## Abstract Algebra Questions... Help Please!!

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:H]=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.

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 Recognitions: Homework Help What have you tried? For the first, think about the kernel of phi.
 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 :)

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## Abstract Algebra Questions... Help Please!!

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?

 I believe every group of prime order is also a cyclic group.
 Recognitions: Homework Help Science Advisor Good, now can it have any nontrivial subgroups?
 I think the only subgroup would be the group containing the identity element... meaning there are no nontrivial subgroups...
 Recognitions: Homework Help Science Advisor Yep, the whole group is also considered a subgroup of itself. Now can you figure out what 'subgroups' have to do with 'kernels'?
 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.

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 Quote by rocky926 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.
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.

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 Quote by 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.
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.

 Thanks!!! I think I can handle this one from here on out... now onto the others... haha, Thanks Again!!
 Could someone please tell me what the notation in problem 2, "[G:H]=m" means...Thanks

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