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# 数学代写|抽象代数作业代写ALGEBRA代考|MATH355 External Direct Products

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## 数学代写|抽象代数作业代写ALGEBRA代考|Definition and Examples

In this chapter, we show how to piece together groups to make larger groups. In Chapter 9, we will show that we can often start with one large group and decompose it into a product of smaller groups in much the same way as a composite positive integer can be broken down into a product of primes. These methods will later be used to give us a simple way to construct all finite Abelian groups.
Definition
External Direct Product
Let $G_{1}, G_{2}, \ldots, G_{n}$ be a finite collection of groups. The external direct product of $G_{1}, G_{2}, \ldots, G_{n}$, written as $G_{1} \oplus G_{2} \oplus \cdots \oplus G_{n}$, is the set of all $n$-tuples for which the ith component is an element of $G_{i}$ and the operation is componentwise.
In symbols,
$$G_{1} \oplus G_{2} \oplus \cdots \oplus G_{n}=\left{\left(g_{1}, g_{2}, \ldots, g_{n}\right) \mid g_{i} \in G_{i}\right}$$
where $\left(g_{1}, g_{2}, \ldots, g_{n}\right)\left(g_{1}^{\prime}, g_{2}^{\prime}, \ldots, g_{n}^{\prime}\right)$ is defined to be $\left(g_{1} g_{1}^{\prime}, g_{2} g_{2}^{\prime}, \ldots, g_{n} g_{n}^{\prime}\right)$. It is understood that each product $g_{i} g_{i}^{\prime}$ is performed with the operation of $G_{i}$. Note that in the case that each $G_{i}$ is finite, we have by properties of sets that $\left|G_{1} \oplus G_{2} \oplus \cdots \oplus G_{n}\right|=\left|G_{1}\right|\left|G_{2}\right| \cdots\left|G_{n}\right|$. We leave it to the reader to show that the external direct product of groups is itself a group (Exercise 1).

## 数学代写|抽象代数作业代写ALGEBRA代考|Properties of External Direct Products

The order of an element in a direct product of a finite number of finite groups is the least common multiple of the orders of the components of the element. In symbols,
$$\left|\left(g_{1}, g_{2}, \ldots, g_{n}\right)\right|=\operatorname{lcm}\left(\left|g_{1}\right|,\left|g_{2}\right|, \ldots,\left|g_{n}\right|\right)$$
PROOF Denote the identity of $G_{i}$ by $e_{i}$. Let $s=\operatorname{lcm}\left(\left|g_{1}\right|,\left|g_{2}\right|, \ldots,\left|g_{n}\right|\right)$ and $t=\left|\left(g_{1}, g_{2}, \ldots, g_{n}\right)\right|$. Because $s$ is a multiple of each $\left|g_{i}\right|$ implies that $\left(g_{1}, g_{2}, \ldots, g_{n}\right)^{s}=\left(g_{1}^{s}, g_{2}^{s}, \ldots, g_{n}^{s}\right)=\left(e_{1}, e_{2}, \ldots, e_{n}\right)$,we know that $t \leq s$. On the other hand, from $\left(g_{1}^{t}, g_{2}^{t}, \ldots, g_{n}^{t}\right)=\left(g_{1}, g_{2}, \ldots, g_{n}\right)^{t}=\left(e_{1}, e_{2}, \ldots, e_{n}\right)$ we see that $t$ is a common multiple of $\left|g_{1}\right|,\left|g_{2}\right|, \ldots,\left|g_{n}\right|$.Thus, $s \leq t$.
The next three examples are applications of Theorem 8.1.

• EXAMPLE 4 Examples of groups of order 100 include
$Z_{100} ; Z_{25} \oplus Z_{2} \oplus Z_{2} ; Z_{5} \oplus Z_{5} \oplus Z_{4} ; Z_{5} \oplus Z_{5} \oplus Z_{2} \oplus Z_{2} ; D_{50} ; D_{10} \oplus Z_{5} ; D_{5} \oplus Z_{10} ;$ and $D_{5} \oplus D_{5}$. That these are not isomorphic is an easy consequence of Theorem 8.1.
• EXAMPLE 5 Let $m$ and $n$ be positive integers that are divisible by 5 . We determine the number of elements of order 5 in $Z_{m} \oplus Z_{n}$. By Theorem $8.1$, we need only count the number of elements $(a, b)$ in $Z_{m} \oplus Z_{n}$ with the property that $5=|(a, b)|=\operatorname{lcm}(|a|,|b|)$. Clearly this requires that $|a|=$ 1or 5 and $|b|=1$ or 5 , but not $(a, b)=(0,0)$. Since both $Z_{m}$ and $Z_{n}$ each have a unique subgroup of order 5 there are exactly five choices for $a$ and 5 choices for $b$ and therefore 25 choices for $(a, b)$, including $(0,0)$. So, there are exactly 24 elements in $Z_{m} \oplus Z_{n}$ of order 5 .

The identical argument shows that if $m$ and $n$ be positive integers that are divisible by a prime $p$, then the number of elements of order $p$ in $Z_{m} \oplus Z_{n}$ is $p^{2}-1$.

## 数学代写|抽象代数作业代写ALGEBRA代考| Properties of External Direct Products

$$\left|\left(g_{1}, g_{2}, \ldots, g_{n}\right)\right|=\operatorname{lcm}\left(\left|g_{1}\right|,\left|g_{2}\right|, \ldots,\left|g_{n}\right|\right)$$

• 例4 订单 100 组的示例包括
$Z_{100} ; Z_{25} \oplus Z_{2} \oplus Z_{2} ; Z_{5} \oplus Z_{5} \oplus Z_{4} ; Z_{5} \oplus Z_{5} \oplus Z_{2} \oplus Z_{2} ; D_{50} ; D_{10} \oplus Z_{5} ; D_{5} \oplus Z_{10} ;$ 和 $D_{5} \oplus D_{5}$. 这些不是同构的， 这是定理8.1的简单结果。
• 示例 5 让 $m$ 和 $n$ 是可被 5 整除的正整数。我们确定 5 阶的元素数 $Z_{m} \oplus Z_{n}$.按定理 $8.1$ ，我们只需要计算元素的数量 $(a, b)$ 在 $Z_{m} \oplus Z_{n}$ 具有属性 $5=|(a, b)|=\operatorname{lcm}(|a|,|b|)$. 显然，这需要 $|a|=1$ 或 5 和 $|b|=1$ 或 5 ，但不是 $(a, b)=(0,0)$. 由于两者兼而有之 $Z_{m}$ 和 $Z_{n}$ 每个都有一个唯一的 5 阶子群，正好有五个选择 $a$ 和 5 选择 $b$ 因此有 25 种选择 $(a, b)$ 包括 $(0,0)$.因此，正好有 24 个元素 $Z_{m} \oplus Z_{n}$ 订单 5 .
相同的参数表明，如果 $m$ 和 $n$ 是可被素数整除的正整数 $p$ ，然后是元素的顺序数 $p$ 在 $Z_{m} \oplus Z_{n}$ 是 $p^{2}-1$.

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