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I have been working on this for the past week and still can not get the answer. This is dealing with Groups. THANKS!

Yiorgos S. Smyrlis
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Jewelss
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4 Answers4

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$(ab)^{|a||b|}= (a^{|a|}b^{|a|})^{|b|}$ $ = (e b^{|a|}) ^{|b|}$ $= b^{|a|||b}$ $= (b^{|b|})^{|a|}$ $= e^{|a|}$ $=e$

after the first "=" that is from the fact that $ab=ba$

Liddo
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    I was completely puzzled by this question-until I remembered the absolute value sign in group theory means the order of the element. It's true-if you don't use something for awhile,you lose it.That and the fact the elements commute are the key facts here-I'm not sure,but I think this was given as an exercise in Herstien's classic TOPICS IN ALGEBRA. – Mathemagician1234 Jan 28 '14 at 22:47
  • My just for curiosity, if $|a|=|b|=\infty$, would the proof be exactly the same? – Liddo Jan 29 '14 at 18:29
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Well, because $a$ and $b$ commute, you can say: $$ (ab)^{|a||b|} = \underset{|a||b|-times}{\underbrace{ab\cdots ab}} = \underset{|a||b|-times}{\underbrace{a\cdots a}}\cdot\underset{|a||b|-times}{\underbrace{b\cdots b}} = a^{|a||b|}b^{|a||b|} = (a^{|a|})^{|b|}(b^{|b|})^{|a|} = e^{|b|}e^{|a|} = e $$

Alex R.
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$(ab)^{|a||b|}=a^{|a||b|}b^{|a||b|}=e$ if $ab=ba$.

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my suggestion: prove two lemmas before proving the question.

Lemma 1: if $a,b\in G$ and $ab=ba$, then $(a b)^2 = a^2 b^2$.

Lemma 2: if $a,b\in G$ and $ab=ba$, then $\forall k \in \mathbb{N} (ab)^k = a^k b^k$. (Trivially by Induction)

Corollary: $(ab)^{|a||b|} = ((ab)^{|a|})^{|b|} = ((a)^{|a|} (b)^{|a|})^{|b|} = e ((b)^{|a|})^{|b|} = ((b)^{|b|})^{|a|}= e$.

Igor Caetano Diniz
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