# Thread: Find order of this element

1. ## Find order of this element

Let G be a finite group that has the properties that it contains 2 elements a and b such that:

1. $a^5=e$
2. $aba^- ^1 =b^2$
3. a ≠ e
4. b ≠ e

Find O(b).

2. Note that $aba^{-1}=b^2 \Rightarrow (aba^{-1})^2 = ab^2a^{-1}=b^4 \Rightarrow a(aba^{-1})a^{-1} = a^2ba^{-2}=b^4$

Can you continue?

3. I'll give it another shot and report back

4. Originally Posted by Defunkt
Note that $aba^{-1}=b^2 \Rightarrow (aba^{-1})^2 = ab^2a^{-1}=b^4 \Rightarrow a(aba^{-1})a^{-1} = a^2ba^{-2}=b^4$

Can you continue?
I understand what you did here, but I'm not really seeing how I'll be able to get rid of all the b's just so it equals e...

5. Originally Posted by dynas7y
I understand what you did here, but I'm not really seeing how I'll be able to get rid of all the b's just so it equals e...
Try applying the same 'process' I did to $a^2ba^{-2}$

6. Ok, I've tried applying the process for others such as $b^1^0$, or really any n where $b^n$ can be broken down into factors of $b^2$ and I think I'm just not seeing it. When attempting to find something such as $b^5$, if I broke that down into $b$ and $b^4$, then the order would matter since the group is not necessarily abelian, right? I may just be missing something or it's not catching my eye but I don't see right now how I'm going to get $b^n=e$, for some n. Any additional "hints" or points in the right direction are appreciated

7. What I meant was that you do this: $a^2ba^{-2}=b^4 \Rightarrow a^2b^2a^{-2}=b^8$ $\Rightarrow a^2(aba^{-1})a^{-2} = a^3ba^{-3} = b^8$.

Can you continue in this way until you reach $a^5ba^{-5}$ ?

You can also prove the more general theorem: if $aba^{-1} = b^i$ then $a^mba^{-m} = b^{(i^m)}$

8. Originally Posted by Defunkt
What I meant was that you do this: $a^2ba^{-2}=b^4 \Rightarrow a^2b^2a^{-2}=b^8$ $\Rightarrow a^2(aba^{-1})a^{-2} = a^3ba^{-3} = b^8$.

Can you continue in this way until you reach $a^5ba^{-5}$ ?

You can also prove the more general theorem: if $aba^{-1} = b^i$ then $a^mba^{-m} = b^{(i^m)}$
So continuing until I reach $a^5ba^{-5}$ we obtain:

$a^4ba^{-4}=b^1^6 \Rightarrow a^4b^2a^{-4}=b^3^2$ $\Rightarrow a^4(aba^{-1})a^{-4} = a^5ba^{-5} = ba^{-5} = b^3^2$.

Now what?

9. If $a^5=e$ , what is $a^{-5}$ ?

10. Originally Posted by Defunkt
If $a^5=e$ , what is $a^{-5}$ ?
I'm not quite sure, but wouldn't $a^{-5}=e$ as well?

11. I think I get it in that if $a^{-5}=e$, then we have $b=b^{32}$, which gives us $e=b^{31}$. Thus, the order of b must be a divisor of 31, meaning it is either 1 or 31. However it can't be 1 since b is a non-identity element by the given, thus $o(b)=31$.

However, can you please explain why $a^5=e$ gives $a^{-5}=e$? Thanks.

12. Since $a^5a^{-5}=a^{-5}a^5=e$ and the inverse of the identity is the identity itself.