Abstract Algebra

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Prove that a group in which every element except the identity element has order 2 is abelian.

...



for ...

I can't seem to get

thanks...

[QuantumJG]

Well I'm doing group theory next semester so I 'should' know the answer to this by next semester.

[/0]

Cool, I think group theory is gonna be pretty interesting. I've started reading Michael Artin's 'Algebra', which we're using next semester. Do you know which text you're using?

[QuantumJG]

Our prescribed text is only the lecture notes. I'm also looking forward to this subject (despite looking at the exam and perceiving this to be my hardest maths subject), I'm also doing electromagnetism and optics and I'm also doing vector calculus (should help with electromagnetism and my previous physics subjects have kind of suffered from not having this background).

[humph]

Prove that a group in which every element except the identity element has order 2 is abelian.

...



for ...

I can't seem to get

thanks...

Hint: show that for any (possibly nonabelian) group , for all .

Borger is a pretty good lecturer. Weird that he's changing texts though - the set text used to be Fraleigh. Dummit and Foote is supposed to be a decent abstract algebra text too.

[Ahmad]

Hint:

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Thanks!,









But

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How would one go about proving block multiplication for general matrices? It just seems to be such a massive task, to prove it for every shape of matrix and given the complexity of the matrix multiplication formula, but it's an exercise in the book. :|
Does it require some kind of induction on matrices, how would this be done?

Borger is a pretty good lecturer. Weird that he's changing texts though - the set text used to be Fraleigh. Dummit and Foote is supposed to be a decent abstract algebra text too.

Cool... I hope Borger doesn't endlessly copy proofs from the book like someone I won't mention...
I just discovered that Michael Artin is the son of Emil Artin, who was a famous number theorist. He also lectures at MIT, which has a reputation for good books xD

[humph]

How would one go about proving block multiplication for general matrices? It just seems to be such a massive task, to prove it for every shape of matrix and given the complexity of the matrix multiplication formula, but it's an exercise in the book. :|
Does it require some kind of induction on matrices, how would this be done?

Induction certainly isn't the way to go. You know that you can actually write matrix multiplication in terms of summations for each entry, but it's quite ugly. I reckon the way to go would be to think of matrix multiplication in terms of the dot product, or the product of a matrix with a vector (as this gives you a scalar and a vector respectively, as opposed to a matrix), and decompose it in that way. But that might be a bit too obscure a hint.

Borger is a pretty good lecturer. Weird that he's changing texts though - the set text used to be Fraleigh. Dummit and Foote is supposed to be a decent abstract algebra text too.

Cool... I hope Borger doesn't endlessly copy proofs from the book like someone I won't mention...
I just discovered that Michael Artin is the son of Emil Artin, who was a famous number theorist. He also lectures at MIT, which has a reputation for good books xD

Oh, who's the one who copies proofs from the book?
Yeah I've gotten the two Artins confused a couple of times. Both are pretty well-reknowned for their work in algebra (and Galois theory in particular).

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Thanks humph, I might return to that problem a bit later though...
I thought my Analysis professor did a lot of proof copying... but then, perhaps I'm just not used to that style of teaching. Analysis proofs give me headaches.


Anyway...

With questions like:
"Prove that the inverse of an element in a subgroup is the same as the inverse of that element in "
Are the proofs meant to be as trivial as this:
Suppose . Let and .
Then since , too.
So you have ... and by taking inverses


"Show by example that the product of elements of finite order in a nonabelian group need not have finite order."

Hmmm... I thought for a while permutations groups might work... but now I'm sure they won't, since for permutations and , there exists a permutation , which has finite order.

Pretty much the only other non-abelian group I know is ... probably not worth it?

[humph]

Thanks humph, I might return to that problem a bit later though...
I thought my Analysis professor did a lot of proof copying... but then, perhaps I'm just not used to that style of teaching. Analysis proofs give me headaches.

Analysis proofs do indeed induce headaches. Not sure about proof copying, a lot of the time there's only so many ways you can prove some statement, so it's not so much stealing as proving it the natural way...

Anyway...

With questions like:
"Prove that the inverse of an element in a subgroup is the same as the inverse of that element in "
Are the proofs meant to be as trivial as this:
Suppose . Let and .
Then since , too.
So you have ... and by taking inverses

Yeah, that's pretty much it. You should mention that the identity in a subgroup is the same as that in the group, and the last step isn't taking inverses, but simply by the uniqueness of inverses.

"Show by example that the product of elements of finite order in a nonabelian group need not have finite order."

Hmmm... I thought for a while permutations groups might work... but now I'm sure they won't, since for permutations and , there exists a permutation , which has finite order.

Pretty much the only other non-abelian group I know is ... probably not worth it?

It's clearly not going to work with permutation groups, because they have finite order, so every element in them must have finite order. So you'll need some sort of nonabelian infinite group. Matrix groups are really the obvious examples, unfortunately - you probably want , as everything interesting that happens with nonabelian infinite groups should happen in this group. Mind you, it could take a bit of effort to find the example that you want (and I really can't be bothered...).

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Thanks again humph
(I didn't mean 'copying' like that, I guess what I meant was we could just as well get a lot of proofs from the book, but examples would be helpful)

TI-89 to the rescue!

    ,     

[humph]

Thanks again humph
(I didn't mean 'copying' like that, I guess what I meant was we could just as well get a lot of proofs from the book, but examples would be helpful)

TI-89 to the rescue!

    ,     

How do you actually prove the last statement? (Hint: look up nilpotent matrices. This is covered very briefly in MATH1115.)

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I guess I would proceed by induction.

It appears that



Where are Fibonacci numbers

Then I would use induction




I'm not sure how you would use nilpotent matrices however... since none of the matrices here seem to be nilpotent.

[zzdfa]

Another way you could prove that is to write and then use the binomial theorem to expand that.