§ Class equation, Pgroup structure
§ Centralizer
 The centralizer of a subset $S$ of a group $G$ is largest subgroup of $G$ which is the center of $S$. It's defined as $C_G(S) \equiv \{ g \in G : \forall s \in S, gs = sg \}$. This can be written as $C_G(S) \equiv \{ g \in G : \forall s \in S, gsg^{1} = s \}$.

§ Conjucacy classes and the class equation
 Define $g \sim g'$ if there exists a $h$ such that $g' =kgk^{1}$. This is an equivalence relation on the group, and it partitions the group into conjugacy classes .
 Suppose an element $z \in G$ is in the center (Zentrum). Now, the product $kzk^{1} = z$ for all $k \in G$. Thus, elements in the center all sit in conjugacy classes of size $1$.
 Let $Z$ be the center of the group, and let $\{ J_i \subset G \}$ (J for conJugacy) be conjugacy classes of elements other than the center. Let $j_i \in J_i$ be representatives of the conjugacy classes, which also generate the conjugacy class as orbits under the action of conjugation.
 By orbit stabilizer, we have that $J_i = Orb(j_i) = G/Stab(j_i)$.
 The stabilizer under the action of conjugation is the centralizer! So we have $Orb(j_i) = G/C(j_i)$.
 Thus, we get the class equation: $G = Z + \sum_{j_i} G/C(j_i)$.
§ $p$group
 A $p$ group is a group where every element has order divisible by $p$.
 Claim: a finite group is a $p$group iff it has cardinality $p^N$ for some $N$.
 Forward  $G = p^N$ implies $G$ is a $p$group: Let $g \in G$. The $\langle g \rangle$ divides $G = p^N$ by Lagrange. Hence proved.
 Backward  $p$ divides \langle g \rangle for all $g \in G$ implies $G = p^N$ for some $N$: Write $G$ as disjoint union of cyclic subgroups: $G = \langle g_1 \rangle \cup \langle g_2 \rangle \cup \dots \langle g_n \rangle$. Take cardinality on both sides, modulo $p$. Each of the terms on the RHS $\langle g_i \rangle$ is divisible by $p$, and thus vanish. Thus, $G =_p 0 + 0 + \dots + 0 = 0$ modulo $p$. Hence, $G$ is divisible by $p$.
§ Center of $p$ group
 Let $G$ be a $p$group. We know that $G = Z(G) + \sum_{g_i} Orb(g_i)$, where we are considering orbits under group conjugation.
 See that $Orb(g_i) = G/Stab(g_i)$. The quantity on the right must be a power of $p$ (since the numerator is $p^N$). The quantity must be more than $1$, since the element $g_i$ is not in the center (and thus is conjugated nontrivially by some element of the group).
 Thus, $Orb(g_i)$ is divisible by $p$.
 Take the equation $G = Z(G) + \sum_{g_i} Orb(g_i)$ modulo $p$. This gives $0 =_p Z(G)$. Hence, $Z(G) \neq \{ e \}$(Since that would give $Z(G) =_p 1 \neq 0$). So, the center is nontrivial.
§ Cauchy's theorem: order of group is divisible by $p$ implies group has element of order $p$.
 Abelian case, order $p$: immediate, must be the group $Z/pZ$ which has generator of order $p$. Now induction on group cardinality.
 Abelian case, order divisible by $p$: Pick an element $g \in G$ and let the cyclic subgroup be generated by it be $C_g$ and let the order of $g$ be $o$ (Thus, $C_g = o$).
 Case 1: If $p$ divides $o$, then there is a power of $g$ with order $p$ (Let $o' \equiv o/p$. Consider $g^{o'}$; this has order $p$).
 Case 2: If $p$ does not divide $o$. Then $p$ divides the order of the quotient $G' \equiv G / C_g$. Thus by induction, we have an element $h C_g \in G / C_g$ of order $p$.
 Let $o$ be the order of $h$ in $G$. Then we have that that $(h C_g)^o = h^o C_g = e C_g$, where the last equality follows from the assumption that $o$ is the order of $h$. Thus we can raise $h C_g$ to $o$ get the identity in $G/C_g$. This implies $p$ (the order of $h G/C_g$) must divide $o$ (the order of $h$).
 Thus, by an argument similar to the previous, there is some power of $h$ with order $p$. (Let $o' \equiv o/p$. Consider $h^{o'}$' this has order $p$)
 General case: consider the center $Z$. If $p$ divides $Z$, then use the abelian case to find an element of order $p$ and we are done.
 Otherwise, use the class equation: $G = Z + \sum_{j_i} Orb(j_i)$.
 The LHS vanishes modulo $p$, the RHS has $Z$ which does not vanish. Thus there is some term $j_i$ whose orbit is not divisible modulo $p$.
 We know that $Orb(j_i) = G/Stab(j_i)$ where the action is conjugacy. Since the LHS is not divisible by $p$, while $G$ is divisible by $p$, this means that $Stab(j_i)$ has order divisible by $p$ and is a subgroup of $G$.
 Further, $Stab(j_i)$ is a proper subgroup as $Orb(j_i)$ is a proper orbit, and is thus not stabilized by every element of the group.
 Use induction on $Stab(j_i)$ to find element of order $p$.
§ Subgroups of pgroup
 Let $G$ be a finite $p$ group. So $G = p^N$. Then $G$ has a normal subgroup of size $p^l$ for all $l \leq N$.
 Proof by induction on $l$.
 For $l = 0$, we have the normal subgroup $\{ e \}$.
 Assume this holds for $k$. We need to show it's true for $l \equiv k + 1$.
 So we have a normal subgroup $N_k$ of size $p^k$. We need to establish a subgroup $N_l$ of size $p^{k+1}$.
 Consider $G/N_k$. This is a $p$group and has cardinality $p^{Nk}$. As it is a $p$group, it has nontrivial center. So, $Z(G/N_k)$ is nontrivial and has cardinality at least $p$.
 Recall that every subgroup of the center is normal. This is because the center is fixed under conjugation, thus subgroups of the center are fixed under conjugation and are therefore normal.
 Next, by Cauchy's theorem, there exists an element $z$ of order $p$ in $Z(G/N_k)$. Thus, there is a normal subgroup $\langle z \rangle \subset G/N_k$
 We want to pull this back to a normal subgroup of $G$ of order $\langle z \rangle \cdot N_k = p^{k+1}$.
 By correspndence theorem, the group $\langle z \rangle \cdot N_k$ is normal in $G$ and has order $p^{k+1}$. Thus we are done.