Definition The equivalence relation $a \sim b$ iff $\exists g, a^g = b$ partitions a group into conjugacy classes. Each elements of the center of a group is its own conjugacy class.
Proposition The number of cosets of a centralizers is the same as the number of elements of a conjugacy classes.
proof: Let $C$ be a conjugacy class so that $C^g = C$ for all $g$. Let $a \in C$ then the orbit of $a$ under the conjugacy action fills up the whole of $C$. The stabilizer of this action is equal to the centralizer of $a$ so by orb-stab we have $|C| = |G:C_G(a)|$.
Lemma Prime power order implies not centerless.
proof: Let $G$ act on itself by conjugation, clearly $Z(G)$ is invariant with respect to this action. We have the conjugacy class equation, where the sum runs over conjugacy class representatives $$|G| = |Z(G)| + \sum_{g}|G:C_G(g)|$$ with $C_G(g) = \{x \in G\mid \forall x \in G, xg = gx \}$ being the centralizer of $g$. Using the fact that $C_G(g)$ is never the whole group (otherwise $g$ commutes with everything, so it would be in the center instead) we deduce the lemma $\mod p$.
Lemma A nonabelian group can never have a nontrivial cyclic quotient.
proof: Let $G/N$ be generated by $gN$ so that every coset is of the form $g^i N$ and so every element of the group is of the form $g^i n$. Then the group is abelian since $$g^i n g^j n' = g^{i+j} n n' = g^j n' g^i n.$$
Theorem $|G|=p$ then $G=C_p$.
proof: Cauchy's theorem gives an element of order $p$, it must generate the whole group.
Theorem $|G|=p^2$ then $G=C_{p^2}$ or $C_p^2$.
proof: We know from counting conjugacy classes that $|Z(G)|$ is $p$ or $p^2$ and it can't be $p$ by the lemma (because $|G/Z(G)|=p$), so the group is abelian.
Theorem $|G|=p^3$ then $G$ is abelian or $|Z(G)|=p$
proof: In the non-abelian case $|Z(G)|$ must be $p$ or $p^2$, but $p^2$ cannot occur by the lemma since then $|G/Z(G)|=p$ would be cyclic.
Classification The nonabelian groups of order $2^3$ are $D_8$ and $Q$.
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