# Abstract Algebra Questions Help Please

1. Apr 30, 2007

### rocky926

Any and all help on these problems would be greatly appreciated. Thank you in advance to any who offer help .

1. Let φ:G->H be a group homomorphism, where G has order p, a prime number. show that φ is either one-to-one or maps every element of G to the identity element of H.

2. Show that if H is a normal subgroup of G (with operation multiplication) and [G]=m, then for every g in G, g^m is in H.

3. Every symmetry of the cube induces a permutation of the four diagonals connecting the opposite vertices of the cube. This yields a group homomorphism φ from the group G of symmetris of the Cube to S4 (4 is a subscript). Does φ map G onto S4? Is φ 1-1? If not, describe the symmetries in the kernel of φ. Determine the order of G.

2. Apr 30, 2007

### StatusX

What have you tried? For the first, think about the kernel of phi.

3. May 1, 2007

### rocky926

I know the kernal is the set K where the elements of K are the elements of G that when put into phi return the identity element of H, but I am not sure how to calculate this.
For example if I had φ:Z(mod 24) -> Z(mod 81), what would the kernal be??
I dont understand how to go about finding the kernal of this... Thank you for your help :)

4. May 1, 2007

### Dick

In general, you can't say much about the kernel until you know what the map is. But to go back to your problem, a group of prime order is a very special and simple kind of group. What kind? For example, does it have any non-trivial subgroups?

5. May 1, 2007

### rocky926

I believe every group of prime order is also a cyclic group.

6. May 1, 2007

### Dick

Good, now can it have any nontrivial subgroups?

7. May 1, 2007

### rocky926

I think the only subgroup would be the group containing the identity element... meaning there are no nontrivial subgroups...

8. May 1, 2007

### Dick

Yep, the whole group is also considered a subgroup of itself. Now can you figure out what 'subgroups' have to do with 'kernels'?

9. May 1, 2007

### rocky926

well since the kernel of a group homomorphism is a subgroup of the group on the left, in this case G, and since G has no nontrivial subgroups.. then the kernel of phi must be the entire group G or just he identity element.

10. May 1, 2007

### Chris Hillman

Comment on the "meaning" of the statement

The homomorphic image of a group is a simplified model of that group, in which we identify the cosets of the kernel with the elements in the image. So the result stated in the exercise says that there are no nontrivial simplified models of groups of prime order, because the only homomorphic images are either isomorphic copies (not simplified at all!) or the trivial group (absurdly oversimplified!). This might remind you of a fact about divisors of prime numbers.

11. May 1, 2007

### Dick

So if the kernel is the entire group then you are all done. Now you just have to show that if the kernel is only the identity then the map is one-to-one.

12. May 1, 2007

### rocky926

Thanks!!! I think I can handle this one from here on out... now onto the others... haha, Thanks Again!!

13. May 1, 2007

### rocky926

Could someone please tell me what the notation in problem 2, "[G]=m" means...Thanks

14. May 1, 2007

### Dick

I believe it means that the order of the factor group G/H is the integer m.

15. May 1, 2007

### rocky926

Could someone give me a hint as to where to start on either problem 2 or problem 3.... I am really having trouble with this chapter in the text.... Thanks!

16. May 1, 2007

### StatusX

For 2, do you know what a quotient group is? Also, note that the order of any element in a group divides the order of the group. There's a few steps to fill in, but it'd be hard to give you any more hints without giving it away.

17. May 1, 2007

### rocky926

Im reading about the factor/quotient groups right now, but honestly am having a lot of trouble grasping this concept right now.... Thanks for these hints though.. Im going to try and figure out the rest of the steps.. If you could offer any other help I would greatly appreciate it. :)

18. May 1, 2007

### Dick

Do you know that the quotient of a group by a normal subgroup is a group? BTW, rather than post a number of questions in a single query I think you'd do better to post them in shorter bites.

19. May 1, 2007

### rocky926

Im sorry Dick im not sure what you mean by that. Ive been having a lot of trouble with the quotient groups... and am still trying to get a hold on the basic concept.

20. May 1, 2007

### Dick

It's like equivalence classes. Do you know them? If H is a normal subgroup of G then the right cosets and the left cosets are equal and there are #(G)/#(H) of them (where # just means number of elements in a set). And they form a group under the operation (xH)*(yH)=(x*y)H. Since this isn't the Physics and Math Lecture Forums, I'll just suggest that you try and review them and post further questions if you get confused, ok? This should remind you of a homomorphism. Because it is.

Last edited: May 1, 2007