Proving the Necessity of Commutativity for Automorphisms in Groups

[johnnyboy2005]

So they ask to show that each element of a group can be mapped to it's inverse is an automorphism only if the group is abelian.

i can figure it out why each element is abelian to it's inverse, but i am short of understaning why every element must be commutable. any tips/ hints? thanks

 

[matt grime]

"abelian to its inverse"? what does that mean?

I can't think of a tip that doesn't just give you the answer., so we'll have to go for the standard tips for ANY question.

What does it mean if inversion is an automorphism. ie what is the definition of automorphism, heck, just show it's not a homomorphism. write out this condition: if f is a homomorphism then f(xy)=f(x)f(y). Clearly this will show where the problem is.,

 

[johnnyboy2005]

okay, i get that, but it says show this is true if and only if G is abelian.

 

[matt grime]

Erm, yes, well, haver you written out what this means exactly when f(x)=x^{-1}? Because that is all you have to do.

 

[math-chick_41]

just show 1-1 onto and op.
1-1, easy
onto, easy take x=y^-1
op. f(xy)=(xy)^-1=(y^-1)(x^-1)=(x^-1)(y^-1)=f(x)f(y)

 

[matt grime]

In what way are you claiming that is a proof of the stated question? (nb we're not supposed to just hand out answers, especially if the OP has not written out their work and indicated where they're stuck and why)

 

[johnnyboy2005]

this is what I'm saying so far

since everyelement in G is mapped to it's inverse, and since the following is true alpha(ab)-->(ab)^-1 --->(ba)^-1-->alpha(ba) then G must be abelian for this to be true. ...make sense or am i missing something?

 

[matt grime]

I think you have the right idea, though what the --> means is puzzling me, and that sequence, well, I'm not sure how to interpret it: if it is a homomorphism, then it follows that

(xy)^{-1]}=x^{-1}y^{-1}

however we know that (xy)^{-1}= y^{-1}x^{-1}

so...?

People really ought to use more words when answering questions.