0
$\begingroup$

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?

$\endgroup$
13
  • $\begingroup$ 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
2
$\begingroup$

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

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.