homomorphism proof

[Hansel13]

Let G be a group. Define the function f: G-->G by f(g) = g[sup:1uyyaesy]-1[/sup:1uyyaesy]

Prove: if f is a homomorphism, then G is commutative.

I completely missed this problem on the exam, but I want to be able to redo it for extra credit. I don't even know where to start, can anyone help?

 

[daon]

You assume that your given function f is a homomorphism from G to itself. Then it is asking you to prove that any two elements of G commute with one another.

i.e. show that for each x and each y in g, xy=yx. Since it is a homomorphism, it is opperation preserving. That is a good place to start.

 

[Hansel13]

OK, so...
f(a)f(b)=a[sup:v9plbdn9]=1[/sup:v9plbdn9]b[sup:v9plbdn9]-1[/sup:v9plbdn9]
and
f(b)f(a)=b[sup:v9plbdn9]=1[/sup:v9plbdn9]a[sup:v9plbdn9]-1[/sup:v9plbdn9]

And since it's operation preserving:
f(a)f(b) = f(a*b)=?
f(b)f(a) = f(b*a)=?

This is kinda where I get stuck. How does knowing operation preservation help me figure out what f(a*b) or f(b*a)??

thanks

 

[daon]

The image of an element g is the inverse of g.

\(\displaystyle f(ab) = (ab)^{-1} = b^{-1}a^{-1}\)

 

[Hansel13]

Ha, i'm so dumb! OK so..

f(a)f(b)=a[sup:6zpgsjff]-1[/sup:6zpgsjff]b[sup:6zpgsjff]-1[/sup:6zpgsjff]
f(b)f(a)=b[sup:6zpgsjff]-1[/sup:6zpgsjff]a[sup:6zpgsjff]-1[/sup:6zpgsjff]

And since it's operation preserving:
f(a)f(b) = f(a*b) = (ab)[sup:6zpgsjff]-1[/sup:6zpgsjff] = b[sup:6zpgsjff]-1[/sup:6zpgsjff]a[sup:6zpgsjff]-1[/sup:6zpgsjff]
f(b)f(a) = f(b*a) = (ba)[sup:6zpgsjff]-1[/sup:6zpgsjff] = a[sup:6zpgsjff]-1[/sup:6zpgsjff]b[sup:6zpgsjff]-1[/sup:6zpgsjff]

So we see:

f(a)f(b)=a[sup:6zpgsjff]-1[/sup:6zpgsjff]b[sup:6zpgsjff]-1[/sup:6zpgsjff] = b[sup:6zpgsjff]-1[/sup:6zpgsjff]a[sup:6zpgsjff]-1[/sup:6zpgsjff]
&
f(b)f(a)=b[sup:6zpgsjff]-1[/sup:6zpgsjff]a[sup:6zpgsjff]-1[/sup:6zpgsjff] = a[sup:6zpgsjff]-1[/sup:6zpgsjff]b[sup:6zpgsjff]-1[/sup:6zpgsjff]

Which shows G is commutative.

Thanks

 

[daon]

Good, you got it. One caveat though. You showed for all \(\displaystyle a,b \in G, \,\, \,\, a^{-1}b^{-1} = b^{-1}a^{-1}\). This actually does prove that G is commutative (since every element is an inverse and hence may be represeted by one), but it isn't very clear--and certainly won't be obviously equivilant to someone with little experience reading it.

What you have is equivilant to saying \(\displaystyle f(a)f(b)=f(b)f(a) \implies f(ab)=f(ba)\) for all a,b in G. If you can show G is 1-1 then you get the result \(\displaystyle ab=ba\) which is really what you want. The fact that G is 1-1 is hidden somewhere here in this post (note also that it is surjective but that is not necessary here) .

 

[DrMike]

Ha, i'm so dumb! OK so..

f(a)f(b)=a[sup:3agrrakj]-1[/sup:3agrrakj]b[sup:3agrrakj]-1[/sup:3agrrakj]
f(b)f(a)=b[sup:3agrrakj]-1[/sup:3agrrakj]a[sup:3agrrakj]-1[/sup:3agrrakj]

And since it's operation preserving:
f(a)f(b) = f(a*b) = (ab)[sup:3agrrakj]-1[/sup:3agrrakj] = b[sup:3agrrakj]-1[/sup:3agrrakj]a[sup:3agrrakj]-1[/sup:3agrrakj]
f(b)f(a) = f(b*a) = (ba)[sup:3agrrakj]-1[/sup:3agrrakj] = a[sup:3agrrakj]-1[/sup:3agrrakj]b[sup:3agrrakj]-1[/sup:3agrrakj]

So we see:

f(a)f(b)=a[sup:3agrrakj]-1[/sup:3agrrakj]b[sup:3agrrakj]-1[/sup:3agrrakj] = b[sup:3agrrakj]-1[/sup:3agrrakj]a[sup:3agrrakj]-1[/sup:3agrrakj]
&
f(b)f(a)=b[sup:3agrrakj]-1[/sup:3agrrakj]a[sup:3agrrakj]-1[/sup:3agrrakj] = a[sup:3agrrakj]-1[/sup:3agrrakj]b[sup:3agrrakj]-1[/sup:3agrrakj]

Which shows G is commutative.

Thanks

Yes! But you actually proved it twice... You could have just said :

Ha, i'm so dumb! OK so..

f(a)f(b)=a[sup:3agrrakj]-1[/sup:3agrrakj]b[sup:3agrrakj]-1[/sup:3agrrakj]

And since it's operation preserving:
f(a)f(b) = f(a*b) = (ab)[sup:3agrrakj]-1[/sup:3agrrakj] = b[sup:3agrrakj]-1[/sup:3agrrakj]a[sup:3agrrakj]-1[/sup:3agrrakj]

So we see:

f(a)f(b)=a[sup:3agrrakj]-1[/sup:3agrrakj]b[sup:3agrrakj]-1[/sup:3agrrakj] = b[sup:3agrrakj]-1[/sup:3agrrakj]a[sup:3agrrakj]-1[/sup:3agrrakj]

Which shows G is commutative.

Or, re-written it like this :

ab=f(a[sup:3agrrakj]-1[/sup:3agrrakj])f(b[sup:3agrrakj]-1[/sup:3agrrakj])=f(a[sup:3agrrakj]-1[/sup:3agrrakj]b[sup:3agrrakj]-1[/sup:3agrrakj])=f((ba)[sup:3agrrakj]-1[/sup:3agrrakj])=ba.