# Horrible problem about abelian groups

1. Sep 12, 2011

### Oster

G is a group and for all elements a,b in G,
(ab)^i = (a^i)(b^i) holds for 3 consecutive positive integers.
Show that G is abelian.

I know how to prove that if (ab)^2 = (a^2)(b^2) then G is abelian. I was thinking that you could reduce the given equality integer by integer till 2 or something like that, but all I have been able to do so far is use the cancellation laws to get some gibberish. HAALP!

2. Sep 12, 2011

### micromass

Staff Emeritus
OK, we can do this all formally, but it's best to do with an example. Let's assume that

aaaabbbb = abababab

aaaaabbbbb = ababababab

aaaaaabbbbbb = abababababab

Let's deal with the first two equalities first. Substitute abababab=aaaabbbb into aaaaabbbbb.
Can you use this to prove that abababab = babababa ??

3. Sep 12, 2011

### Oster

baabaa black sheep

4. Sep 12, 2011

### Oster

starting with the 2nd equation, using the left and right cancellation laws on a and b you get
aaaabbbb =babababa. Using this is with the first equation yields babababa=abababab

5. Sep 12, 2011

### micromass

Staff Emeritus
OK, now do the same with the second and the third equation.

6. Sep 12, 2011

### Oster

3rd equation => aaaaabbbbb=bababababa
plugging into the 2nd gives bababababa=ababababab
then plug in babababa=abababab to cancel and get ab=ba
=D
thank you@!

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