Fantastic properties of Z/2Z

Question

Recently I gave a lecture to master's students about some nice properties of the group with two elements $\mathbb{Z}/2\mathbb{Z}$. Typically, I wanted to present simple, natural situations where the only group satisfying the given constraints is $\mathbb{Z}/2\mathbb{Z}$ (also $\mathbb{Z}/2\mathbb{Z}$ as a ring or as a field could qualify, but I'd prefer to stick to the group if possible). Here are some examples of theorems that I proved to the students :

1. Let $G$ be a nontrivial group with trivial automorphism group. Then $G$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.
2. Let $G$ be a nontrivial quotient of the symmetric group on $n>4$ letters (nontrivial meaning here different from 1 and the symmetric group itself). Then $G$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.
3. Let $k$ be an algebraically closed field and let $k_0$ be a subfield such that $k/k_0$ is finite. Then $k/k_0$ is Galois and $G=\text{Gal}(k/k_0)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$. (Moreover $k$ has characteristic $0$ and $k=k_0(i)$ where $i^2=-1$.) This is a theorem of Emil Artin and I actually did not prove it because my students did not have enough background in field theory.
4. Let $k$ be a field with the following property: there exists a $k$-vector space $E$ of finite dimension $n>1$ and an isomorphism $E\simeq E^*$ between the space and its linear dual which does not depend on the choice of a basis, i.e. is invariant under $\text{GL}(E)$. Then $k=\mathbb{Z}/2\mathbb{Z}$, $n=2$ and the isomorphism $E\simeq E^*$ corresponds to the nondegenerate bilinear form given by the determinant.

I am looking for some more fantastic apparitions of $\mathbb{Z}/2\mathbb{Z}$. Do you know some?

Answer 1

A nice theorem is: $\{\pm 1\}$ is the only group that can act freely on a sphere of even dimension. In contrast: There are infinitely many groups acting freely on every odd-dimensional sphere.

Answer 2

This is more of a joke than a serious example. Let $K$ be a field, $K_+$ its additive group, and $K_*$ its multiplicative group. Thus $\mathbb{R}_*\cong \mathbb{R}_+\times (\mathbb{Z}/2\mathbb{Z})$. What fields have the "opposite" property, that is, $K_+\cong K_*\times (\mathbb{Z}/2\mathbb{Z})$? Answer: only $\mathbb{Z}/2\mathbb{Z}$.

Answer 3

It is the only non-trivial group whose free square ($G*G$) satisfies a non-trivial identity (or is solvable, or is amenable...)

Edit (Nov 9, 2014), suggested by Sam Nead

... or is virtually cyclic, or is two-ended, or contains no nonabelian free subgroups...

Answer 4

(In the density model) a random group is either $\mathbb{Z} / 2\mathbb{Z}$ or the trivial group. (I learnt this from Danny Calegari but I believe this is origionally due to Gromov.)

More precisely, fix $k \geq 2$ and $D \geq 1/2$. For each $n$ let $X_n$ be the set of reduced words in $x_1^{\pm 1}, \ldots, x_k^{\pm 1}$.

Now consider the group: $$ G_n = \langle x_1, \ldots, x_k | r_1, \ldots, r_l \rangle $$ where each $r_i$ is chosen randomly (independently) from $X_n$ and $l = |X_n|^{D} \approx (2k - 1)^{nD}$. Then $\mathbb{P}(G_n \cong \mathbb{Z / 2Z} \textrm{ or } \mathbb{1}) \to 1$ as $n \to \infty$.

The proof of this is quite slick and relies on the "Random Pigeon Hole Principle": That because of the number of pair of relators are so large you are likely to get a number of almost equal relators, that differ at a single letter, which kills off a generator.

Answer 5

Just a comment. You are probably looking for "fantastic properties" stated in terms of group theory, "if an (abstract) group has such and such properties then it is $Z/2Z$". However various representations of this group also have "fantastic properties". I mean first of all the group of automorphisms $C/R$ which consists of complex conjugation and identity. Several important subjects, like "real algebraic geometry" are based on these properties. Once I was strongly tempted to call my paper "Some applications of representation theory of the group of 2 elements", but my co-author convinced me that this title would be too radical.

The paper I mentioned is: Wronski map and Grassmannians of real codimension 2 subspaces, Computational Methods and Function Theory, 1 (2001) 1-25.

Answer 6

It is the only finite group with exactly two conjugacy classes and it is the only non-trivial group with trivial automorphism group.

EDIT: For the second one, it turns out you need the axiom of choice!

Answer 7

An example from differential geometry: If M is a compact, even-dimensional Riemannian manifold of positive sectional curvature, then its fundamental group is either 1 or Z/2Z.

equivalently: a group acting freely on a compact, even-dimensional Riemannian manifold of positive sectional curvature is 1 or Z/2Z.

This is a variant of Synge's Theorem.

Answer 8

The automorphism group of the category of categories is $\mathbb{Z}/2$.

That is, the group of invertible functors $F: \mathrm{Cat} \to \mathrm{Cat}$ is $\mathbb{Z}/2$.

Answer 9

$Z/2Z$ is abelian group which is canonically isomorphic to its dual abelian group. (Since dual group is again $Z/2Z$ and there is only one automorphism of $Z/2Z$).

Actually it is not the only group with this property, and another one is again related to $Z/2Z$. It is the group $Z/2Z \oplus Z/2Z$: the dual group is group of characters, the kernel of each character contains two elements - identity and another one , so we can set a bijection: character <-> non-identity element in the kernel. This example is the same as item 4, in the original question.

Answer 10

Isn't it remarkable that the fundamental group of the special orthogonal group $SO_n$ is $\mathbb Z/2\mathbb Z$ for $n\ge3$ ?

Answer 11

The largest group which has embeddings into every nonabelian finite simple group is the Cartesian square of this group.

Let $G$ be a group. The holomorph $Hol(G) = G \rtimes Aut(G)$ can be regarded as a subgroup of the symmetric group $S_{|G|}$, by considering the functions $f: G \to G$ sending $x \in G$ to $x^{\alpha} \cdot g$, for $g \in G$ and $\alpha \in Aut(G)$.
This is never a self-normalizing subgroup of $S_{|G|}$ when $G$ is nonabelian, because any anti-automorphism of $G$, including $x \to x^{-1}$, normalizes it but is not in it. So $N_{ S_{ |G|} } (Hol(G))/Hol(G)$ always has a subgroup of order 2 in this case.

The only nontrivial groups lacking proper subgroups are the cyclic groups of prime order. Among these, only a cyclic group of order 2 can be isomorphic to the unique minimal subgroup of two nonisomorphic finite groups of the same order (a cyclic group and a quaternion group, when the order is a power of 2).

Let $G$ be a nonabelian group in which all subgroups are normal. Then $G$ is isomorphic to the Cartesian product of a abelian torsion group in which all elements have odd order, the quaternion group $Q_{8}$, and a group of exponent 2 (by the Axiom of Choice, this last factor is a vector space over the field $\mathbb{Z}/(2)$, as mentioned before).

Notice how the group of exponent 2 above did not need to be required to be abelian? If $G$ is a group and all cyclic subgroups of $G$ are of order 2, it immediately follows that $G$ is abelian.

Answer 12

RH holds if and only if the group of isometries of the complex plane that preserve globally the multiset of non-trivial zeroes of the Riemann Zeta function is isomorphic to $Z/2Z$ (otherwise, it would be isomorphic to $Z/2Z\times Z/2Z$).

Answer 13

Edit after Emil's comment (so my answer is not really good then): It's a group (or any product of it) where addition and substraction seen as a binary (that is, $- = + \circ (Id, i)$ where $i$ is the inverse map) actually coincides. Note that this also means that it's a group where $-$ is actually associative.

Answer 14

The central multiplicative action of $(\mathbb{Z}/2\mathbb{Z})^r$ on the cohomology ring $H^∗(\mathfrak{X}_r(SU(2)))$ is trivial, where $\mathfrak{X}_r(SU(2))$ is the character variety $SU(2)^r/SU(2)$. This ultimately allows for the computation of the Poincaré polynomial of the real moduli spaces $\mathfrak{X}_r(SL(3,\mathbb{R}))$.