I'm requested to list the elements and draw the multiplication table for the group $\langle a, b : |a| = 2 = |b|\rangle$ without any more details. But hence this group is infinite isn't it ?

while listing the elements i found $ = \{ a , b , ab , ba , aba , bab , abab , baba, \dots \}$

and so on ! I see that we must have an information about the order of $ab$ or $ba$?

Am I wrong ? Can anybody help me?

  • $\begingroup$ What do you mean by $|a|$? $\endgroup$ – BIS HD Oct 26 '13 at 18:21
  • $\begingroup$ Since $aa=1=bb$ any element must be a finite sequence of $ab$ or $ba$ and ending with $a$ or $b$. For example, $(ab)^ka$ or $(ba)^kb$. $\endgroup$ – Sigur Oct 26 '13 at 18:22
  • $\begingroup$ @BISHD, order of $a$. $\endgroup$ – Sigur Oct 26 '13 at 18:23
  • $\begingroup$ the order of the element a $\endgroup$ – Enas Oct 26 '13 at 18:23
  • 2
    $\begingroup$ The group is indeed infinite, so I too find it strange that you are being asked to draw the multiplication table for it. $\endgroup$ – Tobias Kildetoft Oct 26 '13 at 18:25

Why the tag "free-groups"? Yours is not a free group but a free product, namely $\;C_2*C_2\;$ = the infinite dihedral group.

By the general theory, $\;C_2*C_2\;$ is an infinite group and the only elements with finite order are those who are conjugates to one of the elements in either factor.

Thus, if we put for the first factor $\;\langle a\rangle=\{1,a\}\;$ , and for the second one $\;\langle b\rangle=\{1,b\}\;$ , the normal form of an element in this group is of the form


each of the two forms above being a finite word in those two letters, and the finite order elements are those of the form $\;g^{-1}ag\;,\;\;g^{-1}bg\;,\;\;g\in C_2*C_2\;$ , for example

$$ababa=(ab)a(ba)=(ab)a(ab)^{-1}\;,\;\;abababa=(aba)b(aba)=(aba)a(aba)^{-1}\;\ldots etc.$$


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