Algebra
Conjugation Collapses In Commutative Groups
Exercise 3, Ch. 2, Sec. 3 - Algebra (Michael Artin): Let \(a,b\) be elements of a group \(G\), and let \(a' = bab^{-1}\). Prove that \(a = a'\) if and only if \(a\) and \(b\) commute.
Proof:
We have \(a = a' = bab^{-1}\). Multiplying on the right by \(b\) we get \(ab = ba(b^{-1}b) = ba\) and so \(a\) and \(b\) commute. This argument is completely reversible so we are done.
A Simple Statement About Conjugate Elements In A Group.
Exercise 2, Ch. 2, Sec. 3 - Algebra (Michael Artin): Prove that the products \(ab\) and \(ba\) are conjugate elements in a group \(G\).
Proof:
Define \(C_{x}:G\rightarrow G\) to be conjugation by the element \(x\). Then \[C_b(ab) = b(ab)b^{-1} = ba(bb^{-1}) = ba\] and \[C_a(ba) = a(ba)a^{-1} = ab(aa^{-1}) = ab.\] Thus \(ab\) and \(ba\) are conjugate elements in \(G\).
An Isomorphism Between The Additive Reals And The Positive Reals With Multiplication
Exercise 1, Ch. 2, Sec. 3 - Algebra (Michael Artin): Prove that the additive group \({\mathbb{R}}^{+}\) is isomorphic to the multiplicative group \(P\) of positive reals.
Proof:
Upper Diagonal Matrix Raised To The N-th Power
Exercise 7, Ch. 1, Sec. 1 - Algebra (Michael Artin): Find a formula for \(
{\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}\right]}^n\) and prove it by induction.
Proof: Defining $$
A = \left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}\right] $$ and considering \(A^2, A^3\) and a few other low powers of \(A\) we come to the inductive hypothesis, $$
{\left[\begin{array}{rrr}
1 & 1 & 1 \\
0 & 1 & 1 \\
\(Z[x]\) is not Euclidian but still a unique factorization domain
Exercise 5, Ch. 2, Sec. 2 - Fundamentals of Number Theory (William J. LeVeque):
If \(D\) is a Euclidean domain, and \(a\) and \(b\) are relatively prime elements of \(D\), then there are \(m,n \in D\) such that \(ma + nb = 1\).
- Show that 2 and \(x\) are relatively prime elements of \(Z[x]\).
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