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。

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, $f *(g * h)=(f * g) * h$
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.
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 $G$.
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.
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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