\(G\) acting on itself (i.e., \(A = G\)) by conjugation: $$ g \cdot a = gag^{-1} $$
Fact
Conjugation by \(g \in G\) is a group action of \(G\) on itself.
Proof
Define \(\cdot\) so that \(g \cdot a = gag^{-1}\).
For any \(g_1, g_2 \in G\) and \(a \in G\), $$ g_1 \cdot (g_2 \cdot a) = g_1 \cdot (g_2ag_2^{-1}) = (g_1g_2)a(g_2^{-1}g_1^{-1}) = (g_1g_2)a(g_1g_2)^{-1} = g_1g_2 \cdot a. $$
For any \(a \in G\), $$ 1 \cdot a = 1a1^{-1} = 1. $$
$$\tag*{$\blacksquare$}$$
Conjugate
Two elements \(a\) and \(b\) of a group \(G\) are said to be conjugate in \(G\) if there is some \(g \in G\) such that \(b = gag^{-1}\) (i.e., if and only if they are in the same orbit of \(G\) acting on itself by conjugation). The orbits of \(G\) acting on itself by conjugation are called the conjugacy classes of \(G\).
More generally, define $$ gSg^{-1} = \{gsg^{-1} \mid s \in S\}. $$ Two subsets \(S\) and \(T\) of \(G\) are said to be conjugate in \(G\) if there is some \(g \in G\) such that \(T = gSg^{-1}\) (i.e., if and only if they are in the same orbit of \(G\) acting on its subsets by conjugation).
Proposition 6
The number of conjugates of a subset \(S\) in a group \(G\) is the index of the normalizer of \(S\), \(|G:N_G(S)|\). In particular, the number of conjugates of an element \(s\) of \(G\) is the index of the centralizer of \(s\), \(|G:C_G(s)|\).
Proof
By Proposition 2, the number of conjugates of \(S\) is \(|G:G_S|\), the index of the stabilizer of \(S\). Since we are considering \(G\) acting on itself by conjugation, by definition of stabilizer and normalizer, $$ G_S = \{g \in G \mid g \cdot S = S\} = \{g \in G \mid gSg^{-1} = S\} = N_G(S). $$
Therefore, the number of conjugates of \(S\) is \(|G:N_G(S)|\).
Additionally, by definition of normalizer and centralizer, $$ N_G(\{s\}) = \{g \in G \mid gsg^{-1} = s\} = C_G(\{s\}). $$
Therefore, the number of conjugates of \(s\) is the index of the centralizer of \(\{s\}\).
*I assume we mean \(C_G(\{s\})\) by \(C_G(s)\)…
$$\tag*{$\blacksquare$}$$
Theorem 7. The Class Equation
Let \(G\) be a finite group and let \(g_1, g_2,…,g_r\) be representatives of the distinct conjugacy classes of \(G\) not contained in the center \(Z(G)\) of \(G\). Then
$$ |G| = |Z(G)| + \sum_{i = 1}^r |G:C_G(g_i)|. $$
Proof
By Proposition 2, orbits are equivalence classes, and thus so are conjugacy classes by definition of conjugacy class. Therefore, by Proposition 0.2, conjugacy classes partition \(G\). Therefore, the order of \(G\) is the sum of the orders of conjugacy classes of \(G\).
By definition of center, for \(x \in G\), \(\{x\}\) is a conjugacy class of size \(1\) if and only if \(x = gxg^{-1}\) for all \(g \in G\) if and only if \(x \in Z(G)\).
It follows that a conjugacy class is either contained in \(Z(G)\) (the conjugacy class is of size \(1\) in this case), or mutually exclusive with \(Z(G)\) (\(\because\) if \(x \in Z(G)\), \(x\) by itself composes a conjugacy class and thus contained in \(Z(G)\)).
Therefore, \(|G| = A + B\), where
- \(A\) denotes the sum of the orders of conjugacy classes contained in the center \(Z(G)\), and
- \(B\) denotes the sum of the orders of conjugacy classes not contained in the center \(Z(G)\).
We just showed that \(\{x\}\) is a conjugacy class of size \(1\) if \(x \in Z(G)\), and thus \(A = |Z(G)|\).
For \(B\), let \(K_1, K_2, …, K_r\) be conjugacy classes not contained in \(Z(G)\) and let \(g_i\) be a representative of \(K_i\). By Theorem 6, \(|K_i| = |G:C_G(g_i)|\).
Hence, $$ |G| = A + B = |Z(G)| + \sum_{i = 1}^r |G:C_G(g_i)|. $$
$$\tag*{$\blacksquare$}$$
Theorem 8
If \(p\) is a prime and \(P\) is a group of prime power order \(p^\alpha\) for some \(\alpha \ge 1\), then \(P\) has a nontrivial center: \(Z(P) \ne 1\).
Proof
By the class equation,
$$ |P| = |Z(P)| + \sum_{i=1}^r |P:C_P(g_i)| $$ where \(g_i\)’s are representatives of the distinct non-central conjugacy classes.
Since \(P\)’s order is \(p^\alpha\), \(p\) divides \(|P|\).
Since \(g_i \notin Z(P)\), by definition of center, there exists \(x \in P\) such that \(g_ix \ne xg_i\). Therefore, by definition of centralizer, \(C_P(g_i) \ne P\).
In general, all distinct cosets of a subgroup \(N\) in a group \(G\) are of the same size and they form a partition of \(G\). Therefore, given that \(P\) is of order \(p^\alpha\) and that \(C_P(g_i) \ne P\), \(p\) must divide \(|P:C_P(g_i)|\) for each \(i\).
Therefore, \(p\) must divide \(|Z(P)|\) and thus \(Z(P) \ne 1\). $$\tag*{$\blacksquare$}$$