Is the Image of a Homomorphism f(G) Always a Subgroup of H?

[steroidjunkie]

1. Show that image of homomorphism f of group G into group H is a subgroup of H.



2. f(G) \equiv { f(g) | g \in G } \subset H

The Attempt at a Solution

The problem is I don't know how to start. So if I could get a hint it would be great...

 

[LCKurtz]

1. Show that image of homomorphism f of group G into group H is a subgroup of H.



2. f(G) \equiv { f(g) | g \in G } \subset H

The Attempt at a Solution

The problem is I don't know how to start. So if I could get a hint it would be great...

You need to show f(G) is nonempty and if h₁ and h₂ are in f(G), then h₁*h₂^-1 is in f(G), where * is the group operation in H.

 

[Hurkyl]

The problem is I don't know how to start. So if I could get a hint it would be great...

Definitions are usually a good place to start. Most "simple" problems like this are trivial exercises in naive set theory after the translation.

 

[steroidjunkie]

Thank you for your guidelines. This is what I came up with: (click on the picture to show it and again to see it clearly)

 

[LCKurtz]

Thank you for your guidelines. This is what I came up with: (click on the picture to show it and again to see it clearly)

It will be much easier to have a discussion if you put your equations here. Use the X² button for superscripts. So it is easy to write a^-1.

You might start by answering this: What is the test for a subset M of a group G to be a subgroup?

What do homomorphisms do to the identity and to inverses?

 

[steroidjunkie]

I think i found the solution. I need to show that f(G) is a nonempty set and that for every g₁ and g₂ element f(G): g₁*g₂^-1 element f(G) by prooving four axioms of group.

And the answer to the second question is: homomorphism of an identity is an identity and homomorphism of an inverse is an inverse.

Thank you to both of you.