http://img10.imageshack.us/img10/1586/mathsw.jpg

http://img19.imageshack.us/img19/3176/maths2.jpg

Can anyone help me with the above problems?

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- Mar 5th 2009, 12:42 PMiain3915Properties of groups
http://img10.imageshack.us/img10/1586/mathsw.jpg

http://img19.imageshack.us/img19/3176/maths2.jpg

Can anyone help me with the above problems? - Mar 5th 2009, 04:13 PMkalagota
3)

$\displaystyle (a^m)^n = \underbrace{a^m+a^m+\cdots a^m}_{n \mbox{ times}} = \underbrace{\underbrace{a+a+\cdots a}_{m \mbox{ times}}+\underbrace{a+a+\cdots a}_{m \mbox{ times}}+\cdots \underbrace{a+a+\cdots a}_{m \mbox{ times}}}_{n \mbox{ times}}$

$\displaystyle = \underbrace{a+a+\cdots a}_{mn \mbox{ times}} = a^{mn} $

6) this has been proven many times here.. but anyways,

let $\displaystyle |ab|=n$ and assume $\displaystyle |ba|=d$. show $\displaystyle n=d$.

suppose $\displaystyle n>d$.. $\displaystyle (ba)^d = e \Rightarrow \underbrace{baba\cdots ba}_{d \mbox{ times}} = e$

$\displaystyle b^{-1}\underbrace{baba\cdots ba}_{d \mbox{ times}}\underbrace{baba\cdots ba}_{n-d \mbox{ times}}b = b^{-1}\underbrace{baba\cdots ba}_{n-d \mbox{ times}}b$

$\displaystyle e=\underbrace{aba\cdots bab}_{n \mbox{ times}}= \underbrace{aba\cdots bab}_{n-d \mbox{ times}}$

and continue....

if $\displaystyle n<d$

$\displaystyle \underbrace{baba\cdots ba}_{d \mbox{ times}} = e$

$\displaystyle \underbrace{ba\cdots b\underbrace{aba\cdots bab}_{n \mbox{ times}}a\cdots ba}_{d \mbox{ times}} = e$

$\displaystyle babae\cdots ba=\underbrace{baba\cdots ba}_{d-n \mbox{ times}} = e$

.... - Mar 5th 2009, 04:31 PMiain3915
Thanks for the reply.

I think the equation in 3 needs multiply signs though. - Mar 5th 2009, 04:53 PMkalagota
multiple signs? what the do you mean?

- Mar 5th 2009, 04:58 PMiain3915
Because (a^m)^n does not equal a^m+a^m+a^m+a^m n times

- Mar 5th 2009, 04:59 PMkalagota
ahh... yeah.. i mean, $\displaystyle \underbrace{a^ma^ma^m\cdots a^m}_{n \mbox{ times}}$... that should be it.. sorry for that..

and, all you have to do is to remove that + signs..

but anyways, if i assume that the operator is denoted by +, everything should be fine.. it is just a matter of how you denote your operator.. - Mar 5th 2009, 05:02 PMiain3915
Do you know of an alternative way of doing these proofs without using all the n times sort of things?

It seems a bit ambiguous as it is. - Mar 5th 2009, 05:05 PMkalagota
it's the best proof and you will see that proof in many references.. (and i can't think of another for now..)

have you read my notes in the previous post? i edited it..