# 数学代写|抽象代数作业代写abstract algebra代考|Math417

## 数学代写|抽象代数作业代写abstract algebra代考|Order of an Element in a Direct Product

I 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 $|a|=1$ or 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 $p^2-1$.

I EXAMPLE 6 We determine the number of cyclic subgroups of order 10 in $Z_{150} \oplus Z_{50}$. We begin by counting the number of elements of order 10 . For $10=|(a, b)|=\operatorname{lcm}(|a|,|b|)$ we must restrict $|a|$ and $|b|$ to $1,2,5$, and 10 . So, we know that that $a$ must belong in the unique subgroup $\langle 15\rangle$ of order 10 in $Z_{150}$ and $b$ must belong in the unique subgroup $\langle 5\rangle$ of order 10 in $Z_{50}$. Thus, we know $(a, b) \in\langle 15\rangle \oplus\langle 5\rangle$, which is isomorphic to $Z_{10} \oplus Z_{10}$. Since the orders of the elements of $Z_{10} \oplus Z_{10}$ are $1,2,5$, and 10 , we can determine the number of elements of order 10 by subtracting from 100 the number of elements of orders 1, 2, and 5 . From Example 5 we know that $Z_{10} \oplus Z_{10}$ has 24 elements of order 5 and 3 elements of order 2. So we get $100-24-3-1=72$ elements of order 10 in $Z_{150} \oplus Z_{50}$. Because each cyclic subgroup of order 10 has four elements of order 10 and no two of the cyclic subgroups can have an element of order $10 \mathrm{in}$ common, there must be $72 / 4=18$ cyclic subgroups of order 10 . (This method is analogous to determining the number of sheep in a flock by counting the number of legs and dividing by 4.) Imagine how much work it would be to do this problem without our theorems!

## 数学代写|抽象代数作业代写abstract algebra代考|The Group of Units Modulo $n$ as an External Direct Product

What is the advantage of expressing the group $U(n)$ as an external direct product of groups of the form $Z_m$ ? Well, for one thing, we immediately see that $|U(105)|=2 \cdot 4 \cdot 6=48$ and that $U(105)$ and $U(144)$ are isomorphic. Another is that from Theorem $8.1$ we know that the orders of the elements in $U(105)$ are $1,2,3,4,6$ and 12. Moreover, arguing as in Examples 5 and 6, we can determine that $U(105)$ has exactly 16 elements of order 12 , say.

These calculations tell us more. Since $\operatorname{Aut}\left(Z_{105}\right)$ is isomorphic to $U(105)$, we also know that there are 16 automorphisms of $Z_{105}$ of order 12. Imagine trying to deduce this information directly from $U(105)$ or, worse yet, from $\operatorname{Aut}\left(Z_{105}\right)$ ! These results beautifully illustrate the advantage of being able to represent a finite Abelian group as a direct product of cyclic groups. They also show the value of our theorems about $\operatorname{Aut}\left(Z_n\right)$ and $U(n)$. After all, theorems are laborsaving devices. If you want to convince yourself of this, try to prove directly from the definitions that $\operatorname{Aut}\left(Z_{105}\right)$ has exactly 16 elements of order 12 .

Here is a fun example that pulls together our results from Chapter 6 and Chapter 8 .

• EXAMPLE 9 Find $\left|\left(\operatorname{Aut}\left(\operatorname{Aut}\left(\operatorname{Aut}\left(Z_{27}\right)\right)\right)\right)\right|$. To this end we note that $\operatorname{Aut}\left(\operatorname{Aut}\left(\operatorname{Aut}\left(Z_{27}\right)\right)\right) \approx \operatorname{Aut}(\operatorname{Aut}(U(27))) \approx$ $\operatorname{Aut}\left(\operatorname{Aut}\left(Z_{18}\right)\right) \approx \operatorname{Aut}(U(18)) \approx \operatorname{Aut}\left(Z_6\right) \approx U(6) \approx Z_2$.

# 抽象代数代考

## 数学代写|抽象代数作业代写abstract algebra代考|The Group of Units Modulo n作为外部直接产品

$|U(105)|=2 \cdot 4 \cdot 6=48$ 然后 $U(105)$ 和 $U(144)$ 是同构的。另一个是从定理 $8.1$ 我们知道元溸的顺序 $U(105)$ 是 $1,2,3,4,6$ 和 12. 此外，如示例 5 和示例 6 所示，我们可以确定 $U(105)$ 比如说，正好有 16 个 12 阶元箦。

• 例 9 查找 $\mid$ (Aut $\left.\left.\operatorname{Aut}\left(\operatorname{Aut}\left(Z_{27}\right)\right)\right)\right) \mid$. 为此，我们注意到
$\operatorname{Aut}\left(\operatorname{Aut}\left(\operatorname{Aut}\left(Z_{27}\right)\right)\right) \approx \operatorname{Aut}(\operatorname{Aut}(U(27))) \approx$
$\operatorname{Aut}\left(\operatorname{Aut}\left(Z_{18}\right)\right) \approx \operatorname{Aut}(U(18)) \approx \operatorname{Aut}\left(Z_6\right) \approx U(6) \approx Z_2$.

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