Find order of this element

**dynas7y** · October 4th 2010, 06:31 AM

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).

**Defunkt** · October 4th 2010, 06:50 AM

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?

**dynas7y** · October 4th 2010, 06:57 AM

I'll give it another shot and report back

**dynas7y** · October 4th 2010, 07:20 AM

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...

**Defunkt** · October 4th 2010, 08:04 AM

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 $a^2ba^{-2}$

**dynas7y** · October 5th 2010, 12:51 PM

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

**Defunkt** · October 5th 2010, 01:23 PM

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 $a^mba^{-m} = b^{(i^m)}$

**dynas7y** · October 5th 2010, 01:58 PM

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?

**Defunkt** · October 5th 2010, 02:27 PM

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

**dynas7y** · October 5th 2010, 03:44 PM

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?

**dynas7y** · October 5th 2010, 06:44 PM

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.

**Defunkt** · October 6th 2010, 02:00 AM

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

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