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I am having trouble with exercises from Chapter IV of Aluffi's Algebra: Chapter 0. After spending many hours on the following two, I guess I need some help:

Let $G$ be a finite group, and suppose there exists representatives $g_1,g_2,\dots,g_r$ of the $r$ distinct conjugacy classes in $G$, such that $g_ig_j=g_jg_i$. Prove that $G$ is commutative.

Since proving the commutativity of $G$ is the same as proving all conjugacy classes are singletons, I am guessing I need to apply the commutativity of $g_i$'s to show that all conjugacy classes are of the same size, since we already know there is always an element whose conjugacy class is a singleton. To do this, I guess we need to use the class formula for group actions, but I do not know how to apply it.

The second problem is:

$G$ is a finite group acting transitively on a set $S$. If $|S|>2$, then there is a $g\in G$ without fixed points in $S$.

The author gives a hint. Since $S$ is isomorphic to $G/H$ as $G$-sets for some subgroup $H$, we might as well let $S=G/H$. The hint says we should use a result we have proved: A finite group cannot be the union of the conjugacy classes of a proper subgroup. Again I do not know how to use this hint.


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Your second question has been answered by Clayton. As for the first, see:… – user3533 Jan 5 '13 at 15:32
up vote 2 down vote accepted

I will address the second problem. This follows directly from the fact that $G$ cannot be the union of the conjugates of any one of its subgroups. Then, using the fact that the isotropic subgroups are conjugate to each other (since it is transitive), we have that $G$ properly contains the union of all of the isotropic subgroups, i.e., there is an element $x\in G$ that is not in any isotropic subgroup, hence $x$ doesn't fix any element in $S$.

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What is an isotropic subgroup? – Derek Holt Jan 5 '13 at 16:08
@DerekHolt: The isotropic subgroup is the stabilizer subgroup (there are various terms used for it). – Clayton Jan 5 '13 at 17:50

That's interesting because I am more used to deducing that a group cannot be a union of conjugates of a proper subgroup from the fact that every transitive permutation group of degree greater than 1 has a fixed-point-free element. The two are clearly equivalent. The existence of a fixed-point-free permutation follows from the well known result that the average number of fixed points of elements in a permutation groups is equal to the number of orbits, which is 1 for a transitive group. Since the identity element fixes more than the average, some element must fix less than average, i.e. 0.

Your first result follows in a similar way. If the group was not commutative, then for a non-central element $g$, the union of the conjugates of $C_G(g)$ in $G$ would be all of $G$.

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A couple of points worth remarking on: there are infinite groups in which a proper subgroup meets every conjugacy class. Perhaps the simplest example is $G = {\rm GL}(n,\mathbb{C}),$ ($n >1$), with $T$ the subgroup of upper triangular matrices in $G.$ Then every element of $G$ is conjugate (even by a unitary matrix) to an element of $T,$ but $T$ is clearly a proper subgroup.

Also, you can see the first problem as a natural consequence of the second,and the second can be proved without the first. The second follows from the orbit-counting formula, often attributed to Burnside. I give a simple (and standard) proof in the transitive case: so suppose that (finite) $G$ acts transitively on $\Omega$. Count the ordered pairs $(\alpha,g) \in \Omega \times G$ such that $\alpha.g = \alpha$. This is $\sum_{ \alpha \in \Omega} |G_{\alpha}| = |G_{\alpha}|[G:G_{\alpha}]$ (for any choice of $\alpha$ = $|G|$.

On the other hand, if we let ${\rm Fix}(g)$ denote the set of fixed points of $G \in G$. the number of ordered pairs above is just $\sum_{g \in G}|{\rm Fix}(g)|.$ Hence $\sum_{g \in G}|{\rm Fix}(g)| = |G|.$ When $|\Omega| >1,$ we conclude that there s some non-identity element $g \in G$ with $|{\rm Fix}(g)| =0,$ since the identity fixes $|\Omega| >1$ elements. This implies that a proper subgroup $H$ of a finite group $G$ does not meet every conjugacy class, for if $H \neq G$ and $\Omega$ is the set of right cosets of $H$ in $G$, then there is some $x \in G$ such that for each $g \in G,$ we have $Hgx \neq Hg.$ Then $gxg^{-1} \not \in H$ for every choice of $g \in G,$ so no conjugate of $x$ lies in $H$.

Hence if $g \in G \backslash Z(G)$ and $H = C_{G}(g),$ there is some element $x \in G$ such that no conjugate of $x$ commutes with $g.$ Hence a group in which any two conjugacy classes have a pair of mutually commuting elements (one from each class) is Abelian.

Another interesting (and well-known) consequence is that if $A$ is a group of automorphisms of the finite group $G$ and $|A|$ and $|G|$ are relatively prime, then no non-identity element of $A$ fixes every conjugacy class of $G.$ It suffices to consider the case that $A$ is cyclic, and then that $A$ has prime power order. In fact,we may suppose that $A$ has prime order $p$ and that $p$ does not divide $|G|$. Let $a$ be a generator of $A.$ If $a$ fixes every conjugacy class of $G,$ then for each $ x \in G,$ $a$ fixes an element of the conjugacy class of $x,$ since the size of that conjugacy class is $[G:C_{G}(x)],$ hch is primeto $p.$ Hence the subgroup of elements fixed by $a$ meets every conjugacy class of $G,$ so is not proper, a contradiction.

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