Need someone to point me in the right direction

[battletoad]

So i just got back from a Group Theory exam.

 

One question's bugging me.

 

|G|= 6655. Prove that G has nontrivial characteristic subgroups H & K, K a proper subgroup of H.

 

Now i've proved that there exists a unique Sylow 11-subgroup (you guys may confidently assume this to be the case if you don't want to do it yourselves), thus it is normal in G. Let's call this subgroup H.

 

We proceed to show that it is a characteristic subgroup of G.

 

If f is any automorphism in G, the order of H is the same as that of the order of H under f (Hf is a subgroup of G). Since H is the unique Sylow 11-subgroup, H=Hf (if not, because Hf qualifies as a Sylow 11-subgroup of G, H!=Hf contradicts H's uniqueness). Hence H is a characteristic (non-trivial) subgroup of G.

 

Now |H|=11^k where k is some positive integer, k<=3. By Sylow, H has Sylow 11-subgroup M. There may be multiple such Sylow 11-subgroups, denote them M=M1, M2, ... ,Mr for some non-negative integer r, r some power of 11.

 

Let K be the intersection of all these Mi's. Thus K is a subgroup of (possibly equal to, if Mi is the unique Sylow 11-subgroup of H) Mi < H

=> K<H

Any automorphism in G permutes the Mi's among themselves, and since K contains all common elements of these Mi's, K is sent to itself under any such automorphism. Thus K is a characteristic subgroup of G.

 

This last part is what's bugging me.

 

All that is required to show is that K is non-trivial.

 

I did some things involving the order of elements in these Mi subgroups dividing the order of H but didn't find to killer blow to tie it up all together.

 

What i have is that H,K are characteristic subgroups of G, K<H and H nontrivial. Not the K nontrivial part.

 

Appreciated if anyone can add something, even if it is criticism of the proof.

[shyvera]

I would suggesting letting H be the Sylow 11-subgroup (which is certainly characteristic since it’s unique) and K be the centre of H. K is definitely nontrivial (the centre of any p-group where p is a prime is in general nontrivial). Thus if K is a proper subgroup of H, the problem would be solved since the centre of any group is a characteristic subgroup. This leaves the case K = H (i.e. H is Abelian); again this would not be a problem unless H is an elementary Abelian group (which would not have any proper nontrivial characteristic subgroups).

[battletoad]

I would suggesting letting H be the Sylow 11-subgroup (which is certainly characteristic since it’s unique) and K be the centre of H. K is definitely nontrivial (the centre of any p-group where p is a prime is in general nontrivial). Thus if K is a proper subgroup of H, the problem would be solved since the centre of any group is a characteristic subgroup. This leaves the case K = H (i.e. H is Abelian); again this would not be a problem unless H is an elementary Abelian group (which would not have any proper nontrivial characteristic subgroups).

 

Thanks for the input. Totally forgot that the centre of any group is non-trivial if the order of the group is some power of a prime (although i proved it an earlier question during the exam!).

 

As things stand, the manner in which I defined K, is it anyway possible to conclude that it's non-trivial without knowing what we're dealing with (apart from the fact that the initial group is of order 6655)?