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Lemma. Suppose $G$ is a group and $\emptyset \neq S \subset G$. Set $N=N_G(S)$. Let $\{g_i\mid i \in I\}$ be a complete set of right cosets representations for $N$ in $G$. Then the conjugates of $S$ in $G$ are $\{S^{g_i}\mid i \in I\}$ and $S^{g_i}=S^{g_j}$ iff $g_i=g_j$.

What is a complete set of right cosets representations? I just don't get what they actually are. I know you can get an equivalence class out of the conjugate classes.

Also, can someone explain what the class equation of finite group is?

As by the definition he has. Let $G$ be a finite group. Let $x_1,x_2,...,x_k$ be elements of $G$ one from each of the $R$ conjugacy classes.

Set $n_i=|x_i^G|$. Assume notation chosen so as $n_1=...=n_l$ and $n_i>1$ for $i>l$

i) $|G|=\sum_{i=1}^k n_i= \sum_{i=1}^k[G:C_{G}(x_i)]$

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what do you mean by $(i \in I)$? – funktor Jan 10 '12 at 19:06

Remember that if $N$ is a subgroup of $G$, then $N$ induces an equivalence relation on $G$ by $$x\sim y\text{ if and only if }Nx = Ny$$ where $Nx = \{nx\mid n\in N\}$ is the collection of all products of elements of $N$ by $x$. The sets $Nx$ are the "right cosets of $N$".

Every equivalence relation induces a partition on the set, so the equivalence relation $\sim$ above partitions $G$ into equivalence classes; the equivalence classes are precisely the right cosets of $G$. So:

  1. Every element of $G$ lies in $Nx$ for some $x\in G$;
  2. If $Nx\cap Ny\neq\emptyset$, then $Nx=Ny$.

A "representative" of an equivalence class is an element of the equivalence class. So a representative of the right coset $Nx$ is any element of $Nx$; you have $x$, but you also have all elements of the form $nx$ with $n\in N$.

A subset of $G$, $\{g_i\mid i\in I\}$ is called a complete set of right coset representatives of $N$ if and only if:

  1. For every $g\in G$ there exists an $i\in I$ such that $Ng = Ng_{i}$ (that is, every coset has at least one representative in the set); and
  2. If $Ng_i = Ng_j$, then $i=j$ (that is, every coset has at most one representative in the set).

In other words, a complete set of right coset representatives of $N$ is a subset of $G$ that contains one and only one element from each right coset of $N$ in $G$.

The quoted theorem says that if you look at the set of conjugates of $S$ in $G$, you have exactly one conjugate for each right coset of $N$; that is, there is a bijection between the set of right cosets of $N$ in $G$, and the set of conjugates of $S$ in $G$; the bijection is given by first picking a set of right coset representatives (which is in bijection with the set of right cosets, since you are taking one and only one representative from each coset), and then associating the right coset representative $g_i$ with the conjugate $S^{g_i} = \{ g_i^{-1}sg_i \mid s\in S\}$. The fact that $S^{g_i}=S^{g_j}$ if and only if $i=j$ means that this map is both one-to-one and well-defined. To prove it is onto, let $x\in G$ and look at $S^x$; then there exists one (and only one) $i\in I$ such that $Nx = Ng_i$; in particular, $xg_i^{-1}\in N$, so $S^{xg_i^{-1}} = S$. Therefore $$S^{g_i} = (S^{xg_i^{-1}})^{g_i} = S^{xg_i^{-1}g_i} = S^x,$$ so $S^x$ corresponds to $g_i$.

The class equation is closely related to this.

Define an equivalence relation $\equiv$ on $G$ by $$x\equiv y\text{ if and only if there exists }z\in G\text{ such that }x^z = y^z.$$ (That is, $x\equiv y$ if and only if $x$ and $y$ are conjugate in $G$). It is not hard to verify that this is an equivalence relation. Since it is an equivalence relation, it partitions $G$ into equivalence classes.

The equivalence class of $x$ consists exactly of all conjugates of $x$, i.e., $$[x] = \{ x^g \mid g\in G\} = x^G.$$ Now, how many elements does $x^G$ have? There is one for every element of $G$, but there can be repeats. How many repeats?

Well, $$\begin{align*} x^y = x^z &\iff y^{-1}xy = z^{-1}xz\\ &\iff xy = yz^{-1}xz\\ &\iff x(yz^{-1}) = (yz^{-1})x\\ &\iff yz^{-1}\text{ commutes with }x\\ &\iff yz^{-1}\in C_G(x)\\ &\iff C_G(x)y = C_G(x)z\\ &\iff y\text{ and }z\text{ are in the same right coset of }C_G(x). \end{align*}$$ (Where $C_G(x)$ is the centralizer of $x$ in $G$). Thus, you get as many repeats as there are elements in $C_G(x)$. That means that you get as many different conjugates as there are right cosets of $C_G(x)$, and the number of right cosets of $C_G(x)$ is precisely the index of $C_G(x)$. That is, $$|x^G| = [G:C_G(x)]\text{ for each }x\in G.$$

Now consider the partition of $G$ induced by $\equiv$. If you take one and only one element from each equivalence class, and call them $x_1,\ldots,x_n$, then for every $g\in G$ there is one and only one $i$ such that $g\in x_i^G$. That means that $G = \cup x_i^G$, and the union is pairwise disjoint. So $$|G| = \left|\bigcup_{i=1}^k x_i^G\right| = |x_1^G| + |x_2^G| + \cdots + |x_k^G|.$$ But we just say that $|x_i^G| = [G:C_G(x_i)]$, so we get: $$|G| = \sum_{i=1}^k |x_i^G| = \sum_{i=1}^k [G:C_G(x_i)],$$ which is the equation given.

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What is a complete set of right cosets representations? I strongly suspect this is a typo: it should read "let $\lbrace g_i \mid i \in I\rbrace$ be a complete set of right coset representatives" - that is, one element $g_i$ from each distinct right coset $N g_i$ of $N$.

Also, can someone explain what the class equation of finite group is? Given a subgroup $H\subset G$, the index of $H$, written $[G:H]$, is the number of distinct cosets of $H$ in $G$. In the case of finite groups, one can prove that $[G:H]=|G|/|H|$. For example, $[\mathbb{Z}_8:4\mathbb{Z}_8]=4$, since $4\mathbb{Z}_8$ has four cosets $4\mathbb{Z}_8,4\mathbb{Z}_8+1,4\mathbb{Z}_8+2$ and $4\mathbb{Z}_8+3$. The class equation for a group relates the size of the group to the indices of certain centralizers $C_G(x)=\lbrace g\in G\mid gxg^{-1}=x\rbrace$. Specifically, if $G$ has $k$ distinct conjugacy classes, with representatives $x_1,\dotsc,x_k$, then

$$|G|=\sum_{i=1}^k [G:C_G(x_i)]$$

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