My Math Forum Question in abstract algebra

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September 30th, 2017, 11:15 PM   #11
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Quote:
 Originally Posted by mona123 thanks you Could you please help me to show that $(0,0,I_n)\in G'$ cannot be expressed as the product of fewer than $[n/2]$ commutators

No I still don't have the right to left inclusion [I deleted an erroneous proof if you saw it before you replied] and my brain has had enough of this one. At least for tonight.

I think right to left might be the hard part of this exercise. Your question about $I_n$ shows that it's naive to hope that an arbitrary $X$ can be expressed as a commutator. Rather in the general case it's only a product of commutators.

I am not sure how to get started on this. I'm sure the right to left direction would be prerequisite to working on the $I_n$ problem. In other words we know that $(0,0,I_n)\in \{(0,0,X)\}$ but we don't know that $(0,0,I_n)\in G'$ till we have the right to left inclusion.

Last edited by Maschke; October 1st, 2017 at 12:10 AM.

 October 2nd, 2017, 01:25 PM #12 Senior Member   Joined: Jan 2015 From: usa Posts: 101 Thanks: 0 @Maschke, There is no news? you didn't manage to prove the other inclusion?
October 2nd, 2017, 01:43 PM   #13
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Quote:
 Originally Posted by mona123 @Maschke, There is no news? you didn't manage to prove the other inclusion?
I haven't worked on it but I can't imagine how to even get started on that one.

Questions like this would be better off posted on math.stackexchange.com, where you'll find professional mathematicians who can explain how to do a problem like this. On that site you will be asked to show what you've done so far and where you're stuck. It's worth the trouble since that's by far the best place to ask questions like this. Unless ABVictor shows up to help us out!

October 3rd, 2017, 07:13 AM   #14
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Quote:
 Originally Posted by Maschke If we define the subgroup generated by all the commutators, $G' = \langle [x,y] : x, y \in G\}\rangle$, . . .
What did you intend to type there?

October 3rd, 2017, 08:31 AM   #15
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Quote:
 Originally Posted by skipjack What did you intend to type there?
The angle brackets indicate the subgroup generated by the commutators.

Oh I see. I have a curly brace that shouldn't be there. Should be $\langle [x,y] : x, y \in G \rangle$.

Had a bit of a hard time finding a reference.

Quote:
 Originally Posted by Wiki Angle brackets are used in group theory to write group presentations, and to denote the subgroup generated by a collection of elements.
https://en.wikipedia.org/wiki/Bracke...ets_and_groups

Last edited by Maschke; October 3rd, 2017 at 08:39 AM.

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