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抽象代数代写| Lagrange’s theorem

1. Cosets and Lagrange’s theorem

2. Left and right cosets

3. Recall the definition

4. Definition 1

Let $G$ be a group and $H$ a subgroup of $G$. A left coset of $H$ is a subset of $G$ of the form

g H={g h, h \in H}

In the same way, we can define right cosets to be $H g={h g, h \in H}$ for $g \in G$.

5. Proposition 1

Consider the relation

a \sim b \text { if there exists } h \in H \text { such that } a=b h \text {. }

Equivalently, $a \sim b$ if and only if $b^{-1} a \in H$ and if and only if $a \in b H .$ It is an equivalence relation, and the left cosets of $H$ are its equivalence classes.

Proof. To verify that equivalence classes modulo relation $\sim$ coincide with left cosets we need to verify two claims:

  1. If $a \sim b$, i.e., $a^{-1} b \in H$, and $a$ belongs to a coset $g H$, then $b$ belongs the same coset. Indeed, since $a \sim b$, we have that there exists $h \in H$ such that $a=b h$. On the other hand, since $a \in g H$, there exists $h_{1}$ such that $a=g h_{1}$. Substituting $a$, we find

b h=g h_{1} \Longleftrightarrow b=g\left(h_{1} h^{-1}\right)

Thus $b$ belongs to the same coset $g H$, as claimed.

  1. If $a, b$ belong to a left coset $g H$, then we can find two elements $h_{a}, h_{b} \in H$ such that

g h_{a}=a \quad g h_{b}=b

But then $a^{-1} b=\left(h_{a}^{-1} g^{-1}\right)\left(g h_{b}\right)=h_{a}^{-1} h_{b} \in H .$ Thus by definition $a \sim b .$

By the previous remark, we have the following:

6. Proposition 2

Let $H$ be a subgroup of a group $G .$ Then $G$ is the disjoint union of the left cosets of $H .$ In other words, the left cosets of $H$ form a partition of $G$.

7. Remark 1

This property is also true for right cosets. This can be seen by introducing another equivalence relation $\sim^{\prime}$ given by $a \sim^{\prime} b$ if and only if there exists $h \in H$ such that $a=h b$ (or, equivalently, $a b^{-1} \in H$, or $a \in H b$ ). Its equivalence classes are the right cosets. Note moreover that $a \sim b$ if and only if $a^{-1} \sim^{\prime} b^{-1}$, so that $a H=b H$ if and only if $H a^{-1}=H b^{-1} .$

i:{\text { left cosets of } H} \rightarrow{\text { right cosets of } H}

given by $i: a H \mapsto H a^{-1}$, well-defined and injective thanks to the previous remark. It is also surjective since for all $b \in G, \alpha\left(b^{-1} H\right)=H b$. We may conclude the following:

8. Proposition 3

Let $G$ be a group and $H$ a subgroup of $G$. Then the number of left cosets of $H$ is equal to the number of right cosets.

9. Index of a subgroup

10. Definition 2

Let $H$ be a subgroup of a group $G$. The index of $H$ in $G$, denoted by $[G: H]$, is defined to be the number of distinct left cosets of $H$ in $G$. If this number is infinite, then we write $[G: H]=\infty$.

11. Remark 2

By proposition 3, this is the same as the number of distinct right cosets.

12. Example 1

The index of ${0,3}$ in $\mathbb{Z}_{6}$ is 3 .

Question: what is the index of $\langle k\rangle$ for $k \in \mathbb{Z}_{n}$.

13. Example 2

Consider $G=\mathbb{Z}$ and $H=n \mathbb{Z} .$ Observe that in this case, the equivalence relation $\sim$ is exactly the relation of congruence modulo $n$, the cosets being exactly

n \mathbb{Z}, 1+n \mathbb{Z}, \ldots,(n-1)+n \mathbb{Z}

Thus, $[\mathbb{Z}: n \mathbb{Z}]=n$.

In particular, index $[G: H]$ might be finite even if $G$ and $H$ are infinite.

Note that in general, $[G: H]$ may be infinite. For example, a left coset of the trivial group in a group $G$ is of the form ${a}$ for $a \in G$. Thus, if $G$ is infinite, $[G:{e}]$ is infinite.

14. Remark 3

If $H$ is a subgroup of index 1 in $G$, then $H=G$.

15. Example 3: Subgroups of index 2

An important special case is that of subgroups of index 2. Let $G$ be a group and $H$ a subgroup of $G$ such that $[G: H]=2$. This means that we have two left cosets, one of them being $H$ itself, and the other being $G \backslash H$, which should be the equivalence class of all $g \in G \backslash H$, so that $G$ is the disjoint union $G=H \sqcup g H$ for any $g \in G \backslash H$. In exactly the same manner, we have two right cosets, one of them being $H$, and the other being given by $H g$ where $g$ is any element of $G \backslash H .$ Therefore, for all $g \in G \backslash H$, we have

g H=G \backslash H=H g

On the other hand, for all $g \in H$, we have

g H=H=H g

Therefore, we observe that in this case, the right cosets and the left cosets of $H$ are the same.

Problem 1: For the cyclic subgroups $H_{1}$ and $H_{2}$ of $\left(S_{3}, *\right)$ (the groups of symmetries of a triangle) generated by the rotation and the reflection respectively, find its cosets and index.

16. Lagrange’s theorem

17. Proposition 4

Any two cosets $a H$ and $b H$ have the same number of elements.

Proof. We construct a bijection between $a H$ and $b H$, which will prove that these two sets have the same number of elements.

Define a map

f: a H \rightarrow b H

by sending each element $g=a h \in a H$ to $b h=\left(b a^{-1}\right) g \in b H .$ This map is bijection, since it admits an inverse (given by an analogous left multiplication with $a b^{-1}$ ).

Observing that the group $G$ therefore is partitioned into $[G: H]$ subsets which all have $|H|$ elements, we have the following important counting formula:

18. Theorem 1: Counting formula

Let $G$ be a finite group and $H$ a subgroup of $G$. Then

|G|=[G: H]|H| .

This formula makes sense even if some of $|G|,[G: H]$ and $|H|$ are infinite. An important consequence of this is Lagrange’s theorem:

19. Theorem 2: Lagrange

Let $G$ be a finite group and $H$ a subgroup of $G$. Then the size of $H$ divides the size of $G$.

Proof. By counting formula we have

|G|=[G: H]|H|

Since $[G: H]$ is an integer, it implies that $|H|$ divides $|G|$.

20. Corollary 1

Let $G$ be a finite group. The order of any element of $G$ divides the size of $G$.

Proof. Any element $a \in G$ generates a cyclic subgroup $\langle a\rangle \subset G$ of $\operatorname{size}$ ord $(a) .$ By the previous theorem, ord $(a)$ divides $|G|$.

21. Corollary 2

Let $G$ be a finite group with order a prime number $p .$ Then $G$ is cyclic, and any $a \in G$ different from the identity element is a generator.

22. Remark 4

Corollary 2 implies that up to isomorphism, there is only one group of order a prime $p$, namely $\mathbb{Z} / p \mathbb{Z}$. Note that we already knew that all elements of $\mathbb{Z} / p \mathbb{Z}$ except 0 are generators.


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