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1. A quick question on groups
Say a multiplicative group of order 8 has elements {e, b, b2, b3, a, ab, ab2, ab3} and I want to show that the order of ab2 is 4. We're given here that b4 = e, and a2 = b2 = (ab)2.

Would it be okay to say (ab2)2 = (b2)2a2 = b4a2 = a2 = b2. Therefore ab2 must have order 4 as (ab2)4 = ((ab2)2)2 = (b2)2 = b4 = e?

In general, can we assume that (ab)n = bnan, for any natural number n?
Last edited by Ree69; 19-05-2012 at 18:13.
2. Re: A quick question on groups
(Original post by Ree69)
Say a multiplicative group of order 8 has elements {e, b, b2, b3, a, ab, ab2, ab3} and I want to show that the order of ab2 is 4. We're given here that b4 = e, and a2 = b2 = (ab)2.

In general, can we assume that (ab)n = bnan, for any natural number n?
I think this only works if the multiplication is commutative, so you are free to re-order as you wish.

Otherwise I expect you have to work with abbabbabbabb and attempt to use some of the identities you are given to show that this is e, but no smaller repetition works.
3. Re: A quick question on groups
(Original post by Ree69)
Would it be okay to say (ab2)2 = (b2)2a2
Could be missing something, but what's your justification for that step?

In general, can we assume that (ab)n = bnan, for any natural number n?
If it's true for n=1, then it's true for all n, but you'd have to justify it for n=1; if you can. Though in actuality you can't.
4. Re: A quick question on groups
(Original post by Ree69)
.
You can ONLY use the relations given.

Spoiler:
Show

Last edited by Zii; 19-05-2012 at 21:19.
5. Re: A quick question on groups
(Original post by Ree69)
In general, can we assume that (ab)n = bnan, for any natural number n?
Yes, but who said that these are natural numbers? Or, indeed, numbers at all? (Indeed, they can't be! Even integers under modular arithmetic don't satisfy these rules.)

(Original post by Ree69)
Would it be okay to say (ab2)2 = (b2)2a2 = b4a2 = a2 = b2.
Not really; you certainly can't jump straight to . You should use the rules you've been given.

Start with this: . You know that , and so this is equal to . Now use the fact that again and the fact that .
Last edited by nuodai; 19-05-2012 at 21:17.
6. Re: A quick question on groups
(Original post by Zii)
You can ONLY use the relations given.

Spoiler:
Show

Pedantically, doesn't this show that the order of ab^2 divides 4? So shouldn't you do a couple more checks?
7. Re: A quick question on groups
(Original post by ian.slater)
Pedantically, doesn't this show that the order of ab^2 divides 4? So shouldn't you do a couple more checks?
I suppose so yes. Checking that (ab^2)^2 doesn't equal the identity would be enough though
8. Re: A quick question on groups
Ah, I see. I remember and then incorrectly thought I could apply that to any positive integer.

(Original post by ian.slater)
x
(Original post by ghostwalker)
x
(Original post by Zii)
x
(Original post by nuodai)
x
Does anyone know how to prove that the above group is not commutative? I'm trying to set up a proof by contradiction here, but it's not quite coming out.
9. Re: A quick question on groups
(Original post by Ree69)
Does anyone know how to prove that the above group is not commutative? I'm trying to set up a proof by contradiction here, but it's not quite coming out.
Outline only; with gaps:

Assume it is, and evaluate (ab)^2.

This equals b^4, but also b^2

Last edited by ghostwalker; 20-05-2012 at 13:41.
10. Re: A quick question on groups
so

ie.
since since and has order four so thus isn't equal to .
11. Re: A quick question on groups
(Original post by Jake22)
so

ie.
since since and has order four so thus isn't equal to .
Beautiful, cheers.
12. Re: A quick question on groups
(Original post by nuodai)
Yes, but who said that these are natural numbers? Or, indeed, numbers at all? (Indeed, they can't be! Even integers under modular arithmetic don't satisfy these rules.)
Ree69's saying n is a natural number, not a and b
Last edited by Daniel Freedman; 20-05-2012 at 20:19.
13. Re: A quick question on groups
(Original post by Daniel Freedman)
Ree69's saying n is a natural number, not a and b
Is this group Abelian?
14. Re: A quick question on groups
(Original post by ian.slater)
Is this group Abelian?
No. But I'm not saying (ab)^n = b^n a^n holds in this group
15. Re: A quick question on groups
(Original post by Daniel Freedman)
No. But I'm not saying (ab)^n = b^n a^n holds in this group
No it doesn't...n=1 is an easy counter-example.
16. Re: A quick question on groups
(Original post by Ree69)
No it doesn't...n=1 is an easy counter-example.
.... I feel like three people (who aren't me) have misread posts. Maybe I'm mad
17. Re: A quick question on groups
(Original post by Daniel Freedman)
.... I feel like three people (who aren't me) have misread posts. Maybe I'm mad
Yea, I'm pretty sure I misread yours too - I could've sworn you said that (ab)^n = b^n a^n holds.