Showing that a subgroup is subnormal in the original group G

In summary, the chain of subgroups H=H_0 \triangleleft H_1 \triangleleft ...\triangleleft H_m = G, where m\leq n, can be constructed using the class equation and lattice isomorphism theorem.
  • #1
bham10246
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Question: Let [itex]G[/itex] be a group of order [itex]p^n > 1[/itex] where [itex]p[/itex] is prime. If [itex]H[/itex] is a subgroup of [itex]G[/itex], show that it is subnormal in [itex]G[/itex]. That is, I need to show that there is a chain of subgroups [itex]H=H_0 \triangleleft H_1 \triangleleft ...\triangleleft H_m = G[/itex], where [itex]m\leq n[/itex].


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

Since H is a subgroup of G, [itex]|H|=p^k [/itex] where [itex]k \leq n[/itex]. 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
[itex]\bar{1}=\bar{H_0}< \bar{H_1} < ... <\bar{H_m}=G/H[/itex].
By Fourth Isomorphism Theorem, does that mean we have a series of normal subgroups [itex]H=H_0 < H_1 < ...< H_m = G[/itex], where [itex]m\leq n[/itex]?



Firstly, how do we know that H is normal in G? :uhh:
Secondly, is [itex]H_i [/itex] normal in [itex]H_{i+1}[/itex]?

Thanks for your time!
 
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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
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 [itex]N_G(H)[/itex] and H is normal in [itex]N_G(H)[/itex]. But (can someone answer this question that) is it so obvious that after repeating finitely many times, we will eventually get that [itex]N_G...(N_G(N_G(H))) = G[/itex]? 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 [itex]\bar{H}=H/Z(G)[/itex] is properly contained in [itex]N_\bar{G} (\bar{H})[/itex]. 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?
 
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1. Why is it important to show that a subgroup is subnormal in the original group G?

Showing that a subgroup is subnormal in the original group G is important because it provides a deeper understanding of the group's structure and can help in further group theory applications. It also allows for the use of specific theorems and techniques that are only applicable to subnormal subgroups.

2. What is the definition of a subnormal subgroup?

A subnormal subgroup is a subgroup that can be reached by a finite chain of subgroups, each of which is a normal subgroup of the next one in the chain. In other words, a subgroup H is subnormal in a group G if there exists a finite sequence of subgroups {H0, H1, ..., Hn} such that H0 = H, Hn = G, and Hi is a normal subgroup of Hi+1 for all i.

3. How can one show that a subgroup is subnormal in the original group G?

To show that a subgroup H is subnormal in a group G, one can use the following method:

  1. Show that H is a normal subgroup of G.
  2. Find a subgroup K of G such that K is a normal subgroup of H.
  3. Repeat step 2 until the sequence of subgroups eventually reaches G.
  4. Since each subgroup in the sequence is normal in the next one, H is subnormal in G.

4. Are there any alternative ways to show that a subgroup is subnormal in the original group G?

Yes, there are alternative methods that can be used to show that a subgroup is subnormal in the original group G. Some of these methods include using the concept of conjugacy classes, using the lattice of subgroups, and using the concept of normal closure of a subgroup.

5. What are the implications of a subgroup being subnormal in the original group G?

If a subgroup H is subnormal in the original group G, then it follows that H is also a normal subgroup of G. This has several implications such as the existence of a composition series for the group G, the group G being solvable, and the group G having a normal series with all factors being abelian. Additionally, subnormal subgroups have many interesting properties and can be used in various group theory applications.

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