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## 数学代写|抽象代数作业代写abstract algebra代考|Direct Sum of Groups

A useful perspective behind considering classification question involves thinking about how to create new (larger) groups, from smaller ones. The direct sum of two groups is one such method.
Definition 1.4.1 (Direct Sum)
Let $(G, *)$ and $(H, \star)$ be two groups. The direct sum of the groups is a new group $(G \times H, \cdot)$ where the set operation is defined by
$$\left(g_1, h_1\right) \cdot\left(g_2, h_2\right)=\left(g_1 * g_2, h_1 \star h_2\right) .$$
The direct sum is denoted by $G \oplus H$.
We observe that the identity of $G \oplus H$ is $\left(e_G, e_H\right)$, the identity of $G$ paired with the identity of $H$.

The direct sum generalizes to any finite number of groups. For example, the group $\left(\mathbb{R}^3,+\right)$ is the triple direct sum of group $(\mathbb{R},+)$ with itself. The reader should feel free to imagine all sorts of possibilities now, e.g., $\mathbb{Z} / 9 \mathbb{Z} \oplus D_{10}$, $\mathbb{Z} \oplus Z_2 \oplus Z_2, \mathrm{GL}_2\left(\mathbb{F}_7\right) \oplus U(21)$, and so on.

Example 1.4.2. Consider the group $Z_4 \oplus Z_2$. We can list the elements of this group as
$$Z_4 \oplus Z_2=\left{\left(x^m, y^n\right) \mid x^4=e \text { and } y^2=e\right} .$$
(To avoid confusion, we should not use the same letter for the generator of $Z_4$ and of $Z_2$. After all, they are different elements and from different groups.) We point out that $Z_4 \oplus Z_2$ is not another cyclic group. The size is $\left|Z_4 \oplus Z_2\right|=8$. However, in a cyclic group of order 8 generated by some element $z$, only the element $z^4$ has order 2 . However, in $Z_4 \oplus Z_2$, both $\left(x^2, e\right)$ and $(e, y)$ have order 2. Thus $Z_4 \oplus Z_2$ is not cyclic.

It is easy to prove that $Z_4 \oplus Z_2$ is abelian. Let $\left(x^i, y^j\right),\left(x^a, y^b\right) \in Z_4 \oplus Z_2$. Then
$$\left(x^i, y^j\right)\left(x^a, y^b\right)=\left(x^i x^a, y^j y^b\right)=\left(x^{i+u}, y^{j+b}\right)=\left(x^a x^i, y^b y^j\right)=\left(x^a, y^b\right)\left(x^i, y^j\right) .$$
Since $Z_4 \oplus Z_2$ is abelian, it does not behave like $D_4$, which is not abelian. Hence, we have found three groups of order 8 with different properties: $Z_8$, $Z_4 \oplus Z_2$ and $D_4$.

## 数学代写|抽象代数作业代写abstract algebra代考|Classification Theorems

Classification theorems usually require theorems not yet at our disposal, e.g., Lagrange’s Theorem, Cauchy’s Theorem, Sylow’s Theorem, and so on. Nonetheless, this section explores what we can discover for possibilities of small groups using the Cayley table or orders of elements.

Example 1.4.4 (Groups of Order 4). We propose to find all groups of order 4 by filling out all possible Cayley tables. Suppose that $G={e, a, b, c}$ with $a, b$, and $c$ distinct nonidentity group elements. A priori, all we know about the Cayley graph is the first column and first row.
\begin{tabular}{c|ccc}
& $c$ & $a$ & $b$ \
\hline$e$ & $e$ & $a$ & $b$ \
$a$ & $a$ & & \
$b$ & $b$ & & \
$c$ & $c$ & &
\end{tabular}
Note that if a group $G$ contains an element $g$ of order $n$, then $\left{e, g, g^2, \ldots, g^{n-1}\right}$ is a subset of $n$ distinct elements. (See Exercise 1.3.23.) Hence, a group of order 4 cannot contain an element of order 5 or higher.
Suppose that $G$ contains an element of order 4, say, the element $a$. Then $G=\left{e, a, a^2, a^3\right}$. Without loss of generality, we can call $b=a^2$ and $c=a^3$ and the Cayley table becomes the following.
\begin{tabular}{c|cccc}
& $e$ & $a$ & $b$ & $c$ \
\hline$e$ & $e$ & $a$ & $b$ & $c$ \
$a$ & $a$ & $b$ & $c$ & $e$ \
$b$ & $b$ & $c$ & $e$ & $a$ \
$c$ & $c$ & $e$ & $a$ & $b$
\end{tabular}
We recognize this table as corresponding to the cyclic group $Z_4$.
Assume that $G$ does not contain an element of order 4 but contains one of order 3. Without loss of generality, suppose that $|a|=3$. Then $G=\left{e, a, a^2, c\right}$, with all elements distinct. The element $a c$ cannot be equal to $a^k$ for any $k$ for then $c=a^{k-1}$, a contradiction. Furthermore, we cannot have $a c=c$ for then $a=e$, again a contradiction. Hence, a group of order 4 cannot contain an element of order 3.

# 数学代写|抽象代数作业代写abstract algebra代考|Direct Sum of Groups

$$\left(g_1, h_1\right) \cdot\left(g_2, h_2\right)=\left(g_1 * g_2, h_1 \star h_2\right) .$$

$$Z_4 \oplus Z_2=\left{\left(x^m, y^n\right) \mid x^4=e \text { and } y^2=e\right} .$$
(为了避免混淆，我们不应该对$Z_4$和$Z_2$的生成器使用相同的字母。毕竟，他们是不同的元素，来自不同的群体。)我们指出$Z_4 \oplus Z_2$不是另一个循环基团。大小为$\left|Z_4 \oplus Z_2\right|=8$。然而，在一个由元素$z$生成的8阶循环群中，只有元素$z^4$的阶是2。但是，在$Z_4 \oplus Z_2$中，$\left(x^2, e\right)$和$(e, y)$都有2阶。因此$Z_4 \oplus Z_2$不是循环的。

$$\left(x^i, y^j\right)\left(x^a, y^b\right)=\left(x^i x^a, y^j y^b\right)=\left(x^{i+u}, y^{j+b}\right)=\left(x^a x^i, y^b y^j\right)=\left(x^a, y^b\right)\left(x^i, y^j\right) .$$

## 数学代写|抽象代数作业代写abstract algebra代考|Classification定理

.

\begin{tabular}{c|ccc}
& $c$ & $a$ & $b$ \
\hline$e$ & $e$ & $a$ & $b$ \
$a$ & $a$ & & \
$b$ & $b$ & & \
$c$ & $c$ & &
\end{tabular}

& $e$ & $a$ & $b$ & $c$ \
\hline$e$ & $e$ & $a$ & $b$ & $c$ \
$a$ & $a$ & $b$ & $c$ & $e$ \
$b$ & $b$ & $c$ & $e$ & $a$ \
$c$ & $c$ & $e$ & $a$ & $b$
\end{tabular}

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