Showing that a subgroup is subnormal in the original group G

[bham10246]

Question: Let G be a group of order p^n > 1 where p is prime. If H is a subgroup of G, show that it is subnormal in G. That is, I need to show that there is a chain of subgroups H=H_0 \triangleleft H_1 \triangleleft ...\triangleleft H_m = G, where m\leq n.


Analysis: We can easily show by induction on n that there is a series of normal subgroups of G
1=G_0 < G_1 < ... < G_n = G such that [G_{i+1}:G_i]=p.

Since H is a subgroup of G, |H|=p^k where k \leq n. Since H is a p-group, H also has a series of normal subgroups of H.


Perhaps if H is normal in G, then we can consider G/H which has a series of normal subgroups
\bar{1}=\bar{H_0}< \bar{H_1} < ... <\bar{H_m}=G/H.
By Fourth Isomorphism Theorem, does that mean we have a series of normal subgroups H=H_0 < H_1 < ...< H_m = G, where m\leq n?



Firstly, how do we know that H is normal in G?
Secondly, is H_i normal in H_{i+1}?

Thanks for your time!

 

[mathwonk]

i guess you want to show that the normalizer of H in G, is larger than H. that smells to me like the proof that a group of order p^2 is always abelian?

 

[mathwonk]

well i think i got it. look at the center of G. lemma: the center is non trivial.

then there are two cases, either H contains the center, use induction on G/H, or it does not, and then N(H) contains more than H.

does that work? i.e. the induction case?

 

[matt grime]

H_i will have index p in H_{i+1}. For p groups it is elementary to show that this means it is normal - as ever these things are just down to partitioning sets into orbits.

 

[bham10246]

Hi Mathwonk, you are right. If H = G or 1, there's nothing to show. So assume that H is a proper nontrivial subgroup of G. Using the class equation, we know that the center of G is non-trivial.

Case 1. If Z(G) is not contained in H, then H is properly contained in <H, Z(G)> which is contained inside the normalizer of H in G. So H is properly contained inside N_G(H) and H is normal in N_G(H). But (can someone answer this question that) is it so obvious that after repeating finitely many times, we will eventually get that N_G...(N_G(N_G(H))) = G? Or are we done by induction?

Case 2. If Z(G) is contained in H, then consider H/Z(G). Since |H/Z(G)|< |H|, we use induction which says that \bar{H}=H/Z(G) is properly contained in N_\bar{G} (\bar{H}). By Lattice Isomorphism Theorem, we're done.


And Matt Grime, don't we have to prove that given any H=H_i in G, a subgroup H_{i+1} containg H_i exists such that H_i is normal in H_{i+1}?

Do you think we should begin by showing that H is contained in a maximal subgroup M of G (M is maximal subgroup of G if M is a proper subgroup and there does not exist proper subgroups K of G such that M < K < G)? Then show that [G:M] = p and M is normal in G (use Mathwonk's idea using normalizer of M in G). Then since |M|<|G|, use induction to finish constructing the chain?

But how we do show that such maximal subgroup M of G exists such that M contains H?