# Order of elements in a group

• Jun 25th 2010, 12:51 PM
bleys
Order of elements in a group
This probably just a simple manipulation problem but I keep going in circles. It's from Introduction to Algebra by Cameron
A group which contains elements a,b,c,d,e (none equal to the identity) such that ab=c, bc=d, cd=e, de=a, ea=b. Find the orders of a,b,c,d,e.
I found the 'squares' and 'cubes' of the elements, but it starts to get ugly with higher powers. I know I'm probably suppose to be looking for when a pattern repeats for the powers, but I always end up with a horrible expression or back where I started. Any help pointing me in the right direction, please?
• Jun 28th 2010, 06:32 AM
Opalg
Quote:

Originally Posted by bleys
This probably just a simple manipulation problem but I keep going in circles. It's from Introduction to Algebra by Cameron
A group which contains elements a,b,c,d,e (none equal to the identity) such that ab=c, bc=d, cd=e, de=a, ea=b. Find the orders of a,b,c,d,e.
I found the 'squares' and 'cubes' of the elements, but it starts to get ugly with higher powers. I know I'm probably suppose to be looking for when a pattern repeats for the powers, but I always end up with a horrible expression or back where I started. Any help pointing me in the right direction, please?

Interesting problem! I started by expressing everything in terms of a and b:

$c=ab$,
$d = bc=bab$,
$e=cd = ab^2ab$,
$a = de = babab^2ab$,
$b = ea = ab^2aba$.

From the last two of those relations it follows that $a^2 = babab^2aba = bab^2.\qquad({}^*)$

After a good deal of experimentation I came across this:

$edcbab = edcbc = edcd = ede = ea = b$. Multiplying on the right by $b^{-1}$, you see that $edcba = 1$, the identity element of the group. (I have to use 1 for the identity element, because e already represents a group element.)

Therefore, writing the elements in terms of a and b, $1 = edcba = ab^2ab^2abab^2a = ab(bab^2)a(bab^2)a = aba^6$ (from (*)). Thus $b = a^{-7}$ and in particular b commutes with a. You can then easily check that $a^{11} = 1$, and the other elements b, c, d, e also have order 11.
• Jul 3rd 2010, 12:51 PM
bleys
Thank you very much for replying! I also found out $a^2 = bab^2$ but would have never even guessed that second expression for b. Thank you for your help!