(n_q \mid p\) forces \(n_q=1\text{,}\) so there is a unique Sylow \(q\)-subgroup \(Q\) in \(G\text{,}\) and by the second Sylow theorem we have \(Q\normaleq G\text{.}\) Now since \(n_p\mid q\) we have either \(n_p=1\) or \(n_p=q\text{.}\) But since \(p\nmid q-1\) we cannot have \(n_p=q\) since \(n_p\equiv 1 \pmod p\text{.}\) So also \(n_p=1\) and there is a unique normal Sylow \(p\)-subgroup \(P\text{.}\)
Now since \(P\) and \(Q\) have relatively prime order we have \(P\cap Q = \{e\}\) by Lagrange’s theorem. We also have \(|PQ|=pq=|G|\text{,}\) so the map \(\phi:P\times Q \to G\)
\begin{equation*}
(p,q)\mapsto pq
\end{equation*}
is a group isomorphism. Since \(P\) and \(Q\) are groups of prime order, they are cyclic, and then by the Criterion for \(G\times H\) Cyclic we also have \(P\times Q\) is cyclic, so \(G\) is cyclic.
Problem Specs/Notes: This problem needs careful use of the Sylow Theorems and our knowledge of properties of the set \(HK\) when \(H,K\leq G\) for a Success.
Suppose that \(R\) is a ring and that \(a^2=a\) for all \(a\in R\text{.}\) Prove that \(R\) is commutative. Such rings are called Boolean rings after the English mathematician George Boole (1815-1864) who developed the algebra of logic. Give an example of a Boolean ring with four elements and example of an infinite Boolean ring.
A Boolean ring with \(4\) elements is \(\Z_2\times \Z_2\) since \(0^2=0,1^2=1\) in \(\Z_2\) and operations in the product are performed component-wise, and a Boolean ring with infinite elements is
Prove that there is no integral domain with exactly six elements. Can your argument be adapted to show that there is no integral domain with exactly four elements? What about 15 elements? Use these observations to conjecture a general result about the number of elements in a finite integral domain.
We have shown that a finite integral domain has prime characteristic \(p\) for some \(p\text{.}\) In an integral domain \(D\) with order \(pqn\text{,}\) where \(p\) and \(q\) are distinct primes and \(n\geq 1\) is an integer, we know by Cauchy’s theorem that there exist elements \(x\) and \(y\) of additive order \(p\) and \(q\) respectively. Suppose \(\char D = p\text{.}\) Then \(p\cdot y = 0\text{,}\) but this is true only if \(p\) divides the order of \(y\text{,}\)\(q\text{.}\) This contradicts that \(p\) and \(q\) are distinct primes. So \(\char D\) is not \(p\text{.}\) On the other hand, if \(\char D=q\text{,}\) a similar argument swapping \(x\) and \(y\) also gives a contradiction. And if \(\char D\) is any other prime dividing \(|D|\) both yield a contradiction. So \(D\) must have order \(p^r\) for some prime \(p\) and \(r \geq 1\text{.}\) In particular, this rules out the existence of integral domains with 6 or 15 elements, but it cannot show that there is no integral domain with 4 elements. (Indeed, such a domain exists, as \(\Z_2[x]/\langle x^2+x+1\rangle\) is a field with four elements.)
Suppose \(I\) is a non-zero ideal of \(F\text{.}\) Then there exists some non-zero \(x \in I\text{.}\) Since \(I\) is an ideal, \(1=xx^{-1}\) is thus in \(I\text{.}\) So \(I=F\text{.}\)
An integral domain \(D\) is called a principal ideal domain (PID) if every ideal of \(D\) has the form \(\langle a\rangle =\{ad\mid d\in D\}\) for some \(a\) in \(D\text{.}\) Show that \(\Z\) is a principal ideal domain.
Let \(I\) be an ideal of \(\Z\text{.}\) If \(I=\{0\}\) then \(I=\langle 0\rangle\text{.}\) Else let \(d\) be the smallest positive integer in \(I\) (one must exist since if \(I\) contains \(x\lt 0\) it also contains \(-x>0\)). Then we have \(\langle d \rangle \subseteq I\) since \(d\in I\) and \(I\) absorbs multiplication. On the other hand, if \(n\in I\) then by the division algorithm there are unique positive integers \(q\) and \(r\) so that
\begin{equation*}
n=dq+r
\end{equation*}
and \(0\leq r\lt d\text{.}\) Then \(r=n-dq\in I\text{,}\) so since \(d\) was the smallest positive integer in \(I\) we must have \(r=0\text{,}\) i.e. \(n\in \langle d\rangle\text{.}\) Therefore \(I=\langle d\rangle\text{.}\)
Problem Specs/Notes: This problem needs care not to accidentally assume the conclusion for a Success. It will be helpful to find use the division algorithm on this problem.
\(\Ann(A)\) contains 0, since \(0a=0\) for all \(a\in A\text{,}\) so \(\Ann(A)\) is non-empty. Now suppose \(x,y, \in \Ann(A)\text{,}\)\(a\in A\) and \(r\in R\text{.}\) We have
\(N(A)\) contains \(0\text{,}\) since \(A\) contains \(0\) and \(A\subseteq N(A)\text{.}\) Now suppose \(x,y \in N(A)\) with corresponding positive integers \(n\) and \(m\) such that \(x^n, y^m \in A\text{,}\) and \(r\in R\text{.}\) Then since \(R\) is commutative we have
For each \(i\) from \(0\) to \(n+m\text{,}\) either \(i\geq n\) or \((n+m)-i \geq m\text{,}\) as if neither is true then we have \(n+m=i+(n+m)-i\lt n+m\text{,}\) a contradiction. Thus each term of the sum lies in \(A\text{,}\) since either \(x^i\) or \(y^{n+m-i}\) is in \(A\) and so any multiple of their product lies in \(A\) since \(A\) is an idea. So \(x-y \in N(A)\text{.}\) Finally, we have
\begin{equation*}
(rx)^n=r^nx^n \in A
\end{equation*}
since \(x^n \in A\) and \(A\) is an ideal. So \(rx\) is in \(N(A)\)
