# Thread: Prove c is an isomorphism iff G is abelian

1. ## Prove c is an isomorphism iff G is abelian

Let G be any group. Define c:G-->G by c(x)=x^-1 for all x in G. prove that c is an isomorphism iff G is abelian.

Let c be an isomorphism. We want to show G is abelian.
Since c is an isomorphism, c is 1-1, onto, and a homomorphism.
We have c(a)=c(b) implies a=b by 1-1
c(a)=a^-1
c(b)=b^-1
a^-1=b^-1
a^-1a=b^-1a
e=b^-1a
be=bb^-1a
b=a
We have c(x)=y by onto
y=c(x)=x^-1
y=x^-1
yx=x^-1x
yx=e
y^-1yx=y^-1e
x=y^-1 onto
Now c(ab)=c(a)c(b)
c(a)c(b)=a^-1b^-1
Since c(ab)=a^-1b^-1
G must be abelian since (ab)^-1=b^-1a^-1

Now suppose G is abelian. We want to show this implies c is an isomorphism.
Since G is abelian, we know ab=ba
1-1 and onto follow by the same ideas as before
c(a)c(b)
a^-1b^-1
Since c is abelian (ab)^-1=a^-1b^-1
c(a)c(b)=c(ab)

I am really not sure if I did any of those correct

2. Originally Posted by kathrynmath
Let G be any group. Define c:G-->G by c(x)=x^-1 for all x in G. prove that c is an isomorphism iff G is abelian.

Let c be an isomorphism. We want to show G is abelian.
Since c is an isomorphism, c is 1-1, onto, and a homomorphism.
We have c(a)=c(b) implies a=b by 1-1
c(a)=a^-1
c(b)=b^-1
a^-1=b^-1
a^-1a=b^-1a
e=b^-1a
be=bb^-1a
b=a
We have c(x)=y by onto
y=c(x)=x^-1
y=x^-1
yx=x^-1x
yx=e
y^-1yx=y^-1e
x=y^-1 onto
Now c(ab)=c(a)c(b)
c(a)c(b)=a^-1b^-1
Since c(ab)=a^-1b^-1
G must be abelian since (ab)^-1=b^-1a^-1

Now suppose G is abelian. We want to show this implies c is an isomorphism.
Since G is abelian, we know ab=ba
1-1 and onto follow by the same ideas as before
c(a)c(b)
a^-1b^-1
Since c is abelian (ab)^-1=a^-1b^-1
c(a)c(b)=c(ab)

I am really not sure if I did any of those correct
I think you're making this harder than it need be.

If $c$ is an isomorphism than for any $x,y\in G$ there are $x',y'$ which map to them. Then, $xy=c(x')c(y')=c(x'y')=(x'y')^{-1}=(y')^{-1}(x')^{-1}$ but $c(x')=x\implies (x')^{-1}=x$ and simillarly for $y$ so that the above states that $xy=yx$

For your second argument I believe the fact that it's a homomorphism from what you said, now show it's bijective.

3. bijective, we have 1-1 , onto
1-1
Assume c(x)=c(y)
Then x^-1=y^-1
xx^-1=xy^-1
e=xy^-1
ey=xy^-1y
y=x

onto:
y=c(x)=x^-1
y=x^-1 solve for x
xy=xx^-1
xy=e
x=y^-1

4. Originally Posted by kathrynmath
bijective, we have 1-1 , onto
1-1
Assume c(x)=c(y)
Then x^-1=y^-1
xx^-1=xy^-1
e=xy^-1
ey=xy^-1y
y=x

onto:
y=c(x)=x^-1
y=x^-1 solve for x
xy=xx^-1
xy=e
x=y^-1
Correct!