有限群生成问题：子集大小超|G|/p时生成整个群

Hey there! Let's work through this group theory problem step by step, filling in the gaps from where you left off.

设$G$是阶大于1的有限群，$S\subset G$是子集，满足$#S>\frac{1}{p}#G$，其中$p$是$G$的阶的最小素因子，证明$\langle S\rangle=G$。

We'll use a proof by contradiction paired with core finite group properties (Lagrange's theorem, coset decompositions) to tackle this.

Case 1: $p=2$ (Even-Order Groups) #

You started on this line of reasoning, so let's wrap it up properly:
Since $#S > \frac{1}{2}#G$, consider the inverse set $S^{-1} = {s^{-1} \mid s \in S}$. Note that $#S^{-1} = #S > \frac{1}{2}#G$.

If $S$ and $S^{-1}$ were disjoint, their union would have size $#S + #S^{-1} > #G$, which is impossible (both sets are subsets of $G$). But we can go further: take any arbitrary element $x \in G$, and look at the set $xS^{-1}$. This set has the same size as $S$ (since multiplying by a fixed group element is a bijection), so $#(xS^{-1}) > \frac{1}{2}#G$.

Again, $xS^{-1}$ and $S$ can't be disjoint—their combined size would exceed $G$'s order. So there must exist $s_1, s_2 \in S$ such that $x s_1^{-1} = s_2$. Rearranging gives $x = s_2 s_1$, meaning every element of $G$ is a product of two elements from $S$. This directly implies $\langle S\rangle = G$.

General Case (Arbitrary Minimal Prime $p$) #

Let's use subgroup cosets and Lagrange's theorem here. Suppose for contradiction that $\langle S\rangle = H$, where $H$ is a proper subgroup of $G$ (i.e., $H \neq G$).

By Lagrange's theorem, the index $[G:H] = \frac{#G}{#H}$ divides $#G$. Since $p$ is the smallest prime divisor of $#G$, the index $[G:H]$ must be at least $p$—any smaller index would have to be a positive integer less than $p$, which can't divide $#G$ (since $p$ is the smallest prime factor).

So $[G:H] = k \geq p$, which means $#H = \frac{#G}{k} \leq \frac{#G}{p}$. But wait—since $\langle S\rangle = H$, every element of $S$ is contained in $H$, so $#S \leq #H$.

Our given condition is $#S > \frac{1}{p}#G$. Combining this with $#H \leq \frac{#G}{p}$ gives $#S > \frac{#G}{p} \geq #H$, which contradicts $#S \leq #H$.

This contradiction means our assumption that $H$ is a proper subgroup is false. Therefore, $\langle S\rangle = G$.

Alternative Coset Perspective #

If you prefer a more visual angle: $G$ can be split into $k \geq p$ disjoint left cosets of $H$, each of size $#H$. Since $#S > \frac{#G}{p} = \frac{k#H}{p} \geq #H$ (because $k \geq p$), $S$ can't fit entirely within a single coset of $H$. But if $\langle S\rangle = H$, all elements of $S$ must lie in $H$ (one coset)—a clear impossibility. Hence $H = G$.

内容的提问来源于stack exchange，提问作者Václav Mordvinov