Thread: Homomorphisms & the isomorphism thm.

P: G to H is a group homomorphism, L = Image(P) = {h in H | h= P(g) for some g in G}.

A. Show L is a subgroup of H.

Now assume G= under addition and H= .

B. Show that P can't be onto. Which subgroups of are possible for L?
C. Find a specific example of some group homomorphism P from G to H which isn't the trivial map.

my thoughts:
for part A, I had that P(g)P(h) = P(gh) for any g, h in G, P( )= and P( )= P , by the definition of a hom, but was told that this is wrong because I need to discuss L. What am I missing here?
for B, I know the possible subgroups of H are <1>, <(123)>, <(12)>,<(23)>, <(13)>, and H, all of order 1, 2, 3 or 6, which all divide the order of G -- from there I don't know how to show there is no onto hom.
for (C), I have no idea how to find a mapping that will work!

Thank you so much for any help!

Originally Posted by kimberu

P: G to H is a group homomorphism, L = Image(P) = {h in H | h= P(g) for some g in G}.

A. Show L is a subgroup of H.

Is it closed under mult.

Spoiler:

since

Does it contain the identity element?

Spoiler:

and

Does it contain inverses?

Spoiler:

If then and so and and

Now assume G= under addition and H= .

B. Show that P can't be onto. Which subgroups of are possible for L?

Hint:

Spoiler:

If is abelian and is an epimorphism (a surjective homomorphism) then clearly means that for some and so ........so

C. Find a specific example of some group homomorphism P from G to H which isn't the trivial map.

Why don't you think about this one for a little more.

Originally Posted by Drexel28

Hint: If is abelian and is an epimorphism (a surjective homomorphism) then clearly means that for some and so ........so

Does this mean that, accordingly, L = the image is an abelian subgroup, so the only possibilities are <1>, <(123)>, <(12)>,<(23)>, <(13)>, and not all of H, so P is not onto?

Then, for part (C), does the following map work?
P(0) --> e
P(1, 3, 5, 7, 9, 11) ---> (12)
P(2,4,6, 8,10) ----> e

If so, the way I found it was basically by guessing...I guess what I'm asking is, is there a systematic way to find this sort of thing dealing with the orders of elements? I tried to find a map onto the group of <(123)>, but I couldn't figure that one out.

Thank you!!

Originally Posted by kimberu

Does this mean that, accordingly, L = the image is an abelian subgroup, so the only possibilities are <1>, <(123)>, <(12)>,<(23)>, <(13)>, and not all of H, so P is not onto?

Then, for part (C), does the following map work?
P(0) --> e
P(1, 3, 5, 7, 9, 11) ---> (12)
P(2,4,6, 8,10) ----> e

If so, the way I found it was basically by guessing...I guess what I'm asking is, is there a systematic way to find this sort of thing dealing with the orders of elements? I tried to find a map onto the group of <(123)>, but I couldn't figure that one out.

Thank you!!

I'm not quite sure what you're asking. The reason that the map in your example can't be surjective is that it would imply that is abeilan since is.

Originally Posted by Drexel28

I'm not quite sure what you're asking. The reason that the map in your example can't be surjective is that it would imply that is abeilan since is.

That makes sense, thanks. Just to confirm, does that mean that <1>, <(123)>, <(12)>,<(23)>, <(13)> are all possibilities for L, as they are not abelian?