So, I let a,b be elements in such a group. So |a|=n and |b|=m, n and m are finite. But |ab| needs to be infinite, but since |ab|=lcm(n,m), how can that be possible?
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Martin Sleziak
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5The result you quote is not correct. In an Abelian group, the order of $ab$ is $\le$ the lcm of the orders of $a$ and $b$. – André Nicolas Oct 26 '15 at 05:20
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Some related posts: Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite and Examples and further results about the order of the product of two elements in a group – Martin Sleziak Oct 26 '15 at 10:51
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Where does this problem come from? Is it from a book? From an assignment? – Martin Sleziak Oct 26 '15 at 10:52
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Consider group of all permutations of $\mathbb{Z}$; permutation group on $\geq 3$ letters is always non-abelian.
Consider $\sigma(x)=-x+1$ and $\tau(x)=-x+2$, the two permutations of $\mathbb{Z}$.
Show that $\tau\circ\tau=\sigma\circ\sigma=I$, hence order of $\sigma,\tau$ is $2$.
Find $\sigma\circ \tau$.
Is it of finite order?
Groups
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3This simple answer (+1) has a nice relation to mine: Two reflections on parallel lines give a translation. This is then just the one-dimensional discrete case. – filipos Oct 26 '15 at 13:18
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Compose two reflections on the plane and you get a rotation. Reflections have order 2. The rotation can have arbitrary order, including infinite.
filipos
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I'm trying to imagine a rotation of infinite order. I think rotation through an irrational number of degrees, or a rational number of radians would work, right? – Todd Wilcox Oct 26 '15 at 13:43
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1Right. By the way, another interesting case is when the reflections are on parallel lines, which gives a translation (see Groups' answer). – filipos Oct 26 '15 at 15:01
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An irrational number of degrees will work. An irrational number of radians might not. A rational number of radians will work though! – MJD Oct 26 '15 at 16:00
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Take the free group on letters $a,b$ with the sole relations that $a^2=b^2=1$. Now $ab$ has infinite order, since $$abababab\cdots ab \neq 1$$
vadim123
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2@safaqfwrq Do you know how to define groups via generators and relations - that is, are you familiar with notation like "$\langle a, b\vert a^2, b^2\rangle$"? – Noah Schweber Oct 26 '15 at 06:25
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6This is in some sense a restatement of the question: it's true that if $a^2 = 1$ and $b^2 = 1$ then $ab$ has infinite order iff it's true in this universal example. So you have not yet answered the question. You're implicitly using the fact that you think you know how to solve the word problem in this group... – Qiaochu Yuan Oct 26 '15 at 06:58
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Matrices, invertible ones are a group under multiplication and here is a good example
Zelos Malum
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