Let $G$ be a group and let $a,b$ be two elements of $G$. What can we say about the order of their product $ab$?

Wikipedia says "not much":

There is no general formula relating the order of a product $ab$ to the orders of $a$ and $b$. In fact, it is possible that both $a$ and $b$ have finite order while $ab$ has infinite order, or that both $a$ and $b$ have infinite order while $ab$ has finite order.

On the other hand no examples are provided. $(\mathbb{Z},+), 1$ and $-1$ give an example of elements of infinite order with product of finite order. I can't think of any example of the other kind! So:

  • What's an example of a group $G$ and two elements $a,b$ both of finite order such that their product has infinite order?

Wikipedia then states:

If $ab = ba$, we can at least say that $\mathrm{ord}(ab)$ divides $\mathrm{lcm}(\mathrm{ord}(a), \mathrm{ord}(b))$

which is easy to prove, but not very effective. So:

  • What are some similar results about the order of a product, perhaps with some additional hypotheses?
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    $\begingroup$ The product of two planar symmetries with respect to some lines has an infinite order if (and only if) the angle between these lines is an irrational part of the full angle, although each of these symmetries has order two. $\endgroup$
    – Did
    Commented May 25, 2011 at 16:32
  • $\begingroup$ I think that this example shows why in general we cannot say too much about the order of the product: In $S_n$ any permutation is a product of transpositions.... $\endgroup$
    – N. S.
    Commented May 25, 2011 at 16:37
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    $\begingroup$ Yes, Coxeter groups are groups which are generated by elements of order two with arbitrary conditions on the orders of $xy$ where $x$ and $y$ are two of the generators. I've always been fascinated by equivalence of the fact: "Finite simple groups are either cyclic of prime order or are of even order" can be restated as "finite simple groups are either rotation groups of prime order or images of a Coxeter group." $\endgroup$ Commented May 25, 2011 at 17:12
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    $\begingroup$ My definition of Coxeter group was imprecise, see en.wikipedia.org/wiki/Coxeter_group. But my statement about the relationship between Coxeter groups and finite simple groups stands. $\endgroup$ Commented May 25, 2011 at 17:21

8 Answers 8


Simple examples are given by free products, assuming you know the normal form for a free product; otherwise, I'm not really saying much.

If you want a more concrete example, take the $2\times 2$ invertible matrices with coefficients in $\mathbb{Q}$, and $$a = \left(\begin{array}{cc}0&1\\1&0\end{array}\right),\qquad b=\left(\begin{array}{cc}0 & 2\\\frac{1}{2}& 0\end{array}\right).$$ Then $a^2=b^2=1$, but $$ab = \left(\begin{array}{cc}\frac{1}{2} & 0 \\0 & 2\end{array}\right)$$ has infinite order.

If $a$ and $b$ commute, with $\mathrm{ord}(a)=m$ and $\mathrm{ord}(b)=n$, then you can do better than Wikipedia. We have that: $$\frac{\mathrm{lcm}(m,n)}{\mathrm{gcd}(m,n)}\quad\text{divides}\quad \frac{\mathrm{lcm}(m,n)}{|\langle a\rangle\cap\langle b\rangle|}\quad\text{divides}\quad \mathrm{ord}(ab)\quad\text{divides}\quad \mathrm{lcm}(m,n).$$ For example, if for every prime $p$ that divides $\mathrm{lcm}(m,n)$, the highest power of $p$ that divides $m$ is different from the highest power of $p$ that divides $n$, then $\mathrm{ord}(ab)=\mathrm{lcm}(m,n)$.

If you are willing to impose global conditions (conditions on $G$), then for example Easterfield proved in 1940 that if $G$ is a $p$-group of class $c$, $a$ has order $p^{\alpha}$, and $b$ has order $p^{\beta}$, then the order of $ab$ is at most $p^m$, where $$ m = \max\left\{\alpha,\beta+\left\lfloor\frac{c-1}{p-1}\right\rfloor\right\}.$$ From this you can get similar results for a finite nilpotent group (which is necessarily a product of $p$-groups), and hence for the product of two elements of finite order in any nilpotent group (or in fact, in any locally nilpotent group, or even more strongly, in any group in which the 2-generated subgroups are nilpotent).

  • $\begingroup$ The first example is just a rephrasing of the question: it is true that a product of an element of order $2$ and an element of order $3$ need not have finite order if and only if this is true of the free product, which is the universal example. So there still remains something to be proved: why doesn't a relation of the form $(ab)^n$ hold in the free product? $\endgroup$ Commented May 25, 2011 at 16:43
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    $\begingroup$ @Qiaochu: Fair enough; I tend to think of free groups and free products in terms of their normal form (as sets of words, with operation "concatenate and cancel") rather than as defined by their universal property, so to me the fact that the free product of finite groups has elements of infinite order is clear (since there is no cancellation if we repeat a word that "starts" in one group and "ends" in the other). If you think of free products mainly in terms of their universal properties, then it's of course not clear at all why that's the case. $\endgroup$ Commented May 25, 2011 at 16:54
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    $\begingroup$ Ah, yes, matrices! $G$ had to be non commutative and had to contain elements of both finite and infinite order... why didn't I think of matrices? I'm accepting this answer because it's the most complete. I will read the cited paper by Easterfield once Cambridges fixes their website. Thank you for your time! $\endgroup$ Commented May 25, 2011 at 22:09

Product of two elements of infinite order = element of finite order?

Let $G$ be any group containing an element $a$ of infinite order. Then $a$ is clearly not the identity, $e$ of G, since $ord(e) = 1$, which is finite. Nor can $a$ be its own inverse, since that would mean $ord(a) = 2$, also finite.

So there must exist an element $b$ such that $b$ = $a^{-1},\; b \neq a,\;b \neq e$, since for every element $x \in G,\; x^{-1} \in G$, and it follows that $ord(a) = ord(a^{-1}) = ord(b)$. Hence $b$, like $a$, is of infinite order Now clearly, $a * b = a*a^{-1} = e \in G$, and, of course, the order of $e$ is finite. So we have if a set has an element of infinite order, it also has an element of finite order which is the product of two elements of infinite order.

Product of elements of finite order = element of infinite order?

Another even more concrete example of two elements of finite order, the product of which is of infinite order:

Take $G=GL_2(\mathbb{Z})$, and two elements $A$ and $B$ such that $$A = \left(\begin{array}{cc}-1&1\\0&1\end{array}\right),\qquad B=\left(\begin{array}{cc}-1 & 0\\ 0 & 1\end{array}\right).$$ $A$ and $B$ each have order 2, yet there product $AB$ is of infinite order: $$AB = \left(\begin{array}{cc}1&1\\0&1\end{array}\right)$$ To see this, note that $$(AB)^n = \left(\begin{array}{cc}1&n\\0&1\end{array}\right)$$ which is the identity matrix only for $n = 0$.

Also, the product $$BA = \left(\begin{array}{cc}1&-1\\0&1\end{array}\right) = (AB)^{-1}$$ has infinite order. So here again, you have $(AB), (AB)^{-1} \in GL_2(\mathbb{Z})$, both of infinite order, whose product is $I_{2x2}$, of finite order!

  • 1
    $\begingroup$ Not every infinite group has elements of infinite order (for example, a product of countably many copies of a finite group). $\endgroup$ Commented May 25, 2011 at 17:07
  • $\begingroup$ @Qiaochu: Thanks for catching that. I meant to write that every infinite group with elements of infinite order...I'll edit my answer. $\endgroup$
    – amWhy
    Commented May 25, 2011 at 17:18
  • $\begingroup$ Your matrix example is the "affine group" version of the infinite dihedral group I mention: (-1,1;0,1) is the affine transformation x→-x+1, "b", and (-1,0;0,1) is the affine transformation x→-x+0, "a". $\endgroup$ Commented May 25, 2011 at 17:44
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    $\begingroup$ @Jack: Yes, I noticed that after I posted and read your answer. I found your post quite interesting; I'm really fascinated with dihedral groups. BTW: I'm not fast, in terms of constructing a post...I started my response (on paper) when no one had yet responded...I get buried in composing answers, and by the time I post, it always seems others manage to post before me :-S $\endgroup$
    – amWhy
    Commented May 25, 2011 at 18:13
  • $\begingroup$ I like it. I should probably be more clear in my comment. "(+1) Your matrix example is...". When people are trying to understand something simple like this, it is hard to predict what sort of example will click. Your answer looks very different, and so stands a good chance of helping where my answer will not. $\endgroup$ Commented May 25, 2011 at 18:55

Here is a rephrasing of the question which I think aptly illustrates the value of thinking in terms of universal properties. Let $n, m$ be two positive integers. There is a group $C_n * C_m$, the free product of the cyclic groups $C_n, C_m$, with presentation $\langle a, b | a^n, b^m \rangle$. It satisfies the following universal property:

Let $G$ be a group with elements $g, h$ such that $g^n = h^m = 1$. Then there is a unique homomorphism $\phi : C_n * C_m \to G$ such that $\phi(a) = g, \phi(b) = h$.

Depending on how comfortable you are with presenting a group by generators and relations, this might be obvious. Nevertheless, the existence of the free product $C_n * C_m$ is helpful because it allows us to recast a question about an entire class of groups (the class of groups with an element of order $n$ and an element of order $m$) into a question about a single group:

Is it true that $(ab)^k = 1$ for some $k$ in $C_n * C_m$?

Note that there are only two possibilities: either $ab$ has order $k$ for some $k$, which means that $gh$ has order (dividing) $k$ in any group $G$ such that $g^n = 1, h^m = 1$, or $ab$ has infinite order. In particular, it is impossible for $gh$ to always have some finite but arbitrarily high order (since that would necessarily imply that $ab$ has infinite order). This is a special case of a very general theorem in first-order logic called the compactness theorem.

But now it should be intuitively obvious that $ab$ has infinite order. More explicitly, it should be intuitively obvious that the elements of $C_n * C_m$ are elements of the free group $F_2 = \langle a, b \rangle$ on $a, b$ such that $a, a^{-1}$ don't appear $n$ times in a row and $b, b^{-1}$ don't appear $m$ times in a row, with the obvious multiplication (cancel all of the obvious relations), and in this group there are no obvious relations to cancel in the word $(ab)^k$.

Of course, this is not a proof. (I believe it follows from standard properties of free groups, but what I've written is not a proof.) But what I'm trying to get at here is that you shouldn't need to see explicit examples to be convinced that this is intuitively true. Undoubtedly writing down explicit examples (nice quotients of $C_n * C_m$) is probably the easiest way to answer this particular question, but for more general questions about the group generated by two elements of finite order (for example, does $a b a^{-1} b a$ have infinite order in $C_n * C_m$?) it's good to be aware that there is a goal you should be aiming for: to understand the specific group $C_n * C_m$.

As far as writing down explicit examples, here now is a chance to illustrate the value of thinking in terms of representation theory in the general sense. Take two random matrices of orders $n$ and $m$. Their product should have infinite order with probability $1$; in fact, they should generate $C_n * C_m$ with probability $1$. An easy way to generate such matrices is just to conjugate a matrix which obviously has order $n$ or $m$ (for example an appropriate permutation) by a random matrix.

The value of doing this is that there are lots of easy-to-verify conditions which imply that a matrix has infinite order: for example, if it is a $d$-dimensional matrix with trace greater than $d$ in absolute value, then it must have infinite order (exercise). It is also surprisingly easy to guarantee that a $2 \times 2$ matrix has order $n > 2$: you just have to make sure that its characteristic polynomial is $x^2 - 2 \cos \frac{2\pi}{n} x + 1$, or equivalently that its determinant is $1$ and that its trace is $2 \cos \frac{2\pi}{n}$.

Let me illustrate with $n = m = 3$: we just need to find two matrices of determinant $1$ with trace $-1$ whose product has trace greater than $2$ in absolute value. In fact it's easy to generate families of such matrices: just take

$$A = \left[ \begin{array}{cc} a-1 & -(a^2-a+1) \\ 1 & -a \end{array} \right], B = \left[ \begin{array}{cc} b-1 & b^2-b+1 \\ -1 & -b \end{array} \right].$$

We compute that the trace of the product is $(a-1)(b-1) + (a^2 - a + 1) + (b^2 - b + 1) + ab$, which in particular is a large positive number as soon as $a, b$ themselves are large and positive. In fact all I had to do was guarantee that this wasn't a constant polynomial: any nonconstant polynomial takes on arbitrarily large values (exercise).

This same trick allows us to construct $2 \times 2$ matrices $A, B$ of orders $n, m$ such that their product has order $k$ for any triple $(n, m, k)$. The corresponding universal groups are called the von Dyck groups and are related to some interesting geometry.

  • $\begingroup$ I didn't understand the argument based on the compactness theorem: I'm not acquainted with the concepts of model and first-order sentence. But, if I did understand correctly the intuitive argument, I claim that if $m=n=2$ then $aba^{-1}ba$ has finite order (it's two), in all other cases it has infinite order. Writing two copies of $aba^{-1}ba$ with the relations $a^2=1$ and $b^2=1$ I see that this product telescopes; the same does not happen with different relations. $\endgroup$ Commented May 26, 2011 at 11:19
  • $\begingroup$ The trace of a matrix is also the sum of its eigenvalues. A matrix of dimension $d$ has $d$ eigenvalues. If $A$ is a matrix of dimension $d$ with $\abs{\trace{A}} > d$ then one eigenvalue must be larger than $1$ in modulo. Since the eigenvalues of $A^{k}$ are the $k$-th powers of the eigenvalues of $A$ this implies that for any $k$ the matrix $A^{k}$ hasn't all eigenvalues equal to one, a necessary (but not sufficient) condition to be the identity. $\endgroup$ Commented May 26, 2011 at 11:19
  • $\begingroup$ @Jacopo: The argument based on the compactness theorem is as follows: suppose there are groups $G_i$ with elements of orders $n$ and $m$ whose product has order at least $i$. Then by the compactness theorem, there is a group $G$ (say the ultraproduct of the groups above) which has elements of orders $n$ and $m$ whose product has infinite order. It's easier to see with ultraproducts, but in more detail: add a 0-ary function $x$ to the language and axioms dictating that $x$ is the product of two elements of orders $n$ and $m$. Then any finite $\endgroup$ Commented May 26, 2011 at 12:26
  • $\begingroup$ ... collection of the axioms "$x$ has order at least $i$" is consistent, so the entire collection is consistent. $\endgroup$ Commented May 26, 2011 at 12:26

For product of two elements of finite order having infinite order, simple examples can be built as follows.

Take two lines, say through the origin, making an angle $\theta$ with each other. Let $a$ be reflection in one line, $b$ reflection in the other. Both $a$ and $b$ have order $2$. Their product is rotation about the origin through $2\theta$. If $\theta$ is not commensurable with $\pi$, then $ab$ has infinite order.

Added comment: That takes care of infinite order. By taking $\theta$ to be $\pi/n$, we can produce $a$ and $b$ of order $2$ with product of any desired finite order.

  • $\begingroup$ Great example, because it also shows that such products can have any order, which supports the claim that there is no general formula. $\endgroup$ Commented May 25, 2011 at 22:12

Dihedral groups are generated by two elements of order two. The order of their product is arbitrary, and defines which dihedral group you get. For instance, the symmetries of a regular n-gon, for n ≥ 3 a positive integer, is a dihedral group of order 2n generated by two reflections a, b whose product is a rotation of order n.

The infinite dihedral group more directly answers your question:

Let a be the permutation of the integers that takes n to −n, and b the permutation that takes n to 1−n. Then a(b(n)) = −(1−n) = n−1, so ab has infinite order. Of course a(a(n) = −(−(n)) = n, so a has order 2; b(b(n)) = 1−(1−n) = n, so b has order 2.

This is similar to André's answer, except that the lines of reflection are parallel in my example, rather than meeting at an irrational angle. This is similar to Arturo's answer, except I use C2 ∗ C2 instead of C2 ∗ C3.


In the symmetric group $\operatorname{Sym}(E)$, where $E$ is a finite or infinite set, every element can be expressed as the product of two involutions. For example, in $\operatorname{Sym}(\mathbb Z)$, the permutation $f(n)=n+1$ can be expressed as $f(n)=g(h(n))$ where $g(n)=1-n$ and $h(n)=-n$.


Consider $GL_2(\mathbb{R})$. Take, $A = \begin{bmatrix} 1 & 0 \\ -1 & 1\end{bmatrix}, B = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$. Then,

$$A^2 = B^2 = I_2$$

But we have, $BAB = \begin{bmatrix} 1 & -1 \\ 0 & 1\end{bmatrix}$, and it is easy to compute that, $$(BAB)^n = \begin{bmatrix} 1 & -n \\ 0 & 1\end{bmatrix}$$ which can be $I_2$ only for $n=0$. Hence, we have the product of elements of finite order can be of infinite order.

  • $\begingroup$ Your matrix A has infinite order, A^n = [[ 1 0 ],[ -n 1 ]]. Indeed, (BAB)^n = B(A^n)B. $\endgroup$
    – Ben
    Commented May 14 at 16:24

A different sort of example of elements of finite order having an infinite order product:

Consider the permutation $f$ on $\mathbb Z^+$ that sends $2i+1$ to $2i+2$ and $2i+2$ to $2i+1$ for every non-negative integer $i$.

Now consider the permutation $g$ also on $\mathbb Z^+$ that sends $1$ to $1$, and sends $2i+2$ to $2i+3$ and $2i+3$ to $2i+2$ for every non-negative integer $i$.

Note both these permutations have order $2$, however $g \circ f$ sends every odd integer $k$ to $k+2$ and is thus of infinite order.


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