What type of object can belong in a group?
So far I've only seen integers and matrices. Can vectors be in a group? Can operators be in a group?
Someone please help stretch my mind a bit as online notes does not help in this respect. Thanks!
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12$\begingroup$ Absolutely anything. Groups are just sets paired with a function that satisfies certain conditions, and sets can contain anything. $\endgroup$ – Edward Jiang Dec 27 '14 at 5:27
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$\begingroup$ So many good answers in such short amount of time! $\endgroup$ – Olórin Dec 27 '14 at 5:33
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$\begingroup$ That reminds me of a billboard advertisement for a plumber service I saw which read (with symbols) that frownie face + plunger = smiley face, which got me wondering what more you could derive from that information. If it was isomorphic to $(\mathbb{Z}_3,+)$ then it turns out plunger + plunger = frownie face $\endgroup$ – JMoravitz Dec 27 '14 at 5:33
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$\begingroup$ @EdwardJiang well, not literally anything. Proper classes can't be elements, for example. At least not in ZF. $\endgroup$ – rschwieb Dec 27 '14 at 5:35
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1$\begingroup$ Ways to flip a mattress can be in a group. $\endgroup$ – Rahul Dec 27 '14 at 5:47
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Any object at all can be in a group. What makes a set a group is the operations with which one endows it, namely, the multiplication map $$G \times G \to G$$ and inverse map $$G \to G$$ satisfying the usual group axioms.
For example, the vector space axioms imply that for any vector space $(\mathbb{V}, +, \cdot)$, the underlying set $\mathbb{V}$ is an (abelian) group under the given addition operation $+$.
And of course, we can think of elements of a matrix group (for example) $G \subseteq GL(n, \mathbb{R})$ as unary operators that act on $\mathbb{R}^n$, e.g., by left multiplication.
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$\begingroup$ I am somewhat puzzle by the downvote, I'd be keen to know what could be improved by what seems to me to be a wholly straightforward answer. $\endgroup$ – Travis Dec 27 '14 at 5:41
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$\begingroup$ I assume whoever downvoted your answer also downvoted mine as well. $\endgroup$ – Clarinetist Dec 27 '14 at 5:41
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As long as you have a nonempty set of stuff, which we call $G$, and some operator $*$ on the set such that for all $a, b, c \in G$,
- $a * b \in G$;
- $a * (b * c) = (a* b) * c$;
- There is an element $e \in G$ such that $e * a = a * e$;
- For all $d \in G$, there is an element $d^{-1} \in G$ such that $d*d^{-1} = d^{-1}*d = e$;
then you have a group.
One of my favorite topics of research is the application of musical set theory (very different from mathematical set theory, by the way), which involves the group structure of the set of twelve musical notes $S$. One would think that musical notes are far removed from math, but for a very long time, 20th-century music theorists have treated $S$ as "essentially being the same" as the integers mod 12, $\mathbb{Z}_{12}$, creating an isomorphism - and then we write $S \cong \mathbb{Z}_{12}$.
answered Dec 27 '14 at 5:29
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$\begingroup$ So e.g. $c*c=a^\flat$ (since $e$ has to be the identity)? $\endgroup$ – Henning Makholm Dec 27 '14 at 13:33
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$\begingroup$ @HenningMakholm - $(S, *)$ works as in $(\mathbb{Z}_{12}, +)$, with $C = 0$, $C\# = 1$, $D = 2$, etc. So for example, $C * C\# = 1$. This has applications in generating tonal matrices. $\endgroup$ – Clarinetist Dec 27 '14 at 17:43
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Any collection of objects can be made into a group. For example, the set consisting of a Teddy bear and an apple can be made into a group as follows:
Let $G$ be the set $\{$ ʕ•͡ᴥ•ʔ, $\}$. Define a binary operation $+: G \times G \to G$ as follows:
ʕ•͡ᴥ•ʔ $+$ $=$ $+$ ʕ•͡ᴥ•ʔ $=$ ʕ•͡ᴥ•ʔ
and
ʕ•͡ᴥ•ʔ $+$ ʕ•͡ᴥ•ʔ $=$
and
$+$ $=$
The definition of a group is merely a blue print that you can use to spot objects that are groups. That elements of groups are denoted by integers and other collections of numbers is just for convenience:
One reason is that it is easier to write $1$ than it is to write ʕ•͡ᴥ•ʔ.
Another reason is that we expect $1$ and $-1$ etc. to behave in certain familiar ways and the notation delivers this intuition to us at no additional cost.
But these two reasons don't mean that we couldn't write $\mathbb Z / 6 \mathbb Z$ entirely using funny symbols of our own choice.
answered Dec 27 '14 at 6:40
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7$\begingroup$ All characters appearing in this work are fictitious. Any resemblance to real persons, living or dead, is purely coincidental. $\endgroup$ – Rudy the Reindeer Dec 27 '14 at 6:41
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5$\begingroup$ A long as real persons have entered the picture, let me add my favorite example of a group: It has exactly one element, namely me. The operation is then uniquely determined. $\endgroup$ – Andreas Blass Dec 27 '14 at 16:56
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$\begingroup$ @Andreas: Evidently you have a strong sense of identity. I'd encourage you to sing "I'm My Own Inverse" but it wouldn't scan. $\endgroup$ – Rahul Dec 28 '14 at 3:48
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Anything can be in a group, so long as an associative binary operation with identity (and inverses for each element) is defined. If $V$ is a vector space, then the set of vectors in $V$ is a group under addition. Nonzero rationals, reals, and complexes form a group under multiplication. Invertible linear operators on a vector space form a group under composition. Bijective functions from a set to itself also form a group under composition. Given an alphabet of symbols $a$, $b$, $c$, ..., the collection of all "words" using these letters and these letters with superscript "$-1$"s forms a group under concatenation.
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3$\begingroup$ It's worth noting that in the last example, we need cancellation - i.e. if we concatenate $ba$ and $a^{-1}c$, we get $baa^{-1}c$ and then we need to be able to cancel $aa^{-1}$ to get $bc$. Otherwise we just have a semigroup $\endgroup$ – Milo Brandt Dec 27 '14 at 5:45
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Groups are phrased in terms of set theory, and in ZF set theory "being in" is a relation that holds between two sets.
So, the elements of any set (being a group has nothing to do with it) can potentially be any sets whatsoever.
This matters little in everyday mathematics, though. It is usually safe to think of elements like "points" in geometry.
