Hi, this question is for my modern algebra class. the question is if a1,a2,a3...,an belong to a group what is the inverse of a1,a2,a3...,an? Justify. So what i did was just (a1,a2,a3...an)( a1^-1,a2^-1,a3^-1...an^-1)=e. Is that all need to show? it seems too easy. Any help would be appreciated! thanks!
if a1,a2,a3...,an belong to a group what is the inverse?
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pka
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Well this time you wrote it out with no commas in the last expression. In that case, it must be a product and \(\displaystyle \left( {a_1 a_2 a_3 \cdot \cdot \cdot a_n } \right)^{ - 1} = \left( {a_n } \right)^{ - 1} \left( {a_{n - 1} } \right)^{ - 1} \cdot \cdot \cdot \left( {a_2 } \right)^{ - 1} \left( {a_1 } \right)^{ - 1}.\)
you're right pka. I guess I didn't read the problem carefully the first time. I was wondering can you say (a1a2a3...an)^-1= (a1)^-1(a2)^-1(a3)^-1...(an-1)^-1,(an)^-1. I was wondering why you wrote it in backwords order? Also I'm having trouble with n-1 why did you put that there, does it need to be there. thanks!
pka
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In any group we have \(\displaystyle \left( {ab} \right)^{ - 1} = b^{ - 1} a^{ - 1} .\)
You do realize that in general groups are not commutative?
In commutative groups, Abelian groups, that order does not matter.
However, you cannot assume this is an Abelian group.
You do realize that in general groups are not commutative?
In commutative groups, Abelian groups, that order does not matter.
However, you cannot assume this is an Abelian group.