Question

help plz abelian group

Answer 1

Let G be a group such that\[(a*b)^{2} = a ^{2}*b ^{2 } \] for all \[a,b \in G\] Show that <G,*> is abelian

Answer 2

Woops!! do you have a classmate here?

Answer 3

mayb.who asked it

Answer 4

http://openstudy.com/study#/updates/51336849e4b0034bc1d7b906

Answer 5

Let's summarize it
\[ (a*b)^2 = (a*b)*(a*b)\\
(a*b)*(a*b) = a^2*b^2 = (a*a)*(b*b)\\
a^{-1}*(a*b)*(a*b) = a^2*b^2 = a^{-1}*(a*a)*(b*b)\\
(a^{-1}*a)*b*(a*b) = a^2*b^2 = (a^{-1}*a)*a*(b*b)\\
(e)*b*(a*b) = a^2*b^2 = (e)*a*(b*b) \\
b*(a*b) = a*(b*b)\\
b*(a*b)*b^{-1} = a*(b*b)*b^{-1}\\
b*a*(b*b^{-1}) = a*b*(b*b^{-1})\\
b*a = a*b \]
and to finalize ... I am still very unsure.

Answer 6

why do say a^2 = a*a

Answer 7

yeah that's the assumption. still if it's not then replace it with multiplication.

Answer 8

ok i'll try to digest this

Answer 9

still ... i think a*a = a^2 the way operations are generalized.

Answer 10

a^2 could be 2a if it's multiplicative group.

Answer 11

sorry ..*additive group

Answer 12

okay.am kinda getting a hang of it

Answer 13

meaning a*a = a^2 is not really (a.a)

Answer 14

i guess not \( a\times a \) but \(a*a\)

Answer 15

yes

Answer 16

thank u

Answer 17

I am not so sure ... if i be sure, i'll notify you.

Answer 18

okay