The group operation on G*G is multiplication.

[Justabeginner]

Homework Statement

Let G be a group, A = G * G. In A, Let T = {(g, g)|g ε G}. Prove that T is isomorphic to G.

Homework Equations

The Attempt at a Solution

A is abelian. Therefore, G * G is abelian. T is a subgroup of G.

I am not sure if my above inferences are even correct. Can someone guide me as to the thought process on this please? Thank you.

 

[Quesadilla]

I don't see how you could have inferred any of those, or why you would need to.

You are asked to show that $G$ and $T$ are isomorphic. There is an obvious candidate for an isomorphism, so you should just verify that it actually is one.

 

[LCKurtz]

And for that matter, you haven't told us what G*G means.

 

[Justabeginner]

And for that matter, you haven't told us what G*G means.

A = G * G, as in G cross G.
I do not understand how T is directly isomorphic to G.

 

[vela]

Homework Statement

Let G be a group, A = G * G. In A, Let T = {(g, g)|g ε G}. Prove that T is isomorphic to G.

Homework Equations

The Attempt at a Solution

A is abelian. Therefore, G * G is abelian. T is a subgroup of G.

How'd you get A is abelian?

If A is abelian, then obviously GxG is abelian since A=GxG.

How can T be a subgroup of G when it's not a subset of G? It's a subset of A, right?

How did you define the group multiplication for A?

I am not sure if my above inferences are even correct. Can someone guide me as to the thought process on this please? Thank you.

 

[pasmith]

Homework Statement

Let G be a group, A = G * G. In A, Let T = {(g, g)|g ε G}. Prove that T is isomorphic to G.

Homework Equations

The Attempt at a Solution

A is abelian.

There is nothing in the question to suggest this. If G is not abelian then A will not be abelian.

Therefore, G * G is abelian. T is a subgroup of G.

Actually T is a subgroup of A.

I am not sure if my above inferences are even correct. Can someone guide me as to the thought process on this please?

You need to find a bijection \phi : G \to T such that \phi(g)\phi(h) = \phi(gh) for every g \in G and h \in G.

It would be good to start by writing out the group operation of A, and see what happens when you restrict it to T.

 

[LCKurtz]

And for that matter, you haven't told us what G*G means.

A = G * G, as in G cross G.
I do not understand how T is directly isomorphic to G.

What is the group operation on G*G? You have to know that before you can even talk about an isomorphism.