MHB Showing that a group is non-abelian

[Guest2]

$4$. If G is a group in which $(a \cdot b)^i = a^i \cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show that $G$ is abelian.

I've done this one. The next one says:

$5$. Show that the conclusion of Problem $4$ does not follow if we assume the relation $(a \cdot b)^i = a^i \cdot b^i$ for just two consecutive integers.

All I can show is that this implies $a^i b = ba^i.$ Any suggestions are appreciated.

 

[Deveno]

What you have written is not true.

Here are two statements that are true:

1) If $G$ is a group such that $(ab)^k = a^kb^k$ for $3$ consecutive integers, $G$ is abelian.

2) If $G$ is a group such that $(ab)^k = a^kb^k$ for $2$ consecutive integers, $G$ may not be abelian.

If #2) is what you meant, then what you are being asked to do is exhibit a counter-example, that is, a non-abelian group such that 2) holds (note that #2) is true of every abelian group, by your previous post).

 

[Guest2]

Thank you, Deveno. I tried to rephrase the question; statement of the problem was in fact that the conclusion of 1) does not follow if we take just two consecutive integers instead. I'm going to have to think about how to exhibit a counterexample to this.

 

[Guest2]

What you have written is not true.

Here are two statements that are true:

1) If $G$ is a group such that $(ab)^k = a^kb^k$ for $3$ consecutive integers, $G$ is abelian.

2) If $G$ is a group such that $(ab)^k = a^kb^k$ for $2$ consecutive integers, $G$ may not be abelian.

If #2) is what you meant, then what you are being asked to do is exhibit a counter-example, that is, a non-abelian group such that 2) holds (note that #2) is true of every abelian group, by your previous post).

Is this correct? Let $i = k-2, k-1.$ Then by (1) we have have that $(ab)^k = a^kb^k$ implies $G$ is abelian. Also, from my previous post, we have that if $G$ is an abelian group, then for $a, b \in G$ and all integers $n$, $(a \cdot b)^n = a^n \cdot b^n$; and in particular when $n=k.$ Thus $G$ is abelian if and only if $(ab)^k = a^kb^k.$ Thus we have necessary and sufficient condition, so it might not be true for just two consecutive integers.

 

[Deveno]

You can't use 1) in that way, the proof of that requires $3$ consecutive integers, not two.

What you need is to exhibit a non-abelian group $G$ such that $(ab)^i = a^ib^i$ for two consecutive integers $i$, and every $a,b \in G$.

I suggest you pick $G$, and your two values for $i$ carefully. Here is a hint: -1 and 2 would be bad choices.

 

[Guest2]

You can't use 1) in that way, the proof of that requires $3$ consecutive integers, not two.

But I'm using it for three integers. $k-2, k-1$ and $k$.

What you need is to exhibit a non-abelian group $G$ such that $(ab)^i = a^ib^i$ for two consecutive integers $i$, and every $a,b \in G$.

I suggest you pick $G$, and your two values for $i$ carefully. Here is a hint: -1 and 2 would be bad choices.

I take $i=0$ and $i=1$. I get $ab = ab$ which is a counterexample, right?

 

[Deveno]

But I'm using it for three integers. $k-2, k-1$ and $k$.

Unfortunately, you omitted the last $k$ in your prior post-this may have been what you intended, but, alas, my psychic powers of divination are not what they used to be.

I take $i=0$ and $i=1$. I get $ab = ab$ which is a counterexample, right?

Those are good choices for $i$, but you also need some non-abelian group to complete the proof.

$ab = ab$ is an equation, not a proof; to be convincing you need to say something like this:

Let $G$ be a non-abelian group. Then for any two elements $a,b$ (and thus all), we have:

$(ab)^0 = a^0b^0$ because...(insert reason here) and

$(ab)^1 = a^1b^1$ (this might be obvious, but it doesn't hurt to add a little something to the beginning and end)

and $0,1$ are two consecutive integers, therefore...(insert conclusion here).

 

[Guest2]

Unfortunately, you omitted the last $k$ in your prior post-this may have been what you intended, but, alas, my psychic powers of divination are not what they used to be.

(Rofl)

Sorry about that.

Those are good choices for $i$, but you also need some non-abelian group to complete the proof.

$ab = ab$ is an equation, not a proof; to be convincing you need to say something like this:

Let $G$ be a non-abelian group. Then for any two elements $a,b$ (and thus all), we have:

$(ab)^0 = a^0b^0$ because...(insert reason here) and

$(ab)^1 = a^1b^1$ (this might be obvious, but it doesn't hurt to add a little something to the beginning and end)

and $0,1$ are two consecutive integers, therefore...(insert conclusion here).

Thank you very much, this is most helpful!