# Thread: Finding the order of a group element

1. ## Finding the order of a group element

I'm stuck on the following problem: if a and b are in a group G, and a^5=e and b^2=a*b*a^-1, what is the order of b (assuming b is not e). Any ideas where to start? I've tried a bunch of manipulations of the two relationships listed above, but I can't seem to get anywhere.

2. Originally Posted by Jessemath
I'm stuck on the following problem: if a and b are in a group G, and a^5=e and b^2=a*b*a^-1, what is the order of b (assuming b is not e). Any ideas where to start? I've tried a bunch of manipulations of the two relationships listed above, but I can't seem to get anywhere.
Assume $\displaystyle b$ has even order, then find a contradiction (this is relatively easy, just re-arrange the formula you are given).

Next, assume $\displaystyle b$ has odd order, $\displaystyle b^{2i+1}=1$, $\displaystyle b^j \neq 1 \text{ } \forall 1 \leq j < 2i+1$. Then, note that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1}$ and see what pops out.

You should get that the order of $\displaystyle b$ can be one of two things.

I should point out that I didn't use the fact that $\displaystyle a^5=1$, but I can't see a problem in my proof...

3. Originally Posted by Swlabr
Assume $\displaystyle b$ has even order, then find a contradiction (this is relatively easy, just re-arrange the formula you are given).

Next, assume $\displaystyle b$ has odd order, $\displaystyle b^{2i+1}=1$, $\displaystyle b^j \neq 1 \text{ } \forall 1 \leq j < 2i+1$. Then, note that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1}$ and see what pops out.

You should get that the order of $\displaystyle b$ can be one of two things.

I should point out that I didn't use the fact that $\displaystyle a^5=1$, but I can't see a problem in my proof...
@Swlabr - Hi. I follow your first part that we can't have an even order. But have been struggling with following -

Next, assume $\displaystyle b$ has odd order, $\displaystyle b^{2i+1}=1$, $\displaystyle b^j \neq 1 \text{ } \forall 1 \leq j < 2i+1$. Then, note that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1}$ and see what pops out.

Any further help plz?

4. A relatively simple to prove (I'm certain you could prove it on your own!) theorem states that:

If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = aba^{-1}$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

Using this fact makes this question relatively simple.

If you want to see it in a more "algebraic" way:

$\displaystyle aba^{-1} = b^2$

$\displaystyle a^2ba^{-2} = a(aba^{-1})a^{-1} = ab^2a^{-1} = (b^2)^2 = b^4$

$\displaystyle a^3ba^{-3} = a(a^2ba^{-2})a^{-1} = ab^4a^{-1} = (b^2)^4 = b^8$

$\displaystyle a^4ba^{-4} = a(a^3ba^{-3})a^{-1} = ab^8a^{-1} = (b^2)^8 = b^{16}$

$\displaystyle a^5ba^{-5} = a(a^4ba^{-4})a^{-1} = ab^{16}a^{-1} = (b^2)^{16} = b^{32}$

But $\displaystyle a^5=e \Rightarrow b = b^{32} \Rightarrow b^{31} = e$ and since 31 is prime, $\displaystyle o(b) = 31$

5. Originally Posted by aman_cc
@Swlabr - Hi. I follow your first part that we can't have an even order. But have been struggling with following -

Next, assume $\displaystyle b$ has odd order, $\displaystyle b^{2i+1}=1$, $\displaystyle b^j \neq 1 \text{ } \forall 1 \leq j < 2i+1$. Then, note that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1}$ and see what pops out.

Any further help plz?
Well, my thought was that $\displaystyle 1=ab^{2i+1}a^{-1}=ab^{2i}a^{-1}aba^{-1} \Rightarrow ab^{2i}a^{-1}b^2 \Rightarrow b^{-2} = ab^{2i}a^{-1}$ and that this gave us a contradiction. However, it doesn't (I think I thought the $\displaystyle b^{-2}$ was actually a $\displaystyle b$). oops.

6. Originally Posted by Defunkt
A relatively simple to prove (I'm certain you could prove it on your own!) theorem states that:

If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = a^{-1}ba$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

Using this fact makes this question relatively simple.

If you want to see it in a more "algebraic" way:

$\displaystyle aba^{-1} = b^2$

$\displaystyle a^2ba^{-2} = a(aba^{-1})a^{-1} = ab^2a^{-1} = (b^2)^2 = b^4$

$\displaystyle a^3ba^{-3} = a(a^2ba^{-2})a^{-1} = ab^4a^{-1} = (b^2)^4 = b^8$

$\displaystyle a^4ba^{-4} = a(a^3ba^{-3})a^{-1} = ab^8a^{-1} = (b^2)^8 = b^{16}$

$\displaystyle a^5ba^{-5} = a(a^4ba^{-4})a^{-1} = ab^{16}a^{-1} = (b^2)^{16} = b^{32}$

But $\displaystyle a^5=e \Rightarrow b = b^{32} \Rightarrow b^{31} = e$ and since 31 is prime, $\displaystyle o(b) = 31$
Nice. .

@ OP-The only thing that may nee to be said is that the reasong why the text in red above is important is because if $\displaystyle b^n=e$ then $\displaystyle |b|\text{ }|\text{ }n$, but since $\displaystyle 31$ is prime the only positive divisors are $\displaystyle 1,31$ and since $\displaystyle b\ne e$ the only choice is that $\displaystyle |b|=31$

7. Originally Posted by Defunkt
A relatively simple to prove (I'm certain you could prove it on your own!) theorem states that:

If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = a^{-1}ba$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

Using this fact makes this question relatively simple.

If you want to see it in a more "algebraic" way:

$\displaystyle aba^{-1} = b^2$

$\displaystyle a^2ba^{-2} = a(aba^{-1})a^{-1} = ab^2a^{-1} = (b^2)^2 = b^4$

$\displaystyle a^3ba^{-3} = a(a^2ba^{-2})a^{-1} = ab^4a^{-1} = (b^2)^4 = b^8$

$\displaystyle a^4ba^{-4} = a(a^3ba^{-3})a^{-1} = ab^8a^{-1} = (b^2)^8 = b^{16}$

$\displaystyle a^5ba^{-5} = a(a^4ba^{-4})a^{-1} = ab^{16}a^{-1} = (b^2)^{16} = b^{32}$

But $\displaystyle a^5=e \Rightarrow b = b^{32} \Rightarrow b^{31} = e$ and since 31 is prime, $\displaystyle o(b) = 31$

Thanks Defunkt. Guess there was a small typo error in what you wrote. Following should be the correct statement

If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = aba^{-1}$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

I follow it now. Nice question and an equally nice solution. Thanks again

8. Originally Posted by aman_cc
Thanks Defunkt. Guess there was a small typo error in what you wrote. Following should be the correct statement

If G is a group, $\displaystyle a,b \in G : \ \exists i \in \mathbb{N} \ s.t. \ b^i = aba^{-1}$, then $\displaystyle b^{(i^n)} = a^nba^{-n}$

I follow it now. Nice question and an equally nice solution. Thanks again
Thanks, and yes you're right of course, that was a typo :P I'll fix the original post now.