Let a, b be in group G, and let n be an integer. Prove....

[Anood]

Can you please help me with the following question? I think induction is one way to do it and I have to take care of three cases, those being when n is a positive number, when n is a negative number, and when n is zero. But I don't know how to do these. Could somebody please help?

The question is:

Let G be a group. Let a, b in G, and let n be (not necessarliy positive) integer. Prove that, if a and b commute, then (ab)^n = (a)^n (b)^n

 

[pka]

I will give some very general help. Given that ab=ba prove by induction that \(\displaystyle b^n a = ab^n .\) To do note that if:

. . .\(\displaystyle \begin{array}{rcl}
b^K a = ab^K \Rightarrow \quad b^{K + 1} a & = & b^K \left( {ba} \right) \\
& = & b^K \left( {ab} \right) \\
& = & \left( {b^K a} \right)b \\
& = & \left( {ab^K } \right)b \\
& = & ab^{K + 1} \\
\end{array}.\)

Then we have:

. . .\(\displaystyle \begin{array}{rcl}
\left( {ab} \right)^K = a^K b^K \Rightarrow \left( {ab} \right)^{K + 1} & = & \left( {ab} \right)^K \left( {ab} \right) \\
& = & \left( {a^K b^K } \right)\left( {ab} \right) \\
& = & \left( {a^K b^K } \right)\left( {ba} \right) \\
& = & \left( {a^K b^{K + 1} a} \right) \\
& = & \left( {a^{K + 1} b^{K + 1} } \right) \\
\end{array}\)

 

[Anood]

Thanks alot