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I'm trying to prove that every nonidentity element in a free group $F$ has infinite order. I'm really new on free groups and I found this subject really strange I couldn't understand it very well yet, I need a help or a hint to solve this question.

I attached the definition of free groups:

Thanks in advance

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    $\begingroup$ Please tell us which one of the several equivalent definitions of free group you have. $\endgroup$ – DonAntonio May 7 '13 at 22:37
  • $\begingroup$ @DonAntonio I'm using the definition of Hungerford's book, I will edit my question, sorry about the delay $\endgroup$ – user42912 May 8 '13 at 2:53
  • $\begingroup$ I attached a picture of the definition, because I thought it very big to type. Is it ok? $\endgroup$ – user42912 May 8 '13 at 2:58
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    $\begingroup$ Then all of the answers have something for you, @user42912: they all seem to be relying on that definition, more or less. $\endgroup$ – DonAntonio May 8 '13 at 6:10
  • $\begingroup$ A short proof is in math.stackexchange.com/questions/2801683/… $\endgroup$ – user555729 May 30 '18 at 11:56
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A free group $F$ on a set of symbols $S$ is the set of all possible reduced words formed from the symbols and their formal inverses, with multiplication given by concatenation and reduction. A word is reduced if no symbol appears next to its formal inverse in the word.

For a simple example, let's consider the free group on the $2$ symbols $x$ and $y$. The following is a list of all possible reduced words of length less than or equal to $2$:

$$\varnothing,x,y,x^{-1},y^{-1},xx,xy,xy^{-1},yx,yy,yx^{-1},x^{-1}y,x^{-1}x^{-1},x^{-1}y^{-1},y^{-1}x,y^{-1}x^{-1},y^{-1}y^{-1}$$

where $\varnothing$ is the empty word. We multiply by concatenating words, and reducing, that is, removing the subwords $xx^{-1}$, $yy^{-1}$, $x^{-1}x$, and $y^{-1}y$. For example, we have the products $$xy^{-1}\cdot xy=xy^{-1}xy\\xy^{-1}\cdot yy=xy^{-1}yy=xy$$

This product satisfies the group axioms, and we see that $\varnothing$ is the identity of the group.

I recommend playing around a bit with arbitrary elements and their concatenations to get a feel for how elements multiply. Specifically, try to write down an element with finite order, and see what happens.

For a rigorous proof, suppose that for an arbitrary set $S$, we have a reduced word $s_1^{e_1}\ldots s_l^{e_l}$ with finite order, say $n>1$. Here the $s_i\in S$ and $e_i=1$ or $-1$. Then we have

$$s_1^{e_1}\ldots s_l^{e_l}s_1^{e_1}\ldots s_l^{e_l}(\ldots) s_1^{e_1}\ldots s_l^{e_l}s_1^{e_1}\ldots s_l^{e_l}=\varnothing$$

where we write the word next to itself $n$ times. Notice it follows that $s_1^{e_1}s_l^{e_l}=\varnothing$, and that $s_2^{e_2}\ldots s_{l-1}^{e_{l-1}}$ also has order $n$. We continue this process until we reach the middle of the word, noting that $s_i^{e_i}s_{l-i+1}^{e_{l-i+1}}=\varnothing$ for all $i=1,2,\ldots,\lfloor\frac{l+1}{2}\rfloor$. Then we can cancel from the inside out, showing that our original word was in fact $\varnothing$, a contradiction as $n>1$.

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  • $\begingroup$ What you say is a free group is not very accurate...in fact, the very same can be said about all groups. $\endgroup$ – DonAntonio May 7 '13 at 23:14
  • $\begingroup$ Why $s_1^{e_1}s_l^{e_l}=\emptyset$? $\endgroup$ – user42912 May 7 '14 at 14:15
  • $\begingroup$ @user42912: Since $s_1^{e_1}\ldots s_l^{e_l}$ is reduced, but its $n$-fold product is not (it is the empty word), we must be able to reduce the $n$-fold product between some pair of symbols. The only pair of symbols that appear together in the $n$-fold product that are not together in the original word is the pair $s_1^{e_1}s_l^{e_l}$. It follows that $s_1=s_l$ and $e_1=-e_l$. $\endgroup$ – Jared May 7 '14 at 17:23
  • $\begingroup$ Maybe you meant $s_l^{e_l}s_1^{e^1}$ instead of $s_1^{e^1}s_l^{e_l}$? $\endgroup$ – user42912 May 7 '14 at 18:31
  • $\begingroup$ @user42912: Either one. Any element of a group commutes with its inverse, and the point of what I've written is to communicate that $s_1^{e_1}$ and $s_l^{e_l}$ are inverses of each other. $\endgroup$ – Jared May 7 '14 at 18:54
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Intuitively, a free group $F$ on a set of generators $S$ is the most economic way of turning $S$ into a group without introducing any relations among the generators other than what is absolutely necessary in order to obtain a group. If $g\in F$ is a non-identity element, then it can't satisfy the relation $g^n=e$, for $n>0$, since that would be an extra relation which is not necessary for $S$ to be turned into a group. So, all non-identity elements have infinite order.

To prove it, here is a first step. Let $g\in F$. Then $g$ can be represented as a word in the generators $S$ and their formal inverses, and assume it's written in such a way the word can't be reduced (i.e., no adjacent $s$ and $s^{-1}$). Suppose the first letter is $s$. If the last letter is not $s^{-1}$, then the length of any power $g^n$ will be at least the length of the word you started with (since the only way for it to become shorter is for the first and last letter to cancel, which, under the assumption, can't happen). Thus, the only way for any power of $g$ to become the identity (i.e., a word of length $0$), is for the original word to have length $0$, i.e., $g=e$. Now analyze what happens if the first and last letter of $g$ are inverses of each other.

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Let $g \in F$ be a non empty reduced word. Take a reduced subword $h$ of $g$ of maximal length such that $g=hxh^{-1}$ for some reduced word $x \in F$. Because $g$ is reduced, $x$ is non empty, and by maximality, there is no cancellation in the product $xx$. Therefore, $hx^nh^{-1}=g^n$ is a reduced word; hence $g^n \neq 1$.

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  • $\begingroup$ What do you mean by "there is no cancellation in the product $xx$"? $\endgroup$ – user42912 May 7 '14 at 14:35
  • $\begingroup$ Here, $g,x,h$ are viewed as words over a fixed generating set of $F$. So there is no cancellation in the product $xx$ in the sense that the concatenation of the two words is already reduced. $\endgroup$ – Seirios May 8 '14 at 7:57
  • $\begingroup$ Is there always such $h$, which I guess must not be the empty word $()$? $\endgroup$ – rgm May 7 '17 at 14:20
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    $\begingroup$ It may happen that $h$ is empty. If so, one says that $g$ is cyclically reduced. For instance, if you fix a basis $\{a,b\}$ of your free group, then $a$, $a^2b$, $b^{-1}ab^{-1}$ are examples of cyclically reduced words. $\endgroup$ – Seirios May 8 '17 at 7:36
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Let $F$ be a free group over the generator set $X$. Assume $F$ has elements of finite order. As element of $F$, we can write these elements as reduced words with letters from $X\cup X^{-1}$ and their inverses. Let $l(\omega)\in\mathbb N_0$ denote the length of word $\omega$, that is (in your posted definition) the number of sequence terms that are $\ne 1$. Among all elements $\ne1$ of finite order select one element $\alpha$ that has minimal length $l(\alpha)>0$.

Assume that the last letter of that word $\alpha$ is the inverse of the first letter, so $\alpha=x\beta x^{-1}$ with $l(\beta)=l(\alpha)-2$ (note that $x\ne x^{-1}$ for $x\in X\cup X^{-1}$ by definition, so surely $l(\alpha)\ge2$). Because $\beta^n=(x^{-1}\alpha x)^n=x^{-1}\alpha^nx=x^{-1}x=1$ if $\alpha^n=1$, we see that $\beta$ has finite order and is shorter than $\alpha$. By minimality of $\alpha$, we conlcude $\beta=1$, but then $\alpha=xx^{-1}$ was not reduced - contradiction. Therefore the last first and last letter of $\alpha$ are not inverse of each other. This means that no cancellation takes place when multiplying (i.e. concatenating copies of) $\alpha$ repeatedly. Thus $l(\alpha^n)=n l(\alpha)>0$ and $\alpha^n\ne 1$ for all $n\ge 1$. This contradicts the assumption that $\alpha$ has finite order.

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  • $\begingroup$ "...we can these___ as words..." ....?? Write? $\endgroup$ – DonAntonio May 7 '13 at 22:46
  • $\begingroup$ The above perhaps won't help a lot a greenhorn that got the definition of Free Group by means of the universal property... $\endgroup$ – DonAntonio May 7 '13 at 22:47
  • $\begingroup$ @DonAntonio Ack, but the OP posted such a "constructive" definition by now. Actually, I wonder if the proof starting from the universal property can avoid the construction via string concatenation and cacellation altogether; after all one needs sufficiently "rich" groups to exhibit a homomorphism that does not map $\alpha^n$ to 1. $\endgroup$ – Hagen von Eitzen May 8 '13 at 8:50
  • $\begingroup$ I agree with you, @Hagen. In fact, I like the van der Waerden trick to construct a free group and prove all what's required. I only wondered what the OP got as definition, as this could matter. Happily, he seems to have gotten the definition by alphabet, words and stuff. $\endgroup$ – DonAntonio May 8 '13 at 9:04
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This has been satisfactorily answered but I originally came here looking for a proof that appealed to the universal property and without so much explicit computation on the reduced words. I didn't find it so here's my stab.

Let $F$ be free on $X$, let $a\in F$, $|a| = n$. Writing $a$ as a reduced word gives, $a = x^{\delta_1}...x^{\delta_m}$. Let $p$ be a prime s.t. $p>n$ and $p > |\sum \delta_i|$. Consider the map $f:X\rightarrow \mathbb{Z}_p$, $x\mapsto \bar{1}$. By the universal property, $f$ induces a homomorphism $\bar{f}$. We have that $\bar{f}(a) = \sum \delta_i \mod p \neq 0 \mod p$, so the image of $\bar{f}(\langle a \rangle)$ is a non-trivial subgroup of $\mathbb{Z}_p$ of size at most $n$, which is a contradiction since $p$ has no proper divisors.

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  • $\begingroup$ Unfortunately, this is not quite correct. For example, the word $a^{-1}b^{-1}ab$ will be mapped to the identity by your map (more generally, any word with exponent sum zero, or equivalently contained in the derived subgroup, will be mapped to the identity). $\endgroup$ – user1729 Mar 27 '14 at 9:25
  • $\begingroup$ (That said, your general idea does actually work, although I am unsure how deep the proof is. That is to say, free groups are "residually $p$-groups", which means that for every prime $p$ and every element $g\in F$ there exists a homomorphism $\phi: F\rightarrow G$ where $G$ is a $p$-groups and such that $\phi(g)$ is non-trivial.) $\endgroup$ – user1729 Mar 27 '14 at 9:43

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