# A group with exactly half of the elements in one conjugacy class

Suppose $$G$$ is a finite group and there exists a conjugacy class $$S$$ in $$G$$ contains exactly $$|G|/2$$ many elements. What can we say about $$G$$? For example if $$G$$ is a Dihedral group $$D_{n}$$ with $$n$$ an odd integer then all reflections in $$D_{n}$$ are in the same class.

To be more general. Let $$A$$ be a finite Abelian group with odd order. Then, the group $$G=A\rtimes(\mathbb Z/2)$$, where $$\varphi:x\mapsto -x$$ is the nontrivial automorphism of order $$2$$, also satisfies our condition.

An easy observation is that for any $$g\in S$$ we have the centralizer $$|C_G(g)|=2$$. So $$C_G(g)=\{e,g\}$$ in this case, which implies $$g^2=e$$. Now the conjugacy action $$G$$ on $$S$$ gives an injective map $$i:G\to S_{|G|/2}$$. The image $$i(g)$$ for $$g\in G$$ has either $$0,1$$ or $$|G|/2$$ fixed points($$g\not\in S\cup\{e\},g\in S,g=e$$).

Since $$i(g)$$ is of order $$2$$ and has a unique fixed point for $$g\in S$$. So it is a product of $$2$$-cycles in $$S_{|G|/2}$$. This implies that $$|G|/2$$ is a odd number.

Now I have no idea how to proceed.

• The even permutations in $S_n$ would be such a conjugacy class, would it not? Also, I always take an opportunity to try to root out the abomination that is the $D_{2n}$ convention for dihedral groups. It acts naturally on $n$ elements and is a subgroup of $S_n$ (at least for $n\geq3$). "But it has $2n$ elements" is irrelevant, that's not how we name $S_n$ or $A_n$, so it shouldn't be how we name $D_n$. Commented Jul 14 at 10:57
• @Arthur The conjugacy classes in $S_n$ is determined by its cycle structure. So all even permutations does not form a conjugacy class. Also thanks for your comment on dihedral groups! Commented Jul 14 at 11:03
• @cybat Of course, you're right, I'm mixing concepts here. And I'm exaggerating a little about the $D_{2n}$ notation. The thing that's actually bad is that there are two competing conventions in active use. I'm having a bit of fun as I do my tiny part in trying to consolidate all mathematicians to a single side. Commented Jul 14 at 12:36
• Since $Z(G)=\cap_{x \in G} C_G(x)$ you see that either $Z(G)=1$, or $Z(G)=\{1,g\}$, in which case $\#Cl_G(g)=1$, whence $|G|=2$. Commented Jul 14 at 14:34
• Another observation: for all $g \in S$ we have $gS = G\setminus S$. Because of cardinality observations, this is equivalent to $gS \cap S =\emptyset$. Suppose by contradiction that there exists $h \in S$ such that $gh \in S$. Since any element in $S$ squares to the identity, we have $ghgh =e$. Multiplying by $g,h$ on the two sides we get $hg =gh$. But the only element in $S$ commuting with $g$ is $g$ itself, from which $g=h$. This is a contradiction since $gh=e$ is not in $S$. This holds if the group has cardinality at least 4. Commented Jul 14 at 22:06

Your construction in paragraph 2 gives all examples!

Now the conjugacy action $$G$$ on $$S$$ gives an injective map $$i:G\to S_{|G|/2}$$.

I follow you up to here but here I think you've made a small extra assumption here. Namely, why is this map injective? You need to know that $$S$$ contains at least two elements $$s_1, s_2$$ to conclude that $$C_G(s_1) \cap C_G(s_2) = \{ e \}$$ so that the action of $$G$$ by conjugation on $$S$$ has trivial kernel. But, to be pedantic, this condition doesn't hold if $$|G| = 2$$.

Fortunately that case is contained in your construction in the second paragraph. So, assuming now that $$|G| > 2$$, your observation about fixed points implies that the action of $$G$$ on $$S$$ makes $$G$$ a Frobenius group; if $$s \in S$$ is any element then $$C_2 = \{ e, s \}$$ is a Frobenius complement, and $$G \setminus S$$ is its Frobenius kernel, which by Frobenius' theorem is a normal subgroup $$K$$ (it's not at all obvious a priori that $$K$$ is closed under multiplication), which as you've shown has odd order. From here it follows that $$G$$ is a semidirect product

$$G \cong K \rtimes C_2$$

where $$K$$ has odd order and $$s \in C_2$$ acts via an automorphism $$\varphi : K \to K$$ of order $$2$$ which fixes only the identity. Conversely, with these hypotheses the action of $$K$$ on $$s$$ by conjugation has trivial stabilizer and so must produce all of $$G \setminus K$$, so $$G$$ has a conjugacy class of size $$\frac{|G|}{2}$$.

So the remaining problem is to classify pairs $$(K, \varphi)$$ of a finite group of odd order and an automorphism $$\varphi$$ as above. Wikipedia claims that a Frobenius group whose Frobenius complement has even order (which is the case here) must have the property that the Frobenius kernel $$K$$ is abelian; that ought to have a direct proof in this special case but I'm afraid I don't see it. Assuming that:

Proposition: Suppose $$K$$ is a finite abelian group of odd order and $$\varphi : K \to K$$ is an automorphism of order $$2$$ with no non-trivial fixed points. Then $$\varphi(k) = -k$$.

Proof. Since $$K$$ has odd order, multiplication by $$2$$ is invertible, so we can divide by $$2$$. This lets us write every $$k \in K$$ as a sum

$$k = \frac{k + \varphi(k)}{2} + \frac{k - \varphi(k)}{2}$$

of an element fixed by $$\varphi$$ and an element negated by $$\varphi$$; this expresses $$K$$ as a direct sum $$K \cong K_1 \oplus K_{-1}$$ of the eigenspaces of $$\varphi$$ (note that we don't need to assume that $$K$$ is a vector space over a finite field!). Since $$\varphi$$ has no nontrivial fixed points, $$K_1 = \{ e \}$$ is trivial, so $$K = K_{-1}$$ as desired. $$\Box$$

• Without assuming $K$ is abelian, the proof involves a little trick: define $f: K\to K$ given by $f(k)=k^{-1}\varphi(k)$, check that it's injective, so it's surjective by finiteness. One can then deduce that $\varphi$ is inversion so $K$ is abelian. See linearalgebras.com/… for example Commented Jul 14 at 22:27
• You don't need Frobenius's theorem to find $K$. In the regular representation, an element of order $2$ in $G$ is a product of $|G|/2$ transpositions, hence lies outside $A_{|G|}$. Intersection with the alternating subgroup then gives a normal subgroup of index $2$. Commented Jul 15 at 14:31