Can anyone help me with the above problems?
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 $\displaystyle n<d$
$\displaystyle \underbrace{baba\cdots ba}_{d \mbox{ times}} = e$
$\displaystyle \underbrace{ba\cdots b\underbrace{aba\cdots bab}_{n \mbox{ times}}a\cdots ba}_{d \mbox{ times}} = e$
$\displaystyle babae\cdots ba=\underbrace{baba\cdots ba}_{d-n \mbox{ times}} = e$
....
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..