- #1
RJLiberator
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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.