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I am just beginning to learn Group theory. As an example of finite groups our Professor wrote this group with just two elements, given by $$ \left( \begin{array}{cc} 0 & z \\ z^{-1} & 0\end{array} \right),\:\:\: \left( \begin{array}{cc} z & 0 \\ 0 & z^{-1}\end{array} \right) $$ where $z \:\epsilon\:\mathbb{C} $. Here there infinitely as many values $z$ can take. I am quite confused why we call this a finite group. In general I wish to understand how exactly we find if a group is finite or infinite.

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Your professor meant that you choose and fix a $z \in \mathbb{C}$. You have to choose a non zero $z$ to make sense of $z^{-1}$. Now consider those two elements, and call them $a$ and $b$. EDIT: Following Berci's comment, you actually need $z$ to be a root of unity, i.e. $z$ should be such that $z^n=1$. Then the group they generate is finite:

1) Show that both $a$ and $b$ have finite orders, using matrix multiplication.

2) Same way "see" the relation between $ab$ and $ba$ and try to conclude that the group must be a finite group.

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    $\begingroup$ Well, for many complex $z$, both $a$ and $b$ will have infinite order. In order to get finite group, $z$ has to be a root of unity. $\endgroup$ – Berci Feb 10 '14 at 1:34
  • $\begingroup$ @Berci: Yes you are correct. :). Thanks. $\endgroup$ – voldemort Feb 10 '14 at 1:36
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    $\begingroup$ @voldemort ,@Berci : So then, in my case with all generality, the group i have written down is of infinite order right ? $\endgroup$ – user38249 Feb 10 '14 at 3:07
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    $\begingroup$ @user38249: yes in general it's an infinite group. $\endgroup$ – voldemort Feb 10 '14 at 4:31
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To partially address your general question, whether or not a general group is finite can be a very hard problem. Just to whet your appetite, take a look at Burnside's problem, posed in 1902, which asked whether every group all of whose elements could be expressed as products of terms from a finite list, each of which has finite order, is finite. It took until 1964 to prove by counterexample that the answer is actually no! There are groups which are finitely generated and in which every element is of finite order but which are not finite.

Proving that a particular group is finite will often depend on techniques particular to the family to which it belongs or sometimes even ad hoc techniques. Although, most "sensible" examples will admit a "sensible" attack--in your class you may find that if you're asked as an exercise to prove a group is finite it may be easy enough to give a strategy for enumerating all the elements of the group and then prove that the algorithm eventually halts. This works, for example, for cyclic groups, permutation groups, matrix groups over finite fields, dihedral groups, etc. Proving that an infinite group is infinite is often easier, since most "sensible" examples of infinite groups will have at least one element of infinite order, despite the counterexamples to the Burnside problem.

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