Theorem. There does not exist a group whose commutator subgroup is isomorphic to $$S_4$$.

The relevant facts are that $$S_4$$ is a complete group(no outer automorphisms, trivial center) which is not perfect(that is, the commutator subgroup of $$S_4$$ is not $$S_4$$ itself). Any group which has these properties is never a commutator subgroup of anything. Here’s why.

Lemma. If $$K$$ is a complete group and $$K\lhd G$$, then $$G$$ is the direct product $$K\times H$$ of $$K$$ by its centralizer $$H=C_G(K)$$.

In other words, a complete group can be a normal subgroup only in the most trivial fashion: the large group is just a direct product of the normal group by something.

Proof of the lemma. Let $$H=C_G(K)$$ be the centralizer of $$K$$ in $$G$$, namely the set of all elements which commute with all elements of $$K$$. $$H$$ is a normal subgroup of $$G$$, and $$H\cap K=Z(K)=1$$ since $$K$$ has trivial center. Any element $$g\in G$$ induces an automorphisms $$\phi_g$$ of $$K$$ by conjugation: $$\phi_g(k)=g^{-1}kg$$. But $$K$$ has no outer automorphisms, so $$\phi_g$$ must equal some inner automorphism of $$K$$, that is, for some $$k\in K$$, $$\phi_g=\phi_k$$. Now conjugation by $$gk^{-1}$$ does nothing to $$K$$, so $$gk^{-1}=h\in H$$. In other words $$g=kh$$: every element of $$G$$ is expressible as product of an element of $$K$$ and an element of $$H$$. Since $$H$$ and $$K$$ commute, $$G$$ is the direct product of $$H$$ and $$K$$.            $$\Box$$

Proof. Now suppose that $$G$$ is some group such that $$K=G^\prime=[G,G]$$, the commutator subgroup, is such that $$K$$ is complete and non-perfect. By the lamma, $$G=K\times A$$ where $$A\cong G/K$$ is an abelian group. So any element of $$G$$ can be writeen as $$ka$$ with $$k\in K$$ and $$a\in A$$, and moreover, $$ka=ak$$ for any $$k\in K$$, $$a\in A$$.

Consider a commutator $$c=xyx^{-1}y^{-1}$$ in $$G$$. Write $$x=ka$$ and $$y=lb$$. Since $$K$$ and $$A$$ commute, $$c=aba^{-1}b^{-1}xyx^{-1}y^{-1}$$. Since $$A$$ is abelian, the first part vanishes and $$c=xyx^{-1}y^{-1}$$. So any commutator of $$G$$ lies in the commutator sungroup of $$K$$, and it follows that $$G^\prime=K^\prime$$. Since $$K^\prime\ne K$$, $$G^\prime\ne K$$, as well.

It remains to show that $$S_4$$ is complete and non-perfect.

(i) $$S_4$$ has trivial center: this is obvious. No permutation commutes with all other permutations.

(ii) $$S_4$$ has no outer automorphisms. This is true for all $$S_n$$ except $$n=2$$, $$6$$. It’s a standard result.

(iii) $$S_4$$ is not perfect. This is also ovious: for any two permutations $$\sigma$$, $$\tau\in S_n$$, the commutator $$\sigma^{-1}\tau^{-1}\sigma\tau$$ is an even permutation, so the commutator subgroup is contained in the alternating group $$A_n$$. In fact the commutator group eauals $$A_n$$, but we don’t need that here.

This completes the proof the theorem.                                  $$\Box$$

Remark. a simple modification: $$K^\prime\subset G^\prime$$ is clear, and the other direction follows since $$G/K^\prime=K/{K^\prime\times A}$$ is abelian.

Here is a proof not using those well-known facts about $$S_4$$. (though it’s easy to derive them with it)

Proof. $$S_4$$ has exactly $$4$$ Sylow  $$3$$-subgroups

$P_1=\langle(234)\rangle, P_2=\langle(134)\rangle, P_3=\langle(124)\rangle, P_4=\langle(123)\rangle,$

where $$P_i$$ is the only Sylow $$3$$-subgroup of the stabilizer of $$i$$ in $$S_4$$ for $$i=1$$, $$2$$, $$3$$, $$4$$. So $$\sigma P_i\sigma^{-1}=P_{\sigma(i)}$$ for all $$\sigma\in S_4$$ and $$i=1$$, $$2$$, $$3$$, $$4$$.

Assume $$G^\prime=S_4$$ and take $$g\in G$$. The conjugation with $$g$$ permutes $$P_1$$, $$P_2$$, $$P_3$$, $$P_4$$, so we find $$\rho\in G$$ with $$g P_ig^{-1}=P_{\rho(i)}=\rho P_i\rho^{-1}$$ for $$i=1$$, $$2$$, $$3$$, $$4$$. So $$h\colon \rho^{-1}g\in G$$ satisfies $$h P_ih^{-1}=P_i$$ for $$i=1$$, $$2$$, $$3$$, $$4$$.

Then for all $$\sigma\in S_4$$ we have also $$h^{-1}\sigma h\in S_4$$ and

\begin{equation} \begin{split}P_{h^{-1}\sigma h(i)}&=(h^{-1}\sigma h)P_i(h^{-1}\sigma h)^{-1}\\&=h^{-1}\sigma hP_ih^{-1}\sigma^{-1}h\\&=h^{-1}\sigma P_i\sigma^{-1}h\\&= h^{-1} P_{\sigma(i)}h=P_{\sigma(i)}.\end{split} \end{equation}

and therefore $$h^{-1}\sigma h(i)=\sigma(i)$$ for $$i=1$$, $$2$$, $$3$$, $$4$$.

This gives $$h\in C_G(S_4)$$ and $$g=\rho h\in S_4C_G(S_4)$$ for all $$g\in G$$. So $$G=S_4C_G(S_4)$$ and $$|G\colon A_4C_G(S_4)|\leqslant2$$. But then $$A_4C_G(S_4)\trianglelefteq G$$ with abelian factor, so $$S_4=G^\prime\leq A_4C_G(S_4)$$, and by Dedekind we get

$S_4=A_4C_G(S_4)\cap S_4=A_4(C_G(S_4)\cap S_4)=A_4Z(S_4)=A_4$

since $$Z(S_4)=1$$, as $$\sigma\in Z(S_4)$$ would give $$P_i=\sigma P_i\sigma^{-1}=P_{\sigma(i)}$$ and $$i=\sigma(i)$$ for $$i=1$$, $$2$$, $$3$$, $$4$$. Contradiction!

•  结合律: 对于 $$G$$ 中任意元素 $$a$$, $$b$$, $$c$$, 有 $$(ab)c=a(bc)$$;
• 存在(左)单位元: $$G$$ 中有一个 $$e$$, 使得对于 $$G$$ 中任意元素 $$a$$, 有 $$ea=a$$;
• 存在(左)逆元: 对 $$G$$ 中任意元素 $$a$$, 存在 $$G$$ 中元素 $$b$$, 使得有 $$ba=e$$,

Colonel Johnson 在 A mixed non-group, The American Mathematical Monthly, Vol. 71, No. 7, pp. 785,  举了一个例子来说明, 非空集合 $$G$$ 上的二元运算满足结合律, 并且每个元素有左单位元和右逆元, 然而 $$G$$ 不一定是一个群.

$M=\left(\begin{array}{cc}x&y\\x&y\end{array}\right),$

$J=\left(\begin{array}{cc}0&1\\0&1\end{array}\right)$

$\left(\begin{array}{cc}0&\frac1{x+y}\\0&\frac1{x+y}\end{array}\right)$

### 群的早期历史

$K=\Bbb Q(\sqrt2,\sqrt3,\sqrt5,\dotsc,\sqrt{p_n},\dotsc),$

Abstract algebra(抽象代数)是本科生的基础课. 这里列出一些不错的参考书, 也写出评价. 这里, 暂时不涉及更深入的书.

1. 熊全淹, 近世代数

2. 聂灵沼, 丁石孙, 代数学引论

3. 丘维声, 抽象代数基础, 高等教育出版社

4. 赵春来, 徐明曜, 抽象代数, I, II. 北京大学出版社

5.  冯克勤, 李尚志, 章璞, 近世代数引论, 第三版, 中国科技大学出版社

6. 姚慕生, 抽象代数学, 复旦大学出版社, 第二版

7. 孟道骥 , 陈良云, 白瑞蒲, 抽象代数1:代数学基础, 科学出版社

8. 吴品三, 近世代数, 人民教育出版社

9. David S. Dummit and Richard M. Foote, Abstract Algebra, 3rd Edition

10. Joseph Gallian, Contemporary Abstract Algebra, 8th

11. Michael Artin, Algebra, second edition

Let $$H,K$$ be finite subgroups of a group $$G$$. Show that
$|HK|=\frac{|H|\cdot|K|}{|H\cap K|}.$

$\pi\colon (H\times K)\times HK\to HK$

$((h,k),x)\mapsto hxk^{-1}.$

$H\times (G/K)_l\to (G/K)_l$

$(h,xK)\mapsto hxK.$

$|Orb(K)|=\frac{|HK|}{|K|}.$

$\frac{|HK|}{|K|}=\frac{|H|}{|H\cap K|},$

Group action(the action of a group on a set, 群在集合上的作用) 是任何一本像样的群论入门书都会讲到的概念. 这概念是如此的重要, 在几何, 拓扑, 分析, 数论中用处广泛, 更不论代数了.

Let $$G$$ be a group and let $$X$$ be a set. Let $$S(X)$$ be the group of all permutations of $$X$$. An action or an operation of $$G$$ on $$X$$ is a homomorphism

$\pi\colon G\to S(X)$

of $$G$$ into $$S(X)$$.