1. ## Properties of groups

Can anyone help me with the above problems?

2. 3)
$(a^m)^n = \underbrace{a^m+a^m+\cdots a^m}_{n \mbox{ times}} = \underbrace{\underbrace{a+a+\cdots a}_{m \mbox{ times}}+\underbrace{a+a+\cdots a}_{m \mbox{ times}}+\cdots \underbrace{a+a+\cdots a}_{m \mbox{ times}}}_{n \mbox{ times}}$

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

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

let $|ab|=n$ and assume $|ba|=d$. show $n=d$.
suppose $n>d$.. $(ba)^d = e \Rightarrow \underbrace{baba\cdots ba}_{d \mbox{ times}} = e$

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

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

and continue....

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

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

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

....

I think the equation in 3 needs multiply signs though.

4. multiple signs? what the do you mean?

5. Because (a^m)^n does not equal a^m+a^m+a^m+a^m n times

6. ahh... yeah.. i mean, $\underbrace{a^ma^ma^m\cdots a^m}_{n \mbox{ times}}$... 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..

7. 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.

8. 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..