Let \(H \normal G\) be a normal subgroup of a finite group \(G\), where the order of \(H\) is the smallest prime \(p\) dividing \(\abs{G}\). Prove that \(H\) is contained in the center of \(G\).
\(x\in Z(G)\) iff \(\size C_x = 1\), i.e. the size of its conjugacy class is one.
Normal subgroups are disjoint unions of (some) conjugacy classes in \(G\).
In fact, this is a characterization of normal subgroups (i.e. \(H\) is normal iff \(H\) is a union of conjugacy classes in \(G\)).
Why: if \(H\normal G\) then \(gh\inverseof{g} \in H\) for all \(g\), so \(C_h \subseteq H\) and \(\Union_h C_h = H\). Conversely, if \(H = \Union_{h\in H} C_h\), then \(gh\inverseof{g} \in C_h \subseteq H\) and thus \(gH\inverseof{g} = H\).
Orbit stabilizer theorem: \(\size C_g = \size G/ \size K_g\) where \(C_g\) is the centralizer and \(K_g\) is the conjugacy class of \(g\).
- In particular, \(\size C_g\) divides \(\size G\).
Show an element \(x\) is central by showing \(\size C_x = 1\).
Solution due to Swaroop Hegde, typed up + modifications added by DZG.
Let \(p \da \size H\).
Let \(\ts{ C_i }_{i\leq n}\) be the conjugacy classes in \(G\), then \(G = \disjoint_{i\leq n} C_i\)
By the first fact, there is a sub-collection \(\ts{ C_{i_j}}_{j\leq k }\) such that \[ H = \disjoint_{j\leq k} C_{i_j} .\]
The identity is always in a single conjugacy class, so \(C_e = \ts{ e }\).
Since \(e\in H\), without loss of generality, label \(C_{i_1} = \ts{ e }\).
So \[ H = \Disjoint_{j\leq k} C_{i_j} = C_{i_1}{\textstyle \coprod} \displaystyle\Disjoint_{\substack{ j\leq k \\ j\neq 1} } C_{i_j} .\]
Take cardinality in the above equation \[ p = 1 + \sum_{\substack{ j\leq k \\ j\neq 1 }} \size C_{i_j} .\]
So \(\size C_{i_j} \leq p-1\) for all \(j\neq 1\).
Every \(\size C_{i_j}\) divides \(\size G\), but \(p\) was the minimal prime dividing \(\size G\), forcing \(\size C_{i_j} = 1\) for all \(j \neq 1\).
- This rules out \(\size C_{i_j}\) being a prime less than \(p\), but also rules out composites: if a prime \(q\divides \size C_{i_j}\), then \(q<p\) and \(q\divides \size G\), a contradiction.
By fact 3, each \(x\in C_{i_j}\) satisfies \(x\in Z(G)\).
\(\union C_{i_j} = H\), so \(H \subseteq Z(G)\).
Solution
Because \(|H|=p\) is prime, \(H\cong C_p\). Conjugation by \(G\) on \(H\) gives a homomorphism \[ \varphi:G\longrightarrow \operatorname{Aut}(H)\cong C_{p-1}. \] Hence \(|\operatorname{im}\varphi|\) divides both \(|G|\) and \(p-1\). Every prime divisor of \(|\operatorname{im}\varphi|\) therefore divides \(|G|\), so by minimality of \(p\) it is at least \(p\); but it also divides \(p-1\). Thus \(|\operatorname{im}\varphi|=1\).
Therefore conjugation by every \(g\in G\) acts trivially on \(H\). Equivalently, \(ghg^{-1}=h\) for all \(g\in G\) and \(h\in H\), so \(H\subseteq Z(G)\).