Abstract algebra不算是一门简单的学科，这门学科在国内叫做抽象代数，经常有很多学生在学linear algebra或者analysis(advance calculus)的时候觉得并不困难，但是却觉得Abstract algebra很难，这是因为没有找到正确的方法学习Abstract algebra，UpriviateTA有一系列非常擅长Abstract algebra的老师，可以确保您在Abstract algebra取得满意的成绩。
以下是伯克利MATH113的一次assignment.更多的经典案例请参阅以往案例，关于abstract algebra的更多的以往案例可以参阅相关文章。abstract algebra代写请认准UpriviateTA.
Problem 1. Let \((G, \cdot)\) be a group and \(X\) any set. Let \(F\) be the set of functions with domain \(X\) and range \(G\). Define a binary operation \(*\) on \(F\) by \((f * g)(x):=f(x) \cdot g(x)\). Is \((F, *)\) a group? If so, prove that it is. If not, give an axiom which is violated and prove that this is so.
Proof . Yes, \((F, *)\) is a group.
identity The identity element is the function \(I: X \rightarrow G\) which is identically equal to the identity element, \(e,\) of \(G .\) Indeed, for any \(f \in F\) and any \(x \in X\) we have \((I * f)(x)=I(x) \cdot f(x)=e \cdot f(x)=f(x) .\) Hence, \(I * f=f\).
inverse Let \(f \in F\) be any element of \(F .\) Let \(g: X \rightarrow G\) be defined by \(g(x):=\) \((f(x))^{-1}\). Then for any \(x \in X\) we have \((g * f)(x)=g(x) \cdot f(x)=(f(x))^{-1}\). \(f(x)=e=I(x) .\) Hence, \(g * f=I\) so that \(g\) is a left-inverse of \(f\).
associativity Let \(f, g,\) and \(h\) be elements of \(F\). For any \(x \in X\) we have \(f *(g * h)(x)=\) \(f(x) \cdot(g * h)(x)=f(x) \cdot(g(x) \cdot h(x))=(f(x) \cdot g(x)) \cdot h(x)=(f * g)(x) \cdot h(x)=\)
\((f * g) * h(x) .\) Hence, \(f *(g * h)=(f * g) * h\)
identity The identity element is the function \(I: X \rightarrow G\) which is identically equal to the identity element, \(e,\) of \(G .\) Indeed, for any \(f \in F\) and any \(x \in X\) we have \((I * f)(x)=I(x) \cdot f(x)=e \cdot f(x)=f(x) .\) Hence, \(I * f=f\).
inverse Let \(f \in F\) be any element of \(F .\) Let \(g: X \rightarrow G\) be defined by \(g(x):=\) \((f(x))^{-1}\). Then for any \(x \in X\) we have \((g * f)(x)=g(x) \cdot f(x)=(f(x))^{-1}\). \(f(x)=e=I(x) .\) Hence, \(g * f=I\) so that \(g\) is a left-inverse of \(f\).
associativity Let \(f, g,\) and \(h\) be elements of \(F\). For any \(x \in X\) we have \(f *(g * h)(x)=\) \(f(x) \cdot(g * h)(x)=f(x) \cdot(g(x) \cdot h(x))=(f(x) \cdot g(x)) \cdot h(x)=(f * g)(x) \cdot h(x)=\)
\((f * g) * h(x) .\) Hence, \(f *(g * h)=(f * g) * h\)
Problem 2. Let \((G, *)\) be a group and \(a \in G .\) Suppose that \(a * a=a\). Prove or disprove:
\(a\) must be the identity element.
\(a\) must be the identity element.
Proof . \(a=e * a=\left(a^{-1} * a\right) * a=a^{-1} *(a * a)=a^{-1} * a=e\)
Problem 3. Prove or disprove: the set \(\mathbb{Q}(i)=\{a+b i: a, b \in \mathbb{Q}\}\) is a subgroup of \((\mathbb{C},+)\)
Proof . Yes, this is a subgroup. The set \(\mathbb{Q}(i)\) contains \(0=0+0 i\), is closed under addition as for \(a, b, c,\) and \(d\) ratinonal numbers, \((a+b i)+(c+d i)=(a+c)+(b+d) i\) and each of \(a+c\) and \(b+d\) is a rational number, and is closed under taking inverses as for \(a\) and \(b\) rational numbers, \(-(a+b i)=(-a)+(-b) i\) and each of \(-a\) and \(-b\) is rational.
Problem 4. Let \(G\) be the set of \(2 \times 2\) matrices having integer entries and a nonzero determinant. Prove or disprove: \(G\) is a group under matrix multiplication.
Proof . The matrix \(A=\left(\begin{array}{cc}2 & 0 \\ 0 & 2\end{array}\right)\) belongs to \(G\) (as its determinant is \(4 \neq 0\) and all of its entries are integers) but for any \(2 \times 2\) matrix \(B\) with integer entries, we have \(\operatorname{det}(B) \in \mathbb{Z}\) and \(\operatorname{det}(A B)=\operatorname{det}(A) \operatorname{det}(B)=4 \operatorname{det}(B)\) which cannot be equal
to one. Thus, \(A\) does not have an inverse in \(G\).
to one. Thus, \(A\) does not have an inverse in \(G\).
Problem 5. Consider the function \(\sigma:\{0, \ldots, 15\} \rightarrow\{0, \ldots, 15\}\) defined by
$$
x \mapsto\left\{\begin{array}{l}
x+4 \text { if } x<12 \\
x-12 \text { if } x \geq 12
\end{array}\right.
$$
Show that \(\sigma\) is a permutation and describe its orbits.
$$
x \mapsto\left\{\begin{array}{l}
x+4 \text { if } x<12 \\
x-12 \text { if } x \geq 12
\end{array}\right.
$$
Show that \(\sigma\) is a permutation and describe its orbits.
Proof . To see that \(\sigma\) is a permutation it suffices to check that it is onto. Let \(a \in\{0, \ldots, 15\} .\) If \(a<4,\) then \(a=\sigma(12+a) .\) If \(a \geq 4,\) then \(a=\sigma(a-4)\) The orbits are \(\{0,4,8,12\},\{1,5,9,13\},\{2,6,10,14\}\) and \(\{3,7,11,15\} .\) 18. Let \(G\) be the set of all permutations of \(\mathbb{R}\) which move at most finitely many points. That is, \(\sigma: \mathbb{R} \rightarrow \mathbb{R}\) belongs to \(G\) just in case \(\sigma \in S_{\mathbb{R}}\) and \(\{r \in \mathbb{R}: \sigma(r) \neq r\}\) is finite. Prove or disprove: \(G\) is a group under composition.
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