[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.

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?

[ian.slater]

(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.

[ghostwalker]

(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.

[Zii]

(Original post by Ree69)
.

You can ONLY use the relations given.

Spoiler:

Show

[nuodai]

(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 .

[ian.slater]

(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?

[Zii]

(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

[Ree69]

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.

[ghostwalker]

(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

Contradicting the order of b.

[Jake22]

so



ie.
since since and has order four so thus isn't equal to .

[Ree69]

(Original post by Jake22)
so



ie.
since since and has order four so thus isn't equal to .

Beautiful, cheers.

[Daniel Freedman]

(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

[ian.slater]

(Original post by Daniel Freedman)
Ree69's saying n is a natural number, not a and b

Is this group Abelian?

[Daniel Freedman]

(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

[Ree69]

(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.

[Daniel Freedman]

(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

[Ree69]

(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.