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UCLA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. Consider a rigid rod lying on the $x$ -axis from $x=-1$ to $x=1$, with three beads attached. The beads are located at the endpoints (-1,0) and $(1,0),$ and at the center (0,0) . The beads are to be painted using $n$ colors, and two colorings are regarded as equivalent if one can be mapped into the other by a permutation in the group $G=\{I, \sigma\},$ where $\sigma$ is the 180 degree rotation about the vertical axis. Find the number of distinct colorings.
Proof . Label (-1,0) as vertex 1,(0,0) as vertex $2,$ and (1,0) as vertex $3 .$ Then $I=(1)(2)(3)$ and $\sigma=(1,3)(2) .$ Thus the number of distinct colorings is $\frac{1}{2}\left(n^{3}+n^{2}\right)$.
Problem 2. Assume that two colorings of the vertices of a square are equivalent if one can be mapped into the other by a permutation in the dihedral group $G=D_{8}$. If $n$ colors are available, find the number of distinct colorings.
Proof . The group elements are $I, R=(1,2,3,4), R^{2}=(1,3)(2,4), R^{3}=(1,4,3,2)$, $F=(1)(3)(2,4), R F=(1,2)(3,4), R^{2} F=(1,3)(2)(4), R^{3} F=(1,4)(2,3) .$ Thus
the number of distinct colorings is
$$
\frac{1}{8}\left(n^{4}+n+n^{2}+n+n^{3}+n^{2}+n^{3}+n^{2}\right)=\frac{1}{8}\left(n^{4}+2 n^{3}+3 n^{2}+2 n\right)
$$
the number of distinct colorings is
$$
\frac{1}{8}\left(n^{4}+n+n^{2}+n+n^{3}+n^{2}+n^{3}+n^{2}\right)=\frac{1}{8}\left(n^{4}+2 n^{3}+3 n^{2}+2 n\right)
$$
Problem 3. In Problem 2, assume that only two colors are available, white and green. There are 16 unrestricted colorings, but only 6 equivalence classes. List the equivalence classes explicitly.
Proof . If the vertices of the square are 1,2,3,4 in counterclockwise order, then $W G G W$ will mean that vertices 1 and 4 are colored white, and vertices 2 and 3 green. The equivalence classes are
$$
\begin{array}{c}
\{W W W W\},\{G G G G\},\{W G G G, G W G G, G G W G, G G G W\} \\
\{G W W W, W G W W, W W G W, W W W G\} \\
\{W W G G, G W W G, G G W W, W G G W\},\{W G W G, G W G W\}
\end{array}
$$
$$
\begin{array}{c}
\{W W W W\},\{G G G G\},\{W G G G, G W G G, G G W G, G G G W\} \\
\{G W W W, W G W W, W W G W, W W W G\} \\
\{W W G G, G W W G, G G W W, W G G W\},\{W G W G, G W G W\}
\end{array}
$$
Problem 4. Let $G$ be the group of rotations of a regular $p$ -gon, where $p$ is an odd prime. If the vertices of the $p$ -gon are to be painted using at most $n$ colors, find the number of distinct colorings.
Proof . The group $G=\left\{1, R, R^{2}, \ldots, R^{p-1}\right\}$ is cyclic of order $p .$ Since $p$ is prime, each $R^{i}$ $i=1, \ldots, p-1,$ has order $p,$ and therefore as a permutation of the vertices consists of a single cycle. Thus the number of distinct colorings is
$$
\frac{1}{p}\left[n^{p}+(p-1) n\right]
$$
$$
\frac{1}{p}\left[n^{p}+(p-1) n\right]
$$
Problem 5. Use the result of the previous problem to give an unusual proof of Fermat’s little theorem.
Proof . Since the result of the previous problem is an integer, $n^{p}+(p-1) n=n^{p}-n+n p$ is a multiple of $p,$ hence so is $n^{p}-n$. Thus for any positive integer $n, n^{p} \equiv n \bmod p .$ It follows that if $n$ is not a multiple of $p$, then $n^{p-1} \equiv 1 \bmod p$.
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