prove that by mathematical induction

If(G, *) is an abelian group, then for all a, b € G, prove that by mathematical induction -

Mumbai University > Computer Engineering > Sem 3 > Discrete Structures

Marks: 5 Marks

Year: Dec 2013

1 Answer


  1. n=0. we have $(ab)^0=e \hspace{5cm} \text{[by definition}]$

    Also $a^0b^0=ee=e.\\ \therefore (ab)^0=a^0b^0.$

  2. n > 0. If n=1, then $(ab)^1=ab=a^1b^1.$

    Now suppose for n=k, $(ab)^k=a^kb^k$

    Then $(a+b)^{k+1}$ $=(ab)^k(ab)=a^kb^kab=a^kab^kb \hspace{3cm} [\because \text{G is abelian} \Rightarrow b^ka=ab^k] \\ =a^{k+1}b^{k+1}$

    Thus the result is ture for n=k+1 if it was ture for n=k. But it is ture for n =1. Hence by hematical induction for all $n \gt 0, (ab)^n=a^nb^n.$

  3. n < 0. Let n=-m where m is a positive integer.

    then $(ab)^n$ $=(ab)^{-m}=[(ab)^m]^{-1}=(a^mb^m)^{-1}=(b^ma^m)^{-1} \hspace{1.5cm} [\because \text{G is abelian} \Rightarrow a^mb^m=b^ma^m] \\ =(a^m)^{-1}(b^m)^{-1} \hspace{9.3cm}[\because (ab)^{-1}=b^{-1}a^{-1}] \\ =a^{-m}b^{-m}=a^nb^n.$

Note: Since a,b $\in$ G, Therefore $a^m,b^m \in G$

Also $\text{G is abelian} \therefore a^mb^m=b^ma^m$

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