Hey, can anybody help me with this?
If is a finite group with a minimal non-trivial normal subgroup such that is Abelian, then prime such that .
I have no idea where to start with this. I can see that all proper non-trivial subgroups of aren't normal in , but not sure how this helps.
I'm sure there are other ways, but that way seemed the neatest. Basically, a characteristic subgroup is a subgroup which is fixed under automorphisms,
then .
Examples of characteristic subgroups are the two boring subgroups, the center, the commutator subgroup, and the frattini subgroup.
If you take a cyclic group, then every subgroup is characteristic.
Now, you can think of conjugation by an element as an automorphism, and so such groups are normal. However, the characteristic subgroup of a normal subgroup will be normal in the big group, and this is what I would have utilised.
For the first part, if another prime divided the group M then take the largest q-group in M. As your group is abelian this will encompass all the q-ness of your group (q will not divide the order of M/Q). Therefore, you would have a characteristic subgroup, which will thus be normal.
So if I understand you correctly, this is how you prove it:
Let be prime such that and let be given. Then since is Abelian, and so . Now let be given and define . Then and since . Thus and so by uniqueness . Therefore and since is a minimal non-trivial normal subgroup of or . However, , so by Sylow's theorems, and therefore . Hence is an Abelian -group.
Finally, define . Then and as is Abelian, and , so . Therefore . Now let be given. Then , so since and therefore . Thus and so or . However, by Cauchy's theorem such that and thus . Therefore .
Thanks for the help!
Not quite...you are taking a sylow subgroup of M, not of G. There is no reason to say these coincide.
Let . As is abelian, the sylow -subgroups of are normal and unique, so let be the maximal subgroup of of order a power of . Clearly, if , then .
Now, we know that , but we wish to prove that . This is true because if is an automorphism, then because every element has order a power of and no other element in the group has this order, and automorphisms preserve order (does that make sense)? Therefore, .
Next, notice that for , , is an automorphism of . This is bacause is normal in . As this is an automorphism of it must fix . Therefore, for all , and we thus conclude that , as required to get our contradiction.
Note that .