This was a challange question and I am not sure how to do this.. help would be appreciated thxhttp://img17.imageshack.us/img17/5006/83668755.jpg

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- Mar 17th 2009, 08:53 AMrajrRing and Groups
This was a challange question and I am not sure how to do this.. help would be appreciated thxhttp://img17.imageshack.us/img17/5006/83668755.jpg

- Mar 17th 2009, 09:39 AMstapel
$\displaystyle \mbox{a) }\, (a\, +\, b)^2\, =\, (a\, +\, b)(a\, +\, b)\, =\, (a\, +\, b)a\, +\, (a\, +\, b)b$

$\displaystyle =\, a^2\, +\, ba\, +\, ab\, + \, b^2\, =\, a\, +\, ba\, +\, ab\, +\, b\, =\, (a\, +\, b)$

Subtract $\displaystyle (a\, +\, b)$ from either side. See if that leads anywhere helpful...?

$\displaystyle \mbox{b) }\, (ab)^2\, =\, (ab)(ab)\, =\, abab\, =\, a^2 b^2$

$\displaystyle a^{-1}ababb^{-1}\, =\, a^{-1}a^2 b^2 b^{-1}$

$\displaystyle ebae\, =\, eabe$

...where $\displaystyle e$ is the group's identity element. The result follows immediately. (Wink) - Mar 17th 2009, 09:48 AMrajr
thx for the reply and I can understand the second part and for the first part if I takeaway a+b from both sides then I will get ab = -ba but is it ok to say that it is commutative just from that step or do I have to go further any more steps to prove it

I think for the part (ii) I have to do like as follows:

b(a+a) = 0;

therefore a+a =0 is it right

thx once again for ur help(Happy)