Fantastic properties of Z/2Z 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 :


*

*Let $G$ be a nontrivial group with trivial automorphism group. Then $G$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

*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 $\mathbb{Z}/2\mathbb{Z}$.

*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.

*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?
 A: 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.
A: 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 $\mathbb{Z}/2$.
A: $Z/2Z$ 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. 
A: Isn't it remarkable that the fundamental group of the special orthogonal group $SO_n$ is $\mathbb Z/2\mathbb Z$ for $n\ge3$ ?
A: 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.
A: 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 $\mathbb{Z}/2\mathbb{Z}$. 
A: 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.
A: 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 $Z/2Z\times Z/2Z$).
A: 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... 
A: 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.
A: 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!
A: (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.
A: 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.
A: 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 $\mathfrak{X}_r(SL(3,\mathbb{R}))$.
