# Thread: Given a group how to investigate groups that are homomorphic to it

1. ## Given a group how to investigate groups that are homomorphic to it

This is a more general question related to another thread I just posted.

Given a group G I am relatively comfortable looking for other groups which might be isomorphic to it. I'm not great at it, but I have a few ideas I can check out. Doing the same with homomorphisms is giving me a headache.

There is, of course, the trivial homomorphism (ie the one that maps every element of G to the identity), and I'm pretty sure there is homomorphism between a group and any of its subgroups but what does one do when confronted with two groups G and H? As a negative example of what I am asking, how would I show that two groups are not homomorphic (beyond a trivial homomorphism, that is.)

To look for isomorphisms I know to check the number of elements, check to see if there are the same number of subgroups of the same size, etc. ie. the group structures must be the same. That's relatively easy. Knowing that the two groups are candidates I can then go on to construct an isomorphism.

How do you approach this problem using homomorphisms? Or am I barking up the wrong tree here?

Thanks!
-Dan

2. Okay, abhishekkgp pointed out to me that it is a lot harder than I first thought to find a homomorphism of a group with its subgroups. There is a way, but it's more reminiscent of a trivial homomorphism and I was hoping to avoid that. For example, take D4. We can define an endomorphism from D4 to its subgroup C4:
$\begin{array}{c} e \to e \\ C_4 \to C_4 \\ C_2 \to C_2 \\ C_4^{-1} \to C_4^{-1} \\ \sigma _x \to e \\ \sigma _y \to e \\ \sigma _1 \to e \\ \sigma _2 \to e \end{array}$

(I'll define the various reflection planes if anyone cares to know them.) This is a homomorphism, but not a very interesting one.

-Dan

3. Originally Posted by topsquark
This is a more general question related to another thread I just posted.

Given a group G I am relatively comfortable looking for other groups which might be isomorphic to it. I'm not great at it, but I have a few ideas I can check out. Doing the same with homomorphisms is giving me a headache.

There is, of course, the trivial homomorphism (ie the one that maps every element of G to the identity), and I'm pretty sure there is homomorphism between a group and any of its subgroups but what does one do when confronted with two groups G and H? As a negative example of what I am asking, how would I show that two groups are not homomorphic (beyond a trivial homomorphism, that is.)

To look for isomorphisms I know to check the number of elements, check to see if there are the same number of subgroups of the same size, etc. ie. the group structures must be the same. That's relatively easy. Knowing that the two groups are candidates I can then go on to construct an isomorphism.

How do you approach this problem using homomorphisms? Or am I barking up the wrong tree here?

Thanks!
-Dan
There is always (assuming you're talking about finite groups) the order problem. Namely, let $G,H$ be finite groups and $\phi\in\text{Hom}\left(G,H\right)$ note that since $\phi(G)\leqslant H$ we have (by Lagrange's theorem) that $\left|\phi(G)\right|\mid |H|$ but of course since (by the FIT) we have $\left|G\right|=\left|\phi(G)\right|\left|\ker(\phi )\right|$ and so $\left|\phi(G)\right|\mid |G|$. Thus, if for example $\left(|G|,|H|\right)=1$ you have that the only homomorphism $G\to H$ is trivial. As another simple (haha) example suppose $G$ is simple and you had a homomorphism $\phi:G\to H$ you know then that either $\ker\phi=G$ or $\ker\phi=\{e\}$ but if $\ker\phi=\{e\}$ then $G\cong\phi(G)\leqslant H$ and so in particular $|G|\mid|H|$ so that the only homomorphism from a simple group $G$ to another group whose order is not a multiple of $G$'s order is the trivial one. Also, things like being cyclic, or abelian even are preserved under homomorphic image so that if $H$ had no abelian subgroups and $\phi:G\to H$ is a homomorphism and $G$ abelian then you can conclude $\phi$ is trivial. The list goes on and on.

4. the key to determining whether a homomorphism (besides the trivial one, which always exist) exists between 2 groups, depends on the normal subgroups of G.

put another way, a homomorphic image of G IS a factor group of G (ok, isomorphic to one, but usually "up to isomorphism" is all one is interested in).

so if we have a homomorphism φ:G-->H, that means G/ker(φ) is isomorphic to a subgroup of H.

for finite groups that means for φ to be non-trivial, (|G|, |H|) ≠ 1, for example.

another thing you can do is look at the normal subgroups of H, and the normal subgroups of G.

there is a 1-1 correspondence between the normal subgroups of G containing the kernel K of a homomorphism, φ,

and the normal subgroups of φ(G), so if H has 3 normal subgroups, but G only has 2, there isn't any surjective homomorphism

from G ONTO H. of course, this means if G is simple, the possible homomorphic images of G are an isomorphic copy of G, or the trivial group.

yet another thing you can do, is view both groups as permutation groups (if G is finite), where you can look at the cycle structures of G and H

(this is similar to looking at orders of elements, but can reveal more information).

there is something to be said for the notion that a group is defined by the possible homomorphic images of it, or into it.

5. Thanks everyone. Now it's time to play with some of these concepts and get experience...

-Dan

6. Originally Posted by topsquark
...and I'm pretty sure there is homomorphism between a group and any of its subgroups
Sorry, but I haven't actually read the rest of this thread (I'm in a hurry just now) but this HAS to be commented on, and I don't think anyone has already (bus as I said, haven't read the thread properly)...anyway, this is not true. A simple group has no proper homomorphic images. For example, A_5 has lost of subgroups, but no homomorphisms exist from A_5 to any subgroups other than A_5 and 1, the trivial subgroup.

On the other hand, there exists a homomorphism from any given subgroup of a group into the group - clearly every subgroup injects into the group...(it is embedded in it)...so I'm not sure if that is what you mean...

7. Originally Posted by Swlabr
Sorry, but I haven't actually read the rest of this thread (I'm in a hurry just now) but this HAS to be commented on, and I don't think anyone has already (bus as I said, haven't read the thread properly)...anyway, this is not true. A simple group has no proper homomorphic images. For example, A_5 has lost of subgroups, but no homomorphisms exist from A_5 to any subgroups other than A_5 and 1, the trivial subgroup.

On the other hand, there exists a homomorphism from any given subgroup of a group into the group - clearly every subgroup injects into the group...(it is embedded in it)...so I'm not sure if that is what you mean...
Yeah, I found one between D4 and it's subgroup C4 below. I'm not sure if it is what you would call trivial or not, but it's certainly not very useful!

-Dan

8. i am looking at the "homomorphism" you have exhibited, and i don't think it is one. there are 5 elements that map to e, which suggests that the "kernel" is a subgroup of order 5.

this is impossible.

furthermore, if one looks at (σ_x) o (σ _y) (i am assuming these represent reflections about the x-axis and y-axis, respectively),

we have (σ_x) o (σ _y) = C2, but e*e ≠ C2.

nope, not a homomorphism.

in point of fact, the reflections of D4 do not even form a group, much less a normal subgroup.

is there ANY surjective homomorphism D4-->C4? well, if there was, the kernel would have to be a normal subgroup of order 2.

D4 has precisely 1 normal subgroup of order 2, the subgroup generated by the 180 degree rotation (the subgroup generated

by any reflection is NOT normal. you should convince yourself of this).

i believe you have called this element C2. so what is D4/<C2>? for simplicity of notation, i will call <C2> H.

i will also call C4, "r" (somethimes rho is used), and σ_x "s".

then r^2 = C2, r^3 = C4^-1, rs = σ1 (assuming this is the reflection about y = x), r^2s = σ_y,

r^3s = σ2. so D4 = {1,r,r^2,r^3,s,rs,r^2s,r^3s}.

H = <C2> = {1,r^2}. let's look at the orders of cosets of H.

rH = {r,r^3}. (rH)(rH) = r^2H = H, so rH has order 2.

sH = {s, r^2s}. again, (sH)(sH) = H, so sH has order 2. hmm. a cyclic group of order 4 has only ONE element of order 2.

we have (at least) 2, meaning D4/H must be isomorphic to the kelin 4-group.

looks like there isn't ANY homomorphism from D4 onto C4, you were mis-informed.

imagine you were a square. if you were subjected to a transformation of D4, you'd be "back in your parking space", but you'd be facing a (perhaps) different direction.

perhaps you've labelled your home vertices, or the sides, so that you can tell just how you've moved around.

now, suppose someone removed the labels. now, you can't tell if you've been rotated, but you can tell if you're upside-down or not.

this is the effect of taking D4/<C4>. now suppose the celestial crank that "turns" you, is malfunctioning, and always turns 2 quarters of a turn,

instead of one-quarter. the diagonal reflections are no longer possible, but the horizontal and vertical ones are.

this is the effect of taking D4/<C2>.

9. Originally Posted by Deveno
i am looking at the "homomorphism" you have exhibited, and i don't think it is one. there are 5 elements that map to e, which suggests that the "kernel" is a subgroup of order 5.

this is impossible.

furthermore, if one looks at (σ_x) o (σ _y) (i am assuming these represent reflections about the x-axis and y-axis, respectively),

we have (σ_x) o (σ _y) = C2, but e*e ≠ C2.

nope, not a homomorphism.
You are right. I apparently missed that example when I check it. There are probably others.
in point of fact, the reflections of D4 do not even form a group, much less a normal subgroup.

Originally Posted by Deveno
looks like there isn't ANY homomorphism from D4 onto C4, you were mis-informed.
Not misinformed. My intuition seems to have been way off. It seemed simple enough when I thought about it. Then I was advised that this was not actually the case, and the one example I did come up with is apparently not a homomorphism.

Point taken.

-Dan