# 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 :

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?

• Why do you use ${\bf Z}/2{\bf Z} =\{\{2k+1\}_{k \in {\bf Z}},\{2k\}_{k \in {\bf Z}}\}$ and not $G = \{-1,+1\}$ ? – Patrick I-Z Nov 14 '13 at 22:15
• @Patrick: You mean, in order to focus on the group rather than the field? You're right. But as my fourth example shows, I still have some interest for the field $\mathbb{Z}/2\mathbb{Z}$, which I certainly write $\mathbb{F}_2$ in this context, and that's maybe why I wrote things the way I did. – Matthieu Romagny Nov 14 '13 at 22:20
• For the symmetric group $S_n$, the outer automorphism group $\textrm{Out}(S_n)=\textrm{Aut}(S_n) /\textrm{Inn}(S_n)$ is trivial (i.e. any automorphism is an inner one) unless $n=6$. In fact, $\textrm{Out}(S_6)=\mathbf{Z}/2 \mathbf{Z}$. – Francesco Polizzi Nov 14 '13 at 23:00
• @MatthieuRomagny: For a proof that without AC you can have such spaces, see this math.SE post. If you have a vector space that has a basis (and hence dimension) with more than one element, then swapping two basis vectors gives you a nontrivial automorphism. If you know that such a basis exists, then you don't need AC to define this automorphism; you need AC to guarantee the existence of bases. However, I believe that existence of nontrivial automorphisms does not imply existence of bases. – Arturo Magidin Nov 15 '13 at 18:32
• youtube.com/watch?v=UTby_e4-Rhg – Noam D. Elkies Jul 3 '14 at 2:19

## 14 Answers

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.

• I would say that in the examples above it's the objects that are nice, not the group of two elements. – Lev Borisov Nov 14 '13 at 23:48
• This is very nice; do you have a reference? – Matthieu Romagny Nov 15 '13 at 10:55
• Do you mean: the only finite group? – Matthieu Romagny Nov 15 '13 at 11:02
• @MatthieuRomagny: No, the theorem holds for arbitrary groups. A can be found in Hatcher's book on algebraic topology. It is very easy, once you have the machinery: the mapping degree induces a group morphism $d: G\to \{\pm 1\}$. Since a fixed point free map $S^n \to S^n$ has degree $(-1)^{n+1}$ and $n$ is even, the kernel of $d$ is trivial and hence $G\leq\{\pm 1\}$. – Johannes Hahn Nov 15 '13 at 14:12
• @JohannesHahn Very nice answer. What other compact manifold $M$ satisfies this property:$Z_{2}$ is the only group with free action on $M$? – Ali Taghavi Nov 9 '14 at 15:39

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

• Which fields $K$ have $K_{*} \simeq K_{+} \times (\mathbb{Z}/2\mathbb{Z})$? – Sam Hopkins Nov 15 '13 at 2:27
• Sam: for starters, $K$ is an exponential field. Take $i$ to be the inclusion of $K_+$ into the product; then $E = \iota \circ i_{K_+}$ (where $\iota$ is the above isomorphism) is an exponential function. According to wikipedia $E$ has trivial image if the characteristic of $K$ is nonzero. $E$ is also injective in this case, but $K_+$ can't be trivial. Therefore $K$ has characteristic zero. Can anyone take it farther than that? – Ibrahim Tencer Nov 15 '13 at 5:50
• I have asked this question separately: mathoverflow.net/questions/148949 – Sam Hopkins Nov 15 '13 at 6:10

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...

• Let me try, the identity $x^2=e$ ? – Denis Serre Nov 3 '14 at 7:40
• @Denis, of course not. The infinite dihedral group $Z/2Z*Z/2Z$ has elements of infinite order. – Anton Klyachko Nov 3 '14 at 7:46
• There are several nice identities - the simplest after a few moments thought is $[x^2, y^2]$. – Sam Nead Nov 9 '14 at 10:51
• @Sam, I bet your single identity forms a basis of identities of the group. (This means that all other identities are consequences of this one.) – Anton Klyachko Nov 9 '14 at 11:12
• Oh, that looks nice - I'll have to think about that. By the way, I sort of want to edit your answer: $Z_2 * Z_2$ is the only non-trivial free product that is two-ended (thus is virtually cyclic) while all others have infinitely many ends (so contain rank two free groups, etc). – Sam Nead Nov 9 '14 at 13:44

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

• That's interesting; I managed to extend this property to all groups whose elements have finite order, and when nontrivial elements have infinite order there are counterexamples built by Higman, Neumann and Neumann in the paper Embedding Theorems for Groups in J. London Math. Soc. 1949. – Matthieu Romagny Nov 17 '13 at 17:03

(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.

• That is terrific! Do you know a reference where I can read more on this? – Matthieu Romagny Nov 4 '14 at 23:57
• So I guess that the correct reference should be Gromov's "Asymptotic invariants of infinite groups" but it is ~300 pages and I haven't read it. Danny gave a talk at Cornell and in the first ~15 mins he covers the density model and gives a sketch of this result. It's available here: cornell.edu/video/danny-calegari-random-groups-diamonds-glass – Mark Bell Nov 5 '14 at 7:50

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.

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 $\mathbb{Z}/2$.

• Do you have a reference for this? – Ivan Di Liberti Aug 6 '18 at 23:14
• Sorry, I can't find one now. It's pretty easy to prove. – John Baez Aug 8 '18 at 3:39
• In higher category theory there are results as described here: golem.ph.utexas.edu/category/2011/11/… – David Corfield Aug 8 '18 at 8:36
• Can you tell what is the generator of this group? If you're thinking about the functor $F$ that takes a category to its opposite, then I'm not sure it's a functor $F:\mathrm{Cat}\to \mathrm{Cat}$. The problem is that $F$ maps $\mathrm{Fun}(C,D)$ into $\mathrm{Fun}(C^\circ,D^\circ)^\circ$ rather than into $\mathrm{Fun}(C^\circ,D^\circ)$. – Matthieu Romagny Sep 28 '18 at 13:09
• Ah, maybe you mean $\mathrm{Cat}$ as a 1-category? In this case $\mathrm{Fun}(C,D)$ is just a set and the problem does not arise. – Matthieu Romagny Sep 28 '18 at 19:10

$Z/2Z$ 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.

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.

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

• Why is that remarkable? – Todd Trimble Nov 3 '14 at 7:52

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.

• Instead of "by the Axiom of Choice, this last factor is a vector space over the field Z/(2), as mentioned before" I would rather write : "this last factor is a vector space over the field Z/(2), and by the Axiom of Choice it has a basis, as mentioned before". – Matthieu Romagny Aug 28 '18 at 13:09

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 $Z/2Z\times Z/2Z$).

• Can you elaborate on this example or maybe give some reference where this result is explained or implied? – bonif Jan 8 '18 at 11:00
• From the functional equation of $\zeta$, a hypothetical non trivial zero off the critical line gives rise to a second one which is its image under the map $s\mapsto 1-\bar{s}$. This map coincides with identity for zeroes on this line. As the Dirichlet coefficients of $\zeta$ are real, the map $s\mapsto\bar{s}$ maps a non trivial zero of zeta to a non trivial zero of $\zeta$. Hence the two maps $s\mapsto s$ and $s\mapsto\bar{s}$ are the symmetries of the multiset of non trivial zeroes of $\zeta$. Adding a hypothetical zero off the line gives rise to the additional... – Sylvain JULIEN Jan 8 '18 at 12:03
• ...symmetry $s\mapsto 1-\bar{s}$ and hence generates a global symmetry group isomorphic to $\Z/2\Z\times\Z/2\Z$. As I said, the absence of zero off the line forces the maps $s\mapsto s$ and $s\mapsto 1-\bar{s}$ to coincide, reducing the global isometry group to a single copy of $\Z/2\Z$ . – Sylvain JULIEN Jan 8 '18 at 12:06
• Has this been published somewhere? Coming from the theoretical physics camp I can see off the top of my head there are some possible immediate physical analogies with this. – bonif Jan 8 '18 at 12:33
• I strongly doubt it. As far as I know, I'm probably the only person to consider RH might be true investigating the symmetries of $\zeta$ or equivalently of the multiset of non trivial zeroes thereof. I'd be glad to be proven wrong though. – Sylvain JULIEN Jan 8 '18 at 13:47

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.

• Every group of exponent 2 has this property. These are not just products of $\mathbb Z/2\mathbb Z$, you also have to close them under subgroups. Again, this doesn't make the two-element group special, as every nontrivial group of exponent 2 generates the class of all of them using products and subgroups. – Emil Jeřábek Nov 2 '14 at 12:39
• @Emil, every group of exponent two is a direct product of several two-element groups (if you believe the Axiom of Choice). – Anton Klyachko Nov 3 '14 at 2:13
• @AntonKlyachko Every group of exponent two is a direct sum of two-element groups. (Finite direct sums coincide with finite direct products, but infinite direct sums need not be infinite direct products.) – Todd Trimble Nov 3 '14 at 7:21
• @Todd, I understand you but I prefer to use slightly different terminology. See this discussion in comments: mathoverflow.net/q/92972/24165 – Anton Klyachko Nov 3 '14 at 7:42
• @AntonKlyachko Okay, thanks for clarifying. The boldfaced "is" in your response to Emil sounded to me like a correction, when in fact what he wrote was already correct according to his word usage (and resonates in other ways as well). – Todd Trimble Nov 3 '14 at 12:13

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 $\mathfrak{X}_r(SL(3,\mathbb{R}))$.