Proving a function is a bijection and isomorphic

[RJLiberator]

Homework Statement

If G is a group and a ∈ G, let π: G--> G be the function defined as π(g) = ag, for all g ∈G.
a) Show that π is a bijection
b) Show that if π is an isomorphism, then a is the identity element of G.

Homework Equations

I think to show that pi is a bijection we have to show that it is surjective and injective.
To show that it is an isomorphism we have to show that π(xy) = π(x)π(y).

The Attempt at a Solution

[/B]
First, I do part a and show that it is one to one:
Suppose π(g) = π(g')
then ag = ag'
Since a ∈G we know a^(-1) exists in G.
a^(-1)ag = a^(-1)ag'
g = g'
and so π is one to one.

I am not sure how to prove that this is onto... any hints here would help.
For onto can I just say
Let x,y ∈ G and x = a^(-1)y
since a is a part of the group we know that a^(-1) exists.
And thus it follows that π(x) = y

To show part b)
if x,y ∈G, then we observe
π(xy) = axy
π(x)π(y) = axay

Here, a must equal the identity element of G for this function to be an isomorphism.

 

[Samy_A]

Homework Statement

If G is a group and a ∈ G, let π: G--> G be the function defined as π(g) = ag, for all g ∈G.
a) Show that π is a bijection
b) Show that if π is an isomorphism, then a is the identity element of G.

Homework Equations

I think to show that pi is a bijection we have to show that it is surjective and injective.
To show that it is an isomorphism we have to show that π(xy) = π(x)π(y).

The Attempt at a Solution

[/B]
First, I do part a and show that it is one to one:
Suppose π(g) = π(g')
then ag = ag'
Since a ∈G we know a^(-1) exists in G.
a^(-1)ag = a^(-1)ag'
g = g'
and so π is one to one.

I am not sure how to prove that this is onto... any hints here would help.
For onto can I just say
Let x,y ∈ G and x = a^(-1)y
since a is a part of the group we know that a^(-1) exists.
And thus it follows that π(x) = y

To show part b)
if x,y ∈G, then we observe
π(xy) = axy
π(x)π(y) = axay

Here, a must equal the identity element of G for this function to be an isomorphism.

Mainly correct.

Only the last point needs some explanation.
You have that, for π to be an isomorphism, axy = axay, for all x,y ∈ G.
Why does that imply that a is the identity element of G?

 

[RJLiberator]

Hm, well, it implies that a is the identity element of G as it would mean that a=aa
the only thing that works here is the identity element.

 

[RJLiberator]

Maybe you want it more explicit?

axy = axay

y^(-1) exists as y is a part of G so multiple both sides by it

axy*y^(-1) = axay*y^(-1)

ax = axa

a^(-1) exists as a is a part of G
a^(-1)ax = a^(-1)axa
x = xa
so now a has to be the identity element

 

[Samy_A]

Hm, well, it implies that a is the identity element of G as it would mean that a=aa
the only thing that works here is the identity element.

Voila!