# Thread: [SOLVED] Getting crazy with this excerice....

1. ## [SOLVED] Getting crazy with this excerice....

Here's the problem (regarding groups, but nothing too complex):
I'm told that a and b are elements in G, which is a group.
I'm also told that b<>e (where e is the unit element - thus e*a = a, etc...), and that o(a) = 5 (meaning: a^5=e, and 5 is the smallest natural number that satisfies that).
I'm told that: aba^-1 = b^2
I need to find o(b).

This looks pretty innocent at first (and maybe it really is), but I've spent too much time on this question... tried every trick I know...
I've proved, some excercises before, that o(aba^-1)=o(b), so I tried using that a lot, but it didn't really get me much.
Of course I tried to do "^5" to both sides, and many many other things, playing with the inverses etc....
Best I could get, with much effort, is that o(ba) = 5.
I just don't know what to do :-\

Thanks!
Tomer.

2. Originally Posted by aurora

Here's the problem (regarding groups, but nothing too complex):
I'm told that a and b are elements in G, which is a group.
I'm also told that b<>e (where e is the unit element - thus e*a = a, etc...), and that o(a) = 5 (meaning: a^5=e, and 5 is the smallest natural number that satisfies that).
I'm told that: aba^-1 = b^2
I need to find o(b).

This looks pretty innocent at first (and maybe it really is), but I've spent too much time on this question... tried every trick I know...
I've proved, some excercises before, that o(aba^-1)=o(b), so I tried using that a lot, but it didn't really get me much.
Of course I tried to do "^5" to both sides, and many many other things, playing with the inverses etc....
Best I could get, with much effort, is that o(ba) = 5.
I just don't know what to do :-\

Thanks!
Tomer.
that's a good question! i'll prove a more general result first:

Lemma: let $a,b$ be two elements of a group such that $aba^{-1}=b^n.$ then $a^{\ell}ba^{-\ell}=b^{n^{\ell}}, \ \ \forall \ell \in \mathbb{N}.$

Proof: by induction:

for $\ell = 1,$ the claim is just the hypothesis of the lemma. now suppose the claim is true for $\ell.$ then:

$a^{\ell + 1}ba^{-(\ell + 1)}=a(a^{\ell}ba^{-\ell})a^{-1}=ab^{n^{\ell}}a^{-1}=(aba^{-1})^{n^{\ell}}=(b^n)^{n^{\ell}}=b^{n^{\ell + 1}}. \ \ \ \square$
.................................................. .................................................. ................................

now it's easy to solve your problem: since we're given that $aba^{-1}=b^2,$ we can apply the Lemma for

$n =2$ and $\ell = 5.$ thus: $b=a^5ba^{-5}=b^{32}.$ hence $b^{31}=e.$ since 31 is prime, we have $o(b)=31. \ \ \ \square$

3. Originally Posted by NonCommAlg
$a^{\ell + 1}ba^{-(\ell + 1)}=a(a^{\ell}ba^{-\ell})a^{-1}=ab^{n^{\ell}}a^{-\ell}=(aba^{-1})^{n^{\ell}}=(b^n)^{n^{\ell}}=b^{n^{\ell + 1}}. \ \ \ \square$
Do you mean $ab^{n^\ell}a^{-1}$ instead?

4. Originally Posted by ThePerfectHacker
Do you mean $ab^{n^\ell}a^{-1}$ instead?
i meant exactly what i wrote! haha ... still, what part of that line are you refering to?

5. Originally Posted by NonCommAlg
i meant exactly what i wrote! haha ... still, what part of that line are you refering to?
Look at what I quoted in my first post. The stuff in between 2nd and 3rd "=" sign.

6. Originally Posted by ThePerfectHacker
Look at what I quoted in my first post. The stuff in between 2nd and 3rd "=" sign.
Ah, of course! i fixed the typo. thank you!

7. Wow, bless you!
I guess it's solvable, though somehow sutdents are always fearful to prove more than they need .
Thank you so much!
Tomer.

8. I been curious the key fact here was that $31$ is a prime number. What happened if we had $a^4=e$ then $b = a^4ba^{-4} = b^{2^4}$ and so $b^{15}=e$. What can we conclude? It can still be that the order is $15$.