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My lecture note glosses over it really, introduces it and says "well it intuitively makes sense" but I say, nope it doesn't.

Free groups on generators $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ is a group whose elements are words in the symbols $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ subject to the group axioms. The group operation is concatenation.

What do I not understand? Well, to star with, where's the identity? The operation, say I denote it $*$, is $x_1 * x_2=x_1x_2$ yes? How is the identity defined? I mean, $e*x_1=ex_1$ because it's "concatenation" so I cannot conveniently say $e*x_1=x_1$ and ignore the fact I need to "concatenate" it. These are apparently words, symbols not numbers. The inverse doesn't make sense too, $x_1*x_1^{-1}=x_1x_1^{-1}$ and period. Not $x_1*x_1^{-1}=e$. I mean, I don't even know what $e$ is supposed to be in this supposedly group object so I am left puzzled.

I don't see any mathematics here, concatenation, in other words, is just "lining up the symbols in order." It's not like $1 \times 2 \times 10=20$ but $1 \times 2 \times 20=1220$.

And another problem. Doesn't the free group have order infinity? It can't be finite can it? Because, say I start with $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ but it must be closed under concatenation. Well, $x_1*x_2=x_1x_2$ already causes an issue because clearly we just created a new element. A new word $x_1x_2$. Continuing this way, we keep adding the newly created words and reach infinity.

And before someone directs me to it, no, wikipedia's page on free groups didn't help me understand this either.

This bizarre notion is confusing and incomprehensible than ever. Does anyone know the answers to my questions?

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    $\begingroup$ the identity is the "empty" word. $\endgroup$ – Jorge Fernández Hidalgo Apr 10 '16 at 17:07
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    $\begingroup$ you define an equivalence relation on the set of words, and work with the classes. $\endgroup$ – Jorge Fernández Hidalgo Apr 10 '16 at 17:20
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    $\begingroup$ In simple terms, the group operation is defined by: concatenate the words, then repeatedly cancel consecutive pairs $x_i x_i^{-1}$ or $x_i^{-1} x_i$ until you can't cancel any more. For example $(ab)(b^{-1}c) = ac$. Of course, with this approach, you have to prove that you get the same answer no matter how you cancel. $\endgroup$ – Ted Apr 10 '16 at 17:33
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    $\begingroup$ The key phrase you're ignoring in the description you provide is "subject to the group axioms." In other words, a free group $G$ is a group with no relations (i.e. $ab=e$, $a,b \in G$) other than the ones imposed by the group axioms. So it's not exactly true that a free group is just strings of symbols, but it's as close as you can get without losing some group property, hence the name "free." $\endgroup$ – leibnewtz Apr 10 '16 at 19:28
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    $\begingroup$ Please don't accept an answer too quickly John. I'll write something when I have more time, in perhaps a week or so. There's just too many assignments on my shoulders at the moment to give your excellent question the attention it deserves. $\endgroup$ – goblin Apr 11 '16 at 8:33

10 Answers 10

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My lecture note glosses over it really, introduces it and says "well it intuitively makes sense" but I say, nope it doesn't.

Free groups on generators $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ is a group whose elements are words in the symbols $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ subject to the group axioms. The group operation is concatenation.

What do I not understand? Well, to star with, where's the identity?


A free group consists of strings. You form those strings out of an alphabet of symbols of the form $x_i$ and $x_i^{-1}$. The identity is the empty string - the string with no letters in it.


The operation, say I denote it $*$, is $x_1 * x_2=x_1x_2$ yes? How is the identity defined? I mean, $e*x_1=ex_1$ because it's "concatenation" so I cannot conveniently say $e*x_1=x_1$ and ignore the fact I need to "concatenate" it. These are apparently words, symbols not numbers. The inverse doesn't make sense too, $x_1*x_1^{-1}=x_1x_1^{-1}$ and period. Not $x_1*x_1^{-1}=e$. I mean, I don't even know what $e$ is supposed to be in this supposedly group object so I am left puzzled.


They don't do it, but it might help to imagine quotes around the strings. So that '$abc$' * '$de$' = '$abcde$'. Clearly adding nothing to the end of a string gives you back that string: '$xyz^{-1}$' * ' ' = '$xyz^{-1}$'

Don't think of $e$ as the identity because you will be confusing letters in the alphabet of the free group with symbols that represent particular strings made from letters in the alphabet of the free group. I've see at least one book that used the Greek letter $\iota$ (iota) to represent the empty string: $\iota =$ ' '.

For example, when a letter and its inverse are next to each other, they cancel each other out: '$aa^{-1}$' = ' '; or '$aa^{-1}$' = $\iota$.

So, in a free group, you can concatenate two strings and end up with a smaller string.


I don't see any mathematics here, concatenation, in other words, is just "lining up the symbols in order." It's not like $1 \times 2 \times 10=20$ but $1 \times 2 \times 20=1220$.


Yes, this is mathematics. Yes it is not arithmetic. One is much more general than the other.


And another problem. Doesn't the free group have order infinity? It can't be finite can it? Because, say I start with $x_1,...,x_m,x_1^{-1},...,x_m^{-1}$ but it must be closed under concatenation. Well, $x_1*x_2=x_1x_2$ already causes an issue because clearly we just created a new element. A new word $x_1x_2$. Continuing this way, we keep adding the newly created words and reach infinity.


Every member of a free group, except for the empty string $\iota$ = ' ', is of infinite order. No matter how many $x$'s are in the (finite) string '$xxx \dots x$', you are never going to end up with the empty string.


Now I need to contradict myself. I said that a free group consists of strings. That is not true. Because of the existence of inverse letters, $x_i^{-1}$, some strings can be simplified (reduced). It takes a lot of effort, but it can be shown that, no matter what order of operations you use to reduce a string, you will always end up with the same answer. The members of a free group consists of equivalence classes of the form $[xyz]$ where $[xyz]$ is the set of all strings that reduce to the same string as 'xyz'does.

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    $\begingroup$ Regarding the "it is not arithmetic" comment: arithmetic addition is actually the same as this string concatenation operation over an alphabet of size 1. 3 + 4 + 5 would be 111 * 1111 * 11111 giving 111111111111 or 12. 3^-1 is -3, and the empty string is the identity 0. $\endgroup$ – M.M Apr 11 '16 at 4:01
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    $\begingroup$ @M.M - It CAN be used to do arithmetic. But that isn't what they are trying to do with it at this time. $\endgroup$ – steven gregory Apr 11 '16 at 5:08
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    $\begingroup$ @M.M, precisely! $\mathbb Z$ is a free group over one element - infinite cyclic group. $\endgroup$ – Ennar Apr 11 '16 at 8:13
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    $\begingroup$ This one sentence helped me immensely: "So, in a free group, you can concatenate two strings and end up with a smaller string." $\endgroup$ – Todd Wilcox Apr 11 '16 at 15:46
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In algebra, a "free X" (where X is the name of an algebraic structure, such as "group", "semigroup", "magma", etc.) is basically an algebraic structure in which two expressions are equal if and only if the definition of X says that they must be.

For example, the simplest kind of algebraic structure is a magma: a set $S$ equipped with an arbitrary binary operator $\cdot$, which is only required to be closed (i.e. that if $a$ and $b$ are members of $S$, then $a \cdot b$ exists and is a member of $S$). A magma is not required to satisfy any algebraic identities except the trivial one (that two identically written expressions are equal), and a free magma is one that indeed doesn't — two expressions in a free magma (using only the given generators as variables) are equal if and only if they're written in exactly the same way.

Formally, a free magma can be defined as follows:

  • The generators are (distinct) elements of the magma.
  • For any two elements $a$ and $b$ of the magma, the expression $(a \cdot b)$ is an element of the magma.
  • No two expressions constructed in this way are equal, unless they are written identically. In other words, the expression $(a \cdot b)$ is never equal to a generator, and is only equal to the expression $(c \cdot d)$ when $a = c$ and $b = d$.
  • The magma contains no other elements.

(And yes, this definition indeed implies that any (non-empty) free magma must have infinitely many elements.)

Similarly, a semigroup is a magma that obeys the associative law $(a \cdot b) \cdot c = a \cdot (b \cdot c)$, and a free semigroup can be constructed by taking a free magma and identifying any two expressions that can be made identical by applying the associative law.

It turns out that, because of the associative property, expressions in a semigroup can always be written unambiguously without parentheses, simply as ordered sequences of elements separated by the binary operator: any method of reinserting the parentheses will yield equivalent results under the associative law. Thus, another way of constructing a free semigroup is as the set of all finite-length $n$-tuples of the generators, with concatenation as the semigroup operator:

  • For all generators $x$, the 1-tuple $(x)$ is an element of the semigroup.
  • For any two tuples $a = (a_1, \dots, a_m)$ and $b = (b_1, \dots, b_n)$ in the semigroup, the concatenated tuple $a \cdot b = (a_1, \dots, a_m, b_1, \dots, b_n)$ is an element of the semigroup.
  • No two tuples in the semigroup are equal, unless they have the same length and exactly the same elements.
  • The semigroup contains no other elements.

It's common to also include the empty tuple $\epsilon = ()$, which then acts as an identity element, in the free semigroup. Technically, though, what we then have is a free monoid, a monoid being a semigroup with an identity element $\epsilon$ that satisfies the additional law $a \cdot \epsilon = \epsilon \cdot a = a$ for all $a$.

Finally, a group is a monoid with inverses: that is, every element $x$ of a group has an inverse element $x^{-1}$ such that $x \cdot x^{-1} = x^{-1} \cdot x = \epsilon$. Again, a free group can be obtained from a free monoid by adding inverses for the generators, and identifying any expressions that can be made identical by repeatedly removing pairs of adjacent elements that are inverses of each other.

Similarly, we might e.g. consider free abelian groups, constructed by taking a free group and identifying any expressions that can be made identical by applying the commutative law $a \cdot b = b \cdot a$ to reorder their elements (and removing inverse pairs); these turn out to be representable as (finitely supported) functions from the set of generators to the integers, where the value of the function giving the number of times that generator appears in the expression (with inverses counted as $-1$ appearance).

Or we could skip groups entirely, and instead consider e.g. free commutative monoids; these can be represented in the same way as free groups, except that, since we don't have inverses, no generator can appear a negative number of times in an expression. (A free commutative semigroup is the same as a free commutative monoid, except that it has no identity.)

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A free group is indeed infinite, I don't see any problem here.

For the inverse, note that what you said is not the complete definition of the free group. You have to take the words on your generators with the product of concatenation, that's ok, and then you have to quotient by the equivalence relation generated by $x_ix_i^{-1} \sim x_i^{-1}x_i \sim \varepsilon$, where $\varepsilon$ is the empty word, which is the neutral element of the free group.

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    $\begingroup$ The free group on 0 generators is finite, though $\endgroup$ – Hagen von Eitzen Apr 11 '16 at 14:22
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For simplicity, we'll talk about two generators. Any group with two generators can be thought of as being determined by an equivalence relation. Every element of produced by multiplying $a$'s and $b$'s together in some order, and then canceling the things that are equivalent to $e_G$. For example, in $D_4$ we have that the group is generated by two elements, $a,b$ that satisfy $a^4=e$, $b^2=e$, and $abab=e$. Given some string of multiplied symbols, $aaababbaaaababa$ we can shorthand this as $a^3bab^2a^4baba$ and then simplify with our cancellation rules to get $a^2b$.

A free group is a group where the only cancellation rule is $aa^{-1}=e$ for every $a$ in the group. If the generating element has finite order, we can write $a^{-1}=a^{n-1}$ where $n$ is the order of $a$, but otherwise we need to introduce a new symbol to represent it. So, a free group (on two elements) has elements that are strings of $a$, $a^{-1}$, $b$, $b^{-1}$ such that no consecutive pair are inverses of each other. The general case is similar with however many generators you want.

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The definition you were given is pretty terse; let me see if I can give a more thorough one.

(Note that this is one definition of a free group, which doesn't use equivalence classes. The other definition, which does use equivalence classes, has a certain nice theoretical property, but it's harder to understand if you're not familiar with equivalence classes.)

Suppose you have a set $S$. Then the free group on $S$ is defined by the following properties:

  • An element of this group consists of a list of zero or more elements of $S$, each of which is or is not marked by the inverse operator $^{-1}$. As a caveat, an element of $S$ is not allowed to appear twice in a row, first with the inverse operator and then without it, or vice versa. (But an element of $S$ is allowed to appear twice in a row if it has the inverse operator both times, or does not have it either time.) For clarity, I'll write lists inside of brackets, separated by commas.
  • The identity element is the list with no elements: $[]$.
  • Taking the inverse of an element consists of
    • reversing the list,
    • adding an $^{-1}$ mark to each element which originally did not have one, and
    • removing the $^{-1}$ mark from each element which originally did have one.
  • Multiplying two elements together consists of
    • concatenating the lists,
    • removing any occurrence of an element of $S$ twice in a row, once without the $^{-1}$ mark and once with it, and
    • repeating the above step until it no longer applies.

Here are some elements of the free group on the set $\{a, b, c\}$:

  • $[]$
  • $[a]$
  • $[a^{-1}]$
  • $[a,a]$
  • $[a,b,a^{-1},b^{-1}]$

The list $[a,a^{-1}]$ is not an element of the group, because it violates the caveat.

Some examples of the operations:

  • To take the inverse of $[a, b, c, a^{-1}]$, first we reverse the order of the list, getting $[a^{-1}, c, b, a]$. Then we add or remove the $^{-1}$ mark from each element, giving us the final answer, which is $[a, c^{-1}, b^{-1}, a^{-1}]$.
  • To multiply $[a, b, c, a^{-1}]$ and $[a, c^{-1}, a^{-1}]$, first we concatenate the lists, giving us $[a, b, c, a^{-1}, a, c^{-1}, a^{-1}]$. Then we apply the cancellation rule to the $a^{-1}, a$ bit in the middle, giving us $[a, b, c, c^{-1}, a^{-1}]$. Then we apply the cancellation rule again to the $c, c^{-1}$ bit, giving us $[a, b, a^{-1}]$. Now we're finally finished.

Now, to answer your individual questions:

Where's the identity element? It's the empty list, $[]$. This works as the identity element, because if you concatenate the empty list with another list, such as $[a]$, you get back that other list.

How does concatenation of $[x]$ and $[x^{-1}]$ give me the empty word? It doesn't. To multiply these two elements of the free group, you first concatenate them and then apply the cancellation rule, and that gives you the empty word.

Don't free groups have infinite order? Yes, usually. If $S$ is not the empty set, then the free group on $S$ has infinitely many elements. (The free group on the empty set has just one element, the identity element.)

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You're making strings.

What do I not understand? Well, to star with, where's the identity? The operation, say I denote it ∗∗, is x1∗x2=x1x2x1∗x2=x1x2 yes? How is the identity defined?

Identity is empty string. Not space, but empty string.

Here is example.

$ S = \{a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z,a^{-1},b^{-1},c^{-1},d^{-1},e^{-1},f^{-1},g^{-1},h^{-1},i^{-1},j^{-1},k^{-1},l^{-1},m^{-1},n^{-1},o^{-1},p^{-1},q^{-1},r^{-1},s^{-1},t^{-1},u^{-1},v^{-1},w^{-1},x^{-1},y^{-1},z^{-1}\}$

And here are some example expressions(let $\pi$ be identity):

$a = a$

$aa = a * a$

$aaa = a * a * a$

$aaaa = a * a * a * a$

$abbb = a * b * b * b$

$abbb = a * b * c * c^{-1} * b * b$

$abbb = a * b * b * b * \pi * \pi * \pi * \pi $

Doesn't the free group have order infinity? It can't be finite can it?

For any positive integer $n$, there exists a string with more characters than $n$.

Can it be finite? Maybe. Put some restrictions on it.

I don't see any mathematics here, concatenation, in other words, is just "lining up the symbols in order." It's not like 1×2×10=201×2×10=20 but 1×2×20=12201×2×20=1220.

If that's what you think math is, then you're right: There's no math here.

That, however, is not what most of us think math is limited to.

This bizarre notion is confusing and incomprehensible than ever. Does anyone know the answers to my questions?

Just take a deep breath. You'll get it. Somebody else can understand this, and you, like them, are a person. This means you also can understand this. Just give it some time.

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This notion is not bizarre but actually useful.

  1. You can construct a group out of any set. For example, you can add orange and apple in the free group generated by $\{\text{orange,apple}\}$.

  2. The free group $F(S)$ generated by the set $S$ is uniquely characterized as the following property: For every group $H$ and a set function $f\colon S\to H$, there exists a unique homomorphism $F\colon F(S)\to H$ such that $F|S=f$. Therefore, $S$ behaves just like the bases in vector spaces.

  3. Any group is a quotient of a free group. Given any group $G$, consider the free group $F(|G|)$, where $|G|$ is just the underlying set of $G$. Then, there is an epimorphism from $F(|G|)\to G$. Hence, $G$ is a quotient group of $F(|G|)$ by the isomorphism theorems.

  4. You can hence describe any group by presentation. For example, any cyclic group of order $n$ is $\langle a|a^n=1\rangle$.

Finally, the concatenation is not that weird. After all, how would you construct a group out of a bunch of meaningless symbols?

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The set of reduced words on set $X = \{ x_i : i \in I \}$ is the collection of words $w$ on the alphabet with symbols $x_i, x_i^{-1}$ such that there is no $x_i x_i^{-1}$ and no $x_i^{-1} x_i$ in the word. (Here, $x_i^{-1}$ is just a formal inverse, simply a new symbol which doesn't already appear in $X$, and which we label $x_i^{-1}$.)

One may define the free group on set $X = \{ x_i : i \in I \}$ to be a certain subgroup of the symmetric group on the collection of reduced words, namely the one generated by $\chi_{x_i}: x_i^{-1} w \mapsto w$ and $w \mapsto x_i w$ if $w$ does not start with $x_i^{-1}$.

This has the benefit of being a totally concrete and obviously well-defined definition; the work then has to be done to show that it is isomorphic to the more usual definitions.

The identity element is then the identity permutation. The group operation is composition of permutations.

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Most important is the property of a free group: In a free group there are no relationships between elements other than those that you can prove by the axioms of a group.

For example, $aa^{-1} = 1$ follows from the group axioms and is true in every group, and therefore in a free group as well.

ab = 1 is possible and will be true in some groups, but it doesn't follow from the group axioms, and therefore isn't true in a free group unless a and b were formed from the generators in such a way that in the product ab all the multiplications cancel each other out (that is ab contains a generator followed by its inverse or vice versa, you can remove it, then you have again a generator followed by its inverse and so on until you are left with 1). Nothing like $Z_{12}$ where 5 + 7 = 0 (or (1+1+1+1+1)+(1+1+1+1+1+1+1) = 0) can happen in a free group.

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A reduced word $w$ in $F$ is of the form $$w=x_{i_1}^{\varepsilon_1}x_{i_2}^{\varepsilon_2}\cdots x_{i_r}^{\varepsilon_r},$$ with $$x_{s+1}^{\varepsilon_{s+1}}\neq x_s^{-\varepsilon_s},$$ where the $x_s$ are the generators and the $\varepsilon_s=\pm1$

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