# free groups , a question of listing elements and drawing multiplication table

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?

• What do you mean by $|a|$? – BIS HD Oct 26 '13 at 18:21
• 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 $(ba)^kb$. – Sigur Oct 26 '13 at 18:22
• @BISHD, order of $a$. – Sigur Oct 26 '13 at 18:23
• the order of the element a – Enas Oct 26 '13 at 18:23
• The group is indeed infinite, so I too find it strange that you are being asked to draw the multiplication table for it. – 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
$$abababa\ldots\;,\;\;\text{or}\;\;bababa\ldots$$
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
$$ababa=(ab)a(ba)=(ab)a(ab)^{-1}\;,\;\;abababa=(aba)b(aba)=(aba)a(aba)^{-1}\;\ldots etc.$$