# Showing that a subgroup is subnormal in the original group G

1. Aug 2, 2007

### 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? :uhh:
Secondly, is $H_i$ normal in $H_{i+1}$?

Last edited: Aug 2, 2007
2. Aug 2, 2007

### 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?

3. Aug 2, 2007

### 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?

4. Aug 2, 2007

### 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.

5. Aug 2, 2007

### 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?

Last edited: Aug 3, 2007