Hi all ,

Just wanted to see if i was handling this Q correctly

Q in a group G , a^5 = e, aba^-1 = b^2 for some a,b element of G.
find the order of b?

I said aba^-1 = b^2

i) multiplied RHS and LHS by 'a'

to get ab = b^2a

ii) then i) multiplied RHS and LHS by b^-1 (b inverse)

to get a = b^2ab^-1

then put both side to the power of 5

to get a^5 = e

so e = (b^2ab^-1)^5

e = b^10a^5b^-5

e = b^10b^-5 (a^5 turns to e)
= b^5

so order of b = 5 ??????

Not sure at all if this is correct any suggestions would be great.

thanks

chave

2. so e = (b^2ab^-1)^5

e = b^10a^5b^-5
Take care, here you may have assumed that $\displaystyle a$ and $\displaystyle b$ commute.

But, since $\displaystyle aba^{-1}=b^2,$ we get that the commutator $\displaystyle aba^{-1}b^{-1}=b,$ thus $\displaystyle a$ and $\displaystyle b$ don't commute!

Edit: made a mistake in my development, ThePerfectHacker sent a proof just after your post. For commutativity, $\displaystyle (xy)^5=x^5y^5$ is wrong in general.

3. I didn't assume they commute but am i allowed to put both sides
to the power of 5????

4. Originally Posted by chave
Hi all ,

Just wanted to see if i was handling this Q correctly

Q in a group G , a^5 = e, aba^-1 = b^2 for some a,b element of G.
find the order of b?
If $\displaystyle aba^{-1} = b^2$ then $\displaystyle a^k ba^{-k} = b^{2^k}$. Prove this by induction!
Now if $\displaystyle k=5$ then $\displaystyle a^5 ba^{-5} = b^{32} \implies b^{31} = e$.
Therefore, $\displaystyle b=e\text{ or }|b|=31$ since $\displaystyle 31$ is a prime number.

5. Sorry but I don't really understand perfect hacker what you mean here?? can you explain a bit more please?? thanks especially the induction bit and are you saying the order of b is 31????

6. Originally Posted by chave
Sorry but I don't really understand perfect hacker what you mean here?? can you explain a bit more please?? thanks especially the induction bit and are you saying the order of b is 31????
Can you prove the induction?

I said that $\displaystyle b^{32} = b$ which means $\displaystyle b^{31} = e$. Therefore, either $\displaystyle b$ is the idenity element i.e. $\displaystyle b=e$ (which is trivial and probably is not the case in whatever problem you are working on) or the order of $\displaystyle b$ is $\displaystyle 31$.

7. ok i understand the second bit, no i cant get the induction proof??
sorry again for annoying you.

8. Originally Posted by chave
ok i understand the second bit, no i cant get the induction proof??
sorry again for annoying you.
If $\displaystyle k=1$ then $\displaystyle aba^{-1}=b^2$, which is true by hypothesis of problem. Say that $\displaystyle a^kba^{-k}=b^{2^k}$ then raise both sides to the power of $\displaystyle 2$ to get $\displaystyle \left( a^k b a^{-k} \right)^2 = b^{2^{k+1}}$. However, $\displaystyle \left( a^k b a^{-k} \right)^2 = a^k b^2 a^{-k} = a^k (aba^{-1})a^{-k} = a^{k+1}ba^{-(k+1)}$. This completes induction.

9. thanks a lot mate