Finite group is generated by a set of representatives of conjugacy classes.

Question

Could you tell me how to prove that a finite group is generated by a set of representatives of conjugacy classes?

I've read this https://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class but I don't know

why if $H$ is a proper subgroup of a finite group $G$ intersecting every conjugacy class of $G$, then $G = \bigcup_{g \in G} g H g^{-1}$

and why the fact that we can't cover $G$ with sets in $\bigcup_{g \in G/H} g H g^{-1}$ implies that the group is generated by a set of representatives of conjugacy classes?

Maybe you know a simpler proof?

Thank you.

Answer 1

Let $x \in G$. $x$ is in a conjugacy class, $x^G = \{ g^{-1} x g : g \in G \}$, and $H \cap x^G \neq \varnothing$ by assumption, so let $h \in H \cap x^G$. Since $h \in x^G$, there is some $g \in G$ with $g^{-1} x g = h$ and so $x = ghg^{-1} \in gHg^{-1} \subseteq \cup_{g \in G} gHg^{-1}$.

Since $G$ is not in fact such a union (by the Lemma below), there is no such subgroup $H$. Every subgroup that intersects all conjugacy classes must not be proper, it must be all of $G$.

If $H$ is generated a set of representatives of the conjugacy classes, then it intersects every conjugacy class (for instance, in that chosen generator), and so it cannot be proper; instead $H=G$.

Lemma: No finite group is the union of conjugates of a proper subgroup.

Proof: [Standard and included in Hagen's answer] Let $G = \cup_{g \in G} gHg^{-1}$ for some subgroup $H \leq G$. Since $gh H = gH$ and $H(gh)^{-1} = Hh^{-1} g^{-1} = Hg^{-1}$, we don't need to union over all $g \in G$, since $g$ and $gh$ give the same subgroup $gHg^{-1} = ghH(gh)^{-1}$. We only need $[G:H]$ different $g$. Now consider the non-identity elements of $G$. Each one has to be in at least one of the $gHg^{-1}$, but it won't be the identity element of that subgroup. Hence the $|G|-1$ different non-identity elements of $G$ lie in one of the subsets $gHg^{-1} \setminus \{1\}$ of size $|H|-1$. Since there are at most $[G:H]$ of those subsets we get an inequality, $|G| - 1 \leq [G:H](|H|-1)$, but the right hand side is just $|G|- [G:H]$. Simplifying $|G|-1 \leq |G|-[G:H]$ gives $[G:H] \leq 1$, but since $[G:H]$ is a positive integer, this means $[G:H]=1$ and $H=G$. $\square$

Answer 2

Let $S$ be a set containing an element from each conjugacy class, and let $H=\langle S\rangle$. By assumption, for any $a\in G$ there is a conjugate of $a$ in $S\subseteq H$, that is $g^{-1}ag\in H$ for some $g$ or equivalently $a\in gHg^{-1}$ for some $g$.

The rest of the argument on the linked question goes like this: $\bigcup_{g\in G/H}gHg^{-1}\setminus\{1\}$ has at most $|G/H|\cdot (|H|-1)=|G|-|G/H|$ elements. But if the union is all of $G$ (because $H$ contains elements from all conjugacy classes), we must have $|G|-1$ elements after removing $\{1\}$. Hence $|G/H|=1$, i.e. $G=H=\langle S\rangle$.

Answer 3

This is another proof:

Burnside lemma: Let $G$ be a finite groups acting on a finite set $X$, let $r$ be the number of orbits of this action, and for $g \in G$ Fix($g$) be the fixed points of $g$ in $X$, then $$r= \frac{1}{|G|} \sum_{g \in G} Fix(g) $$ A trivial result of this lemma is : if a finite group G acting transitively on a set $X$ that has more than tow elements, then there is $g \in G$ that fixes no elements. now let $H$ a proper subgroup of $G$,then $G/H$ has more than tow elements, consider the translation action of $G$ on $G/H$, then by previous result there is $x \in G$ that fixes no elements, so $x \in G\setminus \cup_{g \in G} stab(gH)$, and because of $stab(gH) = gHg^{-1}$ then $G \neq \cup_{g \in G} gHg^{-1} $.