# Thread: Homomorphisms & the isomorphism thm.

1. ## 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=$\displaystyle Z_{12}$ under addition and H=$\displaystyle S_{3}$.

B. Show that P can't be onto. Which subgroups of $\displaystyle S_{3}$ 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($\displaystyle 1_G$)=$\displaystyle 1_H$ and P($\displaystyle G^{-1}$)= P$\displaystyle (G)^{-1}$, 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!

2. 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:
$\displaystyle \theta(g),\theta(g')\in\theta(G)\implies \theta(g)\theta(g')=\theta(gg')\in\theta(G)$ since $\displaystyle g,g'\in G\implies gg'\in G$

Does it contain the identity element?
Spoiler:
$\displaystyle e_G\in G$ and $\displaystyle \theta(e_G)=e_H$

Does it contain inverses?
Spoiler:
If $\displaystyle \theta(g)\in G$ then $\displaystyle g\in G$ and so $\displaystyle g^{-1}\in G$ and $\displaystyle \theta(g^{-1})\in \theta(G)$ and $\displaystyle \theta(g)\theta(g^{-1})=\theta(gg^{-1})=\theta(e_G)=e_H$

Now assume G=$\displaystyle Z_{12}$ under addition and H=$\displaystyle S_{3}$.

B. Show that P can't be onto. Which subgroups of $\displaystyle S_{3}$ are possible for L?
Hint:
Spoiler:
If $\displaystyle G$ is abelian and $\displaystyle \theta:G\to G'$ is an epimorphism (a surjective homomorphism) then clearly $\displaystyle a,b\in G'$ means that $\displaystyle a=\theta(g),b=\theta(g')$ for some $\displaystyle g,g'\in G$ and so $\displaystyle ab=\theta(g)\theta(g')=\theta(gg')=\theta(g'g)=\th eta(g')\theta(g)=ba$........so

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

3. Originally Posted by Drexel28

Hint: If $\displaystyle G$ is abelian and $\displaystyle \theta:G\to G'$ is an epimorphism (a surjective homomorphism) then clearly $\displaystyle a,b\in G'$ means that $\displaystyle a=\theta(g),b=\theta(g')$ for some $\displaystyle g,g'\in G$ and so $\displaystyle ab=\theta(g)\theta(g')=\theta(gg')=\theta(g'g)=\th eta(g')\theta(g)=ba$........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!!

4. 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 $\displaystyle S_3$ is abeilan since $\displaystyle \mathbb{Z}_{12}$ is.

5. 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 $\displaystyle S_3$ is abeilan since $\displaystyle \mathbb{Z}_{12}$ 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?