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I am searching for some groups, where it is not so obvious that they are groups.

In the lectures script there are only examples like $\mathbb{Z}$ under addition and other things like that. I don't think that these examples are helpful to unterstand the real properties of a group, when only looking to such trivial examples. I am searching for some more exotic examples, like the power set of a set together with the symmetric difference, or an elliptic curve with its group law.

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Automorphism group of a graph? –  Shahab Apr 15 '13 at 15:03
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@Shahab: Well, it is obviously a group. The clue is in the name... (also, hand-waving, as groups are symmetries and all symmetries are groups, then this is obvious...) –  user1729 Apr 15 '13 at 15:07
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Maybe the fundamental group is one you're looking for - or homotopy groups, more generally? Since it is obvious that you can compose based maps, while it still isn't immediate that they form groups - you need to do some work for that. Apart from that, many subgroups of symmetric groups $S_n$ arise in examples and those can be well-hidden groups, I think. This may not be what you're looking for at all, though. –  HSN Apr 15 '13 at 15:07
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Nice question, Dominic, +1. –  1015 Apr 15 '13 at 18:25
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Sometimes a group will be isomorphic to a group which is obviously a group, but the isomorphism itself isn't obvious. Let's keep this in mind. ("Obvious" is a property of a description of a group, not a property of a group.) –  Qiaochu Yuan Apr 15 '13 at 18:28
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21 Answers

Homological algebra. Let $A,B$ be abelian groups (or more generally objects of an abelian category) and consider the set of isomorphism classes of abelian groups $C$ together with an exact sequence $0 \to B \to C \to A \to 0$ (extensions of $A$ by $B$). It turns out that this set has a canonical group structure (isn't that surprising?!), namely the Baer sum, and that this group is isomorphic to $\mathrm{Ext}^1(A,B)$. This is also quite helpful to classify extensions for specific $A$ and $B$, since $\mathrm{Ext}$ has two long exact sequences. For details, see Weibel's book on homological algebra, Chapter 3. Similarily many obstructions in deformation theories are encoded in certain abelian groups.

Combinatorial game theory. A two-person game is called combinatorial if no chance is involved and the ending condition holds, so that in each case one of the two players wins. Each player has a set of possible moves, each one resulting in a new game. There is a notion of equivalent combinatorial games. It turns out that the equivalence classes of combinatorial games can be made into a (large) group. The zero game $0$ is the game where no moves are available. A move in the sum $G+H$ of two games $G,H$ is just a move in exactly one of $G$ or $H$. The inverse $-G$ of a game $G$ is the one where the possibles moves for the two players are swapped. The equation $G+(-G)=0$ requires a proof. An important subgroup is the class of impartial games, where the same moves are available for both players (or equivalently $G=-G$). This extra structure already suffices to solve many basic combinatorial games, such as Nim. In fact, one the first results in combinatorial game theory is that the (large) group of impartial combinatorial games is isomorphic to the ordinal numbers $\mathbf{On}$ with a certain group law $\oplus$, called the Nim-sum (different from the usual ordinal addition). This identification is given by the nimber. This makes it possible to reduce complicated games to simpler ones, in fact in theory to a trivial one-pile Nim game. Even the restriction to finite ordinal numbers gives an interesting group law on the set of natural numbers $\mathbb{N}$ (see Jyrki's answer). All this can be found in the fantastic book Winning Ways ... by Conway, Berlekamp, Guy, and in Conway's On Numbers and Games. A more formal introduction can be found in this paper by Schleicher, Stoll. There you also learn that (certain) combinatorial games actually constitute a (large) totally ordered field, containing the real numbers as well as the ordinal numbers. You couldn't have guessed this rich structure from their definition, right?

Algebraic topology. If $X$ is a based space, the set of homotopy classes of pointed maps $S^n \to X$ has a group structure; this is the $n$th homotopy group $\pi_n(X)$ of $X$. For $n=1$ the group structure is quite obvious, since we can compose paths and go paths backwards. But at first sight it is not obvious that we can do something like that in higher dimensions. Essentially this comes down to the cogroup structure of $S^n$. There is a nice geometric proof that $\pi_n(X)$ is abelian for $n>1$.

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+1. I especially like the example of games! –  Asal Beag Dubh Apr 15 '13 at 18:33
    
Usually groups are required to be sets. Is it clear that either of these examples is one? I know, completely pedantic, but still. –  ronno Apr 16 '13 at 4:14
    
No, therefore I've also written "(large) group". But the restriction to, say, finite impartial games gives a small group, described in Jyrki's answer. –  Martin Brandenburg Apr 16 '13 at 10:21
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The group of exotic differentiable structures on the $n$-sphere in any given dimension is a group under the operation of connected sum, with the standard sphere being the identity element. Not at all obvious that this forms a group! For example, in dimension 7, this group is isomorphic to $\mathbf{Z}/28$.

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+1. This example is just so mind-blowing. More details can be found at map.mpim-bonn.mpg.de/Exotic_spheres. There it is noted that the inverse is just given by reversing the orientation. How hard is it to prove that it is, in fact, an inverse? –  Martin Brandenburg Apr 15 '13 at 22:14
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I was surprised to learn about the elliptic curve groups. You fix constants $a$ and $b$ take the set $S$ of points on the Riemann sphere (that is, the complex plane plus a point at infinity) of the form $$y^2 = x^3 + ax + b.$$ Then define the sum two points $p_1, p_2$ on this curve by taking the straight line through $p_1$ and $p_2$ and finding the third point $p_3 = \langle x_3, y_3\rangle$ where the line intersects $S$. Then $p_3^{-1} = \langle x_3, -y_3\rangle$ is the group sum of $p_1$ and $p_2$. It's not immediately clear that there is necessarily a point $p_3$, but there is, with suitable treatment of tangents and of the point at infinity. It's not immediately clear that the operation is associative, but it is. The point at infinity is the identity element, and the inverse of the point $\langle x, y\rangle$ is $\langle x, -y\rangle $.

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+1. I think the existence of the third point on a line is not too tricky. Associativity, on the other hand, is a nightmare! (At least if you try to prove it just from this chord-and-tangent definition. As in other examples, there is an "obvious" group $\mathrm{Pic}^0$ lurking in the background, with a non-obvious bijection to the set $S$.) –  Asal Beag Dubh Apr 16 '13 at 16:18
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Notice that the elliptic curve group has a nice interpretation as to its group-properties: just map it to the complex torus, associated with some complex lattice, and show that the map is a homomorphism. So the mysterious operations on elliptic curves are just odinary ones on complex tori. I think this is not entirely anhistorical, as this approach could be traced back to Weierstraß, at least to the functions named after him. Even if this is not true, we can trace it back to the days of Euler, especially to the works of Fagnano. Regards. –  awllower Apr 16 '13 at 16:26
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The sum of the two points $p_1$ and $p_2$ is not the third collinear point but rather its inverse. –  John Bentin Apr 16 '13 at 20:22
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This is my favorite answer to this kind of question, but OP did mention this one explicitly. –  alex.jordan Apr 16 '13 at 21:32
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I find it even less obvious that this works even if the components are finite fields instead of real numbers. –  CodesInChaos Apr 18 '13 at 11:48
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I was surprised the first time I saw the group of unit arithmetic functions under Dirichlet convolution. Arithmetic functions are functions $f:\mathbb{N}\rightarrow F$, where $F$ can be any field (but usually $\mathbb{C}$). The operation is $$(f\star g)(n)=\sum_{d\mid n}f(d)g\left(\frac{n}{d}\right).$$ So, here the identity is the function $$\varepsilon(n)=\left\{\begin{array}{lcl}1&:&n=1\\0&:&\text{otherwise}\end{array}\right.$$while inverses are defined recursively, as described here under "Dirichlet inverse." Note that $1/f(1)$ appears in the definition of the inverses, so we must include only arithmetic functions for which $f(1)$ is invertible in $F$ (this is why we say unit arithmetic functions).

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Should the argument of the first factor of the summand be $d$, not $n$? –  John Bentin Apr 15 '13 at 15:33
    
@JohnBentin Yep - fixed that typo. –  Alexander Gruber Apr 15 '13 at 15:37
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Actually arithmetic functions constitute a ring (with pointwise addition, and Dirichlet convolution as multiplication). Here, $f$ is a unit iff $f(1) \neq 0$ (you should add this to your answer). This is also quite similar (replacing $|$ by $\leq$) to the classification of units in rings of formal power series; only the constant term has to be invertible. –  Martin Brandenburg Apr 15 '13 at 15:47
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The ring can in fact be described in more familiar terms as $F[[x_2,x_3,x_5,x_7,\cdots]]$ (which can be thought of as the ring of formal Dirichlet series), where the formal variables are indexed by primes. The monomials $\prod x_{p_i}^{e_i}$ correspond to the indicator aka characteristic functions of the singleton sets $\{\prod p_i^{e_i}\}$. –  anon Apr 16 '13 at 20:32
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Although I was equally surprised when I saw this example, I became less surprised when I realized that it’s just the (completed) semigroup ring over $F$, where the semigroup in question is the set of positive integers under multiplication. If you use the set of nonnegative integers under addition, you get instead $F[[t]]$. –  Lubin Apr 17 '13 at 16:57
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The Brauer group of a field is not obviously a group in two ways: first it's not obvious that the group is closed under its group operation, and then it's still not obvious that inverses exist.

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+1, nice example! As in the other examples it turns out that this is isomorphic to some "known" group (namely 2nd étale cohomology of $\mathbb{G}_m$). –  Martin Brandenburg Apr 15 '13 at 18:40
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Martin, your "it turns out" is quite anhistoric :-) –  Mariano Suárez-Alvarez Apr 15 '13 at 18:45
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That's right! :) Probably one can say: There are some classes of non-obvious explicit groups which have lead to the development of more general theories which spit out obvious groups for free. –  Martin Brandenburg Apr 15 '13 at 18:47
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1- Fourier analysis: the set of non-vanishing absolutely convergent Fourier series is a group under pointwise multiplication.

The neutral element is the constant function equal to $1$. And product stability follows from Cauchy product. These are straightforward. The existence of inverses is less obvious.

Wiener's lemma: if $f(t)=\sum_{\mathbb{Z}}c_ne^{int}$ is absolutely convergent, i.e. $\sum_{\mathbb{Z}}|c_n|<\infty$, and does not vanish, then $\frac{1}{f(t)}$ is also the sum of an absolutely convergent Fourier series.

This is not that hard either. But it was, and remains, striking. You can find a proof here. Less elementary, but much more interesting, the proof given by Gelfand which raised the interest in Banach algebras. Indeed, the absolutely convergent Fourier series form a unital commutative Banach algebra with spectrum $[0,2\pi]$. More precisely, the characters are the point evaluations $f\longmapsto f(t_0)$. The invertibility of non-vanishing elements is then obvious via Gelfand representation.

2- von Neumann algebras: for a type $\rm{II}_1$ factor von Neumann algebra $M$, i.e. an infinite-dimensional noncommutative probability space, we can make sense of $t\times t$ matrices over $M$ for every real $t>0$. This gives rise to another type $\rm{II}_1$ factor $M^t$.

In their seminal work dating back to the 1930's, Murray and von Neumann introduced the fundamental group of a $\rm{II}_1$ factor $M$ $$ \mathcal{F}(M):=\{t>0\;;\;M^t\simeq M\}. $$ It is not hard, but not obvious per se, to see that this is a subgroup of $(\mathbb{R}^+,\;\cdot\;)$.

One of their striking classification results says that, up to isomorphism, there exists a unique approximately finite-dimensional type $\rm{II}_1$ factor $R$. As a consequence, it follows that $$ \mathcal{F}(R)=\mathbb{R}^+. $$ By Connes' groundbreaking work (1976), any amenable type $\rm{II}_1$ factor is isomorphic to $R$. So these also have fundamental group equal to $\mathbb{R}^+$. This includes the group von Neumann algebra $L(\Gamma)$ of any countable amenable group $\Gamma$ with infinite conjugacy classes.

On the other hand, Connes proved in 1980 that the fundamental group of $L(\Gamma)$ is countable when $\Gamma$ has Kazhdan's property (T). But it remained open for some time whether the fundamental group of a $\rm{II}_1$ factor could be trivial.

In a more recent breakthrough, Popa exhibited in 2001 such examples. In particular, he showed that $$ \mathcal{F}(L(\mathbb{Z}^2\rtimes \rm{SL}(2,\mathbb{Z})))=\{1\}. $$ On the opposite side, he also proved in 2003 that any countable subgroup of $\mathbb{R}^+$ arises as the fundamental group of some type $\rm{II}_1$ factor. For a larger class of such groups and open questions, see these slides by Vaes, another important contributor to the theory.

Finally, note that Voiculescu's free probability theory allowed Radulescu to prove that $$ \mathcal{F}(L(F_\infty))=\mathbb{R}^+ $$ for the free group on a countably infinite number of generators $F_\infty$. Unfortunately, these techniques have not permitted to compute the fundamental group of $L(F_n)$ for the free group on $2\leq n<\infty$ generators. Note that the following puzzling long-standing question remains open: $$ L(F_2)\simeq L(F_3)\;? $$

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That is a nice example +1 –  Dominic Michaelis Apr 15 '13 at 18:27
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May be the groups of nimbers would fit the bill? The underlying set is that of non-negative integers $\mathbb{N}$. The group operation (denoted by $+$) is defined recursively as follows $$ a+b=\operatorname{mex}\left(\{a'+b\mid a'<a\}\cup \{a+b'\mid b'<b\}\right). $$ Here $\operatorname{mex}(S)$ is defined for all proper subsets of $\mathbb{N}$ and means the smallest non-negative integer not in the set $S$ (=Minimum EXcluded number). So $0+0=0$ simply because both sets on the r.h.s. are empty. But then $0+1=\operatorname{mex}(\{0\})=1=1+0$, $1+1=\operatorname{mex}(\{1\})=0$, $0+2=\operatorname{mex}(\{0,1\})=2=2+0$, $1+2=3$, $2+2=0$ et cetera.
The operation is well defined, because the sets on the r.h.s. are obviously finite, and hence proper subsets of $\mathbb{N}$, for all $a,b\in\mathbb{N}$.

Now, it turns out that this operation is just the NIM-sum (addition in base two without carry). That is not entirely obvious even though it isn't exceedingly hard to see either.

It turns out that the sets of the form $S_n:=\{x\in\mathbb{N}\mid x<2^n\}$ are subgroups. Furthermore, if $n$ is a power of two, this set also has a multiplication that turns it into a field. The construction is due to Conway. See this wikipage for more.

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This is just a different approach to the group of combinatorial games in Martin's answer. –  Jyrki Lahtonen Apr 15 '13 at 17:17
    
Well it's the same, but a little bit more detailed, since I mentioned the nim-sum on the natural numbers only very briefly. So it's a nice supplement. –  Martin Brandenburg Apr 15 '13 at 17:28
    
@Martin: Yeah. I was writing this while you were editing (I think). –  Jyrki Lahtonen Apr 15 '13 at 17:31
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So actually $(\mathbb{N},\oplus,\otimes)$ is a a field, isomorphic to $\cup_{n \geq 0} \mathbb{F}_{2^{2^n}}$. Under the Galois correspondence between subfields of $\overline{\mathbb{F}_2}$ and closed subgroups of $\widehat{\mathbb{Z}}$ this becomes the group of $2$-adic integers $\mathbb{Z}_2$. So in some sense there is a correspondence between finite impartial games and $2$-adic integers, perhaps best visualized via their power series expansion. –  Martin Brandenburg Apr 15 '13 at 17:42
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Here are some examples:

  1. The ideal class group of a number field $K$. It is not obvious that this is a group, because for example to be able to invert an ideal requires the definition of an invertible ideal and showing that $\mathfrak{p}\mathfrak{p}^{-1} = \mathcal{O}$. This last part is not trivial and if memory serves me right you need to invoke the Nakayama Lemma.

  2. The fundamental group of a topological space $X$ - not trivial to show that the operation of taking products of loops is associative. When I took such a class my lecturer drew some pretty pictures to show homotopies between $f \ast (g \ast h)$ and $ (f \ast g) \ast h$, but I was not entirely convinced.

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1) On the other hand, there is an obvious monoid of fractional ideals, for every integral domain, and Dedekind domains are characterized by the fact that fractional ideals are invertible. And of course a group is a monoid in which every element is invertible. 2) How do you come up with the homotopy without this picture? Anyway, with enough mathematical maturity, the picture is already the whole proof. –  Martin Brandenburg Apr 15 '13 at 15:16
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@MartinBrandenburg Of course we were then asked to use the picture to write down an explicit homotopy. –  user38268 Apr 15 '13 at 15:26
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@MartinBrandenburg: Having been a TA in a course which included a question like the one BenjaLim describes, I can tell you that writing down the explicit homotopy based on the picture is just a bit harder than it seems--the instructor and I both got the same wrong (discontinuous!) answer on our first attempt. The real danger is that you are so sure it's trivial, you don't bother to think about what you are writing down. –  Charles Staats Apr 16 '13 at 3:12
    
@Martin: Regarding 1) Yes, everything you've said is true. Packaging the material this way, what is not obvious is that the ring of integers in a number field is a Dedekind domain! (I'm currently teaching Algebraic Number Theory I, and on the first day of class I stated a version of that as the first of a short list of the basic theorems of the course.) –  Pete L. Clark Jan 8 at 6:08
    
There is a lot to be said for using Moore paths in defining the fundamental group(oid). A Moore path in $X$ is a pair $(f,r)$ such that $f:[0, \infty) \to X$, $r \in [0, \infty)$, and $f$ is constant on $[r, \infty)$. The composition $(f,r) *(g,s)$ is defined if and only if $f(r)=g(0)$ and is of the form $(h,r+s)$. Composition then gives a category of Moore paths on $X$. –  Ronnie Brown Feb 7 at 15:11
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After some study, it might become clear that the following is a group. But it seemed nonobvious the first time I saw it: the collection of all fractional linear transformations from $\mathbb{C}\cup\{\infty\}$ to itself, with formulas $$z\mapsto\frac{az+b}{cz+d}$$ such that $ad-bc\neq 0$ (in order to guarantee that you don't have a constant map), using composition as the group operation. Firstly, it takes at least a full second to believe that the composition of two such things is another such thing. Secondly, at some point you realize that any one fractional linear transformation has infinitely many representations: $z\mapsto\frac{kaz+kb}{kcz+kd}$. So a lot of your early thoughts on the topic are not 100% correct. Thirdly, associativity is no fun to confirm directly. (Again, it's not immediately obvious, but eventually you can see that this is a factor group of the $2\times 2$ general linear matrix group.)

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The easy way to see that this is a group (and the reason you got downvoted I expect) is that this is easily seen to be a group of matrices because the action of PSL(2,C) matrices on the extended complex plane by möbius transformations is actually just the action of PSL(2,C) on CP^1 by linear transformations, and that is easily seen to be a group if you know the properties of matrix multiplication. –  Vhailor Apr 16 '13 at 22:00
    
@Vhailor Hmm if I had a down vote I guess someone removed it. But anyway, OP seems to be taking an introductory course on groups and is being exposed to really simple examples like $\mathbb{Z}$. I doubt someone in that position finds anything about $\operatorname{PSL}(2,\mathbb{C})$, Möbius transformations, or $\mathbb{C}P^1$ to be obvious or easy. I'm even a little tempted to answer OP with the example of a matrix group, for which associativity is not "obvious" to many students at this level. –  alex.jordan Apr 16 '13 at 23:14
    
at basic level the pretty $SL_2({\mathbb{Z}})\cong{\mathbb{Z}}_2*{\mathbb{Z}}_3$ in the initial crossroads of algebra and topology –  janmarqz Jan 8 at 18:47
    
i mean $PSL_2(\Bbb{Z})$ –  janmarqz Jan 8 at 18:53
    
@janmarqz I think maybe you are interpreting my $a,b,c,d$ to be in $\mathbb{Z}$, but I'm describing a much bigger group. –  alex.jordan Jan 9 at 2:04
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It's not obvious that the collection of integers mod $p$, excluding the coset of $0$, form a group under multiplication. In particular, it's not obvious that inverses exist. You typically use the Euclidean algorithm for that.

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I agree; it’s a familiar example, and an elementary one, but not an obvious one. –  Brian M. Scott Apr 17 '13 at 5:25
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@R.. No, the existence of an inverse is required, and the operation needs to be associative. What about a commutative operation on $\mathbb{Z}/5\mathbb{Z}^{\times}$ called $\odot$ where for all $x,y\neq 1$, $x\odot y=2$? Finite set $\checkmark$. Closed under the operation $\checkmark$. Has a unit $\checkmark$. But fails to have an inverse. For a finite set, if you can show that for each $a$ that the map $x\mapsto ax$ is one-to-one, then that would prove the existence of an inverse. But that's not trivial here either. –  alex.jordan Apr 17 '13 at 15:36
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@R.. In case the point hasn't been made, wouldn't your quick proof prove that $\mathbb{Z}/n\mathbb{Z}^{\times}$ is a group even when $n$ is not prime? –  alex.jordan Apr 17 '13 at 15:41
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What I love about this discussion is that this example really is not obvious, yet it is so familiar to so many people that it feels that way. But when pressed, lots of fine logical people can't give a good reason why inverses should exist mod $p$. Only the ones who remember that the Euclidean algorithm provides the proof. –  alex.jordan Apr 17 '13 at 16:53
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@R.. The statement you may be remembering from your intro to group theory class may have been "Any finite subset $H$ of a group $G$ is a subgroup if it contains the identity of $G$ and is closed under the group operation of $G$." –  Ragib Zaman May 18 '13 at 9:19
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Here is an example that is a bit different, namely one that appears as a subgroup of another group, but where it is not obvious that it is a subgroup.

Let $G$ be a finite group with a proper non-trivial subgroup $H$ such that for all $g\in G\setminus H$ we have $H\cap H^g = \{1\}$ (such an $H$ is called a Frobenius complement in $G$ and if $G$ has a Frobenius complement it is called a Frobenius group).

Define $$N = \left(G\setminus\bigcup_{g\in G}H^g\right)\cup\{1\}$$

Then $N$ is a subgroup of $G$, but I am not aware of a proof of this that does not involve character theory (for a proof see for example Theorem 7.2 in Isaacs' Character Theory of Finite Groups).
($N$ is called the Frobenius kernel of $G$ and it is in fact a normal complement of $H$).

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See also mathoverflow.net/questions/63142/…. –  Martin Brandenburg Apr 16 '13 at 13:29
    
Do you want to assume that $G$ is finite? –  Martin Brandenburg Apr 16 '13 at 13:32
    
@MartinBrandenburg Yes, thank you. –  Tobias Kildetoft Apr 16 '13 at 13:44
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Another example of a subgroup that is not obviously a subgroup is given by Frobenius conjecture, for which the only known proof for relies on the classification of finite simple groups. According to Frobenius conjecture, if $n$ divides the order of $G$ and there are exactly $n$ solutions to $x^n = 1$, then the set of solutions forms a subgroup. Let $G$, $H$ and $N$ be as in your answer. You can prove that $N = \{x \in G: x^{[G:H]} = 1\}$ and that $N$ contains exactly $[G:H]$ elements. So your answer can be seen as a particular case of Frobenius conjecture. –  Mikko Korhonen Apr 16 '13 at 14:18
    
Such a subgroup $H$ is called malnormal in geometric (etc.) group theory. –  user1729 Apr 16 '13 at 16:11
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What about the set of line bundles over a manifold? This forms a group, whether your line bundles are real or complex. The difference between them is also very interesting, as one is 2-torsion, and the other can be torsion-free!

Here the group operation is tensor product: showing it is a closed associative operation is fairly easy. The trivial bundle is the identity. Ah, but what are the inverses?

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This is also quite obvious. Over any ringed space $X$, the line bundles form a $2$-group, with inverses given by $\mathcal{L}^{-1}:=\underline{\hom}(\mathcal{L},\mathcal{O}_X)$. The set of isomorphism classes is therefore a group $\mathrm{Pic}(X)$. You could also use cocycles to identify $\mathrm{Pic}(X) \cong \check{H}^1(X,\mathcal{O}_X^*)$, which is obviously a group since $\mathcal{O}_X^*$ is a sheaf of groups. In other words, for the cocycles $s_{ij}$ of $\mathcal{L}$ the corresponding cocycles for $\mathcal{L}^{-1}$ are $s_{ij}^{-1}$. –  Martin Brandenburg Apr 16 '13 at 10:18
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You have a weird definition of obvious! :) –  user641 Apr 16 '13 at 18:01
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@MartinBrandenburg Who doesn't know that? DUH. –  Alexander Gruber Apr 16 '13 at 18:20
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If you don't like ringed spaces: Standard (continuous) operations of vector spaces carry over to vector bundles; in particular there is a dual vector bundle, and there are tensor products. There is always a canonical map $\underline{\hom}(\mathcal{L},1) \otimes \mathcal{L} \to 1$ (where $1$ is the trivial bundle of rank $1$), and for a line bundle this is obviously an isomorphism. –  Martin Brandenburg Apr 17 '13 at 6:30
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Another example is mentionned here:

Let $G$ be a finite group of order $n$ and $S \subset G$ be any subset. Then $$S^n = \{s_1s_2 \cdots s_n \mid s_1, s_2, \dots, s_n \in S\}$$ is a subgroup of $G$.

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I always found the fact that braid groups are groups at all quite interesting. The elements of the group are all the different braids you can make with, say, $n$ strings. The group operation is concatenation. The identity is the untangled braid. But the fact that inverses exist is not obvious.

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Surely inverses are obvious? Everyone knows that you can untie a tangle of string if it started out untangled... –  user1729 Apr 16 '13 at 9:25
    
This is an obvious group (of course it is an interesting group, but this is not the question here). –  Martin Brandenburg Apr 16 '13 at 10:10
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@user1729: It is obvious you can untangle it; what is not immediately obvious - at least not to me - is that you can do so by concatenating another tangled braid. –  user641 Apr 16 '13 at 18:02
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This is now an obvious example of the fact that obviousness is a function of both the observer and the observed. –  Brian M. Scott Apr 17 '13 at 5:24
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@user1729: It's obvious that if one starts with a set of straight ropes whose ends are tied off, tangles it while leaving the ends tied, and then partitions it into two pieces, one piece will be the inverse of the other. What's not so obvious is that all tangles which can be produced by passing the ends of strings through loops may also be formed as half of the above-described tangle-inverse pair. –  supercat Apr 17 '13 at 22:48
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Exotic example can be rubik's cube group with cube moves.

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Again, this is exotic, but it is obviously a group if you take the view that a group is something which acts on something else (in a nice way)... –  user1729 Apr 15 '13 at 15:13
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Well we better hope it is closed ;) Else we have broken it –  Dominic Michaelis Apr 15 '13 at 15:14
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The rubic cube group is a subgroup of permutations of the stickers. –  Martin Brandenburg Apr 15 '13 at 15:25
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It's called rubik's cube –  wim Apr 15 '13 at 16:33
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This is a non-obvious example in that, in a specific sense (which I will let everyone interested google for) the $4\times 4\times 4$ "is not a group". –  Mariano Suárez-Alvarez Apr 15 '13 at 18:41
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It is not obvious from their definition that KK-groups actually have inverses.

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If $G$ is a Lie group, with multiplication operator $m:G\times G\rightarrow G$ and inverse $i:G\rightarrow G$, then $TG$ is also a Lie group, with multiplication operator $Tm:TG\times TG\rightarrow TG$ and inverse $Ti:TG\rightarrow TG$. To see this, use the obvious notation that for $v_g\in T_g G$ and $h\in G$, $v_g\cdot h = T_g R_h(v_g)$, where $R_h:G\rightarrow G$ is right multiplication by $h$, and similarly for $h\cdot v_g$ (these operations are easily shown to be associative). Then for $g,h\in G$ and $\xi,\zeta\in \mathfrak{g}$, the Lie algebra of $G$, $$T_{(g,\,h)}m(\xi\cdot g,\,\zeta\cdot h) = \xi\cdot g\cdot h + g\cdot\zeta\cdot h = (\xi + \mathrm{Ad}_g\zeta)\cdot gh,$$ and so under the bijection between $\mathfrak{g}\times G$ and $TG$ given by $(\xi,\, g)\mapsto \xi\cdot g$, $TG$ is just $\mathfrak{g}\rtimes_{\mathrm{Ad}}G$, the semidirect product of $\mathfrak{g}$ and $G$ with respect to the adjoint action $\mathrm{Ad}$.

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This is trivial, functors preserving products preserve group objects. –  Martin Brandenburg Apr 17 '13 at 6:32
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Not knowing much category theory, I'm afraid I don't understand this. Can you break it down for me. Are you saying if $F$ is a functor on the category of sets which satisfies $F(X\times Y) = F(X)\times F(Y)$, then $F(G)$ is automatically a group if $G$ is? If so, what is the identity in $F(G)$? –  user17945 Apr 17 '13 at 11:02
    
$F$ should also preserve the terminal object (=empty product). Just look up the definition of a group object in a category with products. –  Martin Brandenburg Apr 17 '13 at 23:13
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Again, this might be due to my naivete regarding category theory, but it just seems to me like you are begging the question somewhat, in that you're assuming –  user17945 Apr 18 '13 at 1:25
    
Let me try again: Ok, just looked up the definition. So to put my question in group object terms, one must have a sensible notion of $T 1:\ast\rightarrow TG$, where $\ast$ is a terminal object in the category of sets, and $1$ the unit map. In other words, you need to show an identity exists on $TG$. Again, this might be due to my naivete regarding category theory, but it just seems to me like you are begging the question somewhat, in that you're assuming this is already true. The existence of an identity for $TG$ wrt multiplication $Tm$ is essentially the proof of the result I mentioned. –  user17945 Apr 18 '13 at 1:31
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Well if you want to get more visual, you can go with symmetry groups and wallpaper groups.

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These are automorphism and stabilizer groups, so the group structure is trivial. –  Martin Brandenburg Apr 15 '13 at 16:45
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That the structure is obvious is presumably irrelevant for OP, they are looking for different situations that are described by groups, and that fact isn't evident for a casual onlooker. –  vonbrand Apr 15 '13 at 17:58
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Many people are surprised to learn that even though there are infinitely many ways to decorate the plane with "symmetrical" periodic designs (single color on a background) that there are 17 different types of such designs. This is seen using a group theory approach and the details can be found, among other places, in Conway and Burgiel's book Symmetry of Things. –  Joseph Malkevitch Apr 17 '13 at 17:09
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Similar to how we turn the natural numbers into the additive group of integers, and the group of integers to the multiplicative group of the rationals, let $A$ be a set with an abelian operation and an identity (abelian monoid). For $A \times A$, declare $(a_1, b_1) \tilde{} (a_2, b_2)$ iff there is a $c \in A$ such that $a_1 + b_2 +c = a_2 + b_1 + c$.

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And why do you think this is not obvious? –  Martin Brandenburg Apr 17 '13 at 23:14
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1) It's obvious in retrospect. Many things are. 2) "Obvious" is, obviously, a relative term. 3) You're starting to sound condescending, especially with your comment to Steve D above. –  jd.r Apr 18 '13 at 0:34
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I think it is not obvious in the following sense. The obvious thing to do is the same definition without the "$c$". But -- nonobviously to just about everyone who has met this definition for the first time -- when the monoid is not cancellative this doesn't even define an equivalence relation. The fact that we can fix this distressing state of affairs so easily by "stabilizing" must have been a brilliant observation when it was first made. –  Pete L. Clark Dec 21 '13 at 22:26
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Two examples:

The set of algebraic numbers is a field and it isn't trivial to prove that their sum and multiplication comply to give two groups, in one had. In the other, consider the Grothendieck group's construction.

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Let $k$ be a field, and let $A$ be a finite type $k$-algebra.

Consider the isomorphism classes of dualizing complexes over $A$.

Given two dualizing complexes $R$ and $S$, define their "product" to be the isomorphism class of the Hochschild cohomology complex of their tensor product over $k$:

$R\cdot S := RHom_{A\otimes_k A}(A,R\otimes_{k} S)$.

Then it is not clear that:

  1. The result is a dualizing complex.
  2. That this operation is associative.
  3. That this operation has an inverse.

Yet, all these turn out to be true. See Section 4 of http://arxiv.org/abs/1401.6678

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