Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a $p$-group. Let $H$ be any subgroup of $G$. How to prove that there exists subgroups of $G$ such that $$H = H_0 \lt H_1 \lt H_2\lt \cdots \lt H_n=G$$ such that $|H_{i+1}/H_i|=p$?

I have proved the theorem when H is normal subgroup of G, & when H is trivial subgroup.

Here H is any given fixed subgroup to start with (not necessarily normal)

share|cite|improve this question
Can you clarify your question? By the way, for a $p$-group we know the center of $G$ is non-trivial, so that gives you (if the group is non-abelian) a normal subgroup, perhaps that is what you want ; but at first glance I cannot guess what you're actually looking for. And I removed the tag "galois-theory" because this is completely unrelated. – Patrick Da Silva Apr 13 '12 at 20:33

In any $p$-group $G$, "normalizers grow" (if $H$ is a proper subgroup of $G$, then $N(H) > H$).

Now take any subgroup of $H$ of $G$. Since normalizers grow, we have $$H < N(H) < N(N(H)) < \ldots < G$$ which we need to fill in to a "complete" chain, i.e., a chain where the ratio of successive orders is $p$.

By induction on $|G|$, we can fill in the chain between all successive pairs above, except for the last pair. But the second-to-last group is a normal subgroup of $G$ (since its normalizer is $G$), and using the known result for normal subgroups, we can fill in the last pair as well, and the result follows.

share|cite|improve this answer

Note. This is an easy consequence of the Sylow Theorems as well: given any finite group $G$, a prime $p$, and if $p^n$ is the largest power of $p$ that divides $|G|$, then given any subgroup $H$ of $G$ such that $|H|=p^i$ with $i\lt n$, there exists a subgroup $K$ of $G$ that contains $H$ and has order $p^{i+1}$. That $H$ is normal in $K$ will follow because the index of $H$ in $K$ is the smallest prime that divides $K$. Applying this to the case of $G$ of order $p^n$ gives you the desired conclusion.

But if you are trying to prove this in order to prove the Sylow Theorems (or you don't know this version of the Sylow Theorems), you can prove it directly, for example, as follows:

Let $|G|=p^m$. We proceed by induction on $m$.

If $m=1$, then either $H$ is trivial and we can take $H_0=H$, $H_1=G$; or else $H=G$ and we can take $H_0=G$.

Assume the result is true for any group of order strictly less than $m$. Let $H$ be a subgroup of $G$.

If $Z(G)\cap H$ is nontrivial, then consider $H/(H\cap Z(G))$ in $G/(H\cap Z(G))$. By induction, we can find a subnormal chain $$\frac{H}{H\cap Z(G)} = H_0 \triangleleft H_1\triangleleft\cdots\triangleleft H_n = \frac{G}{H\cap Z(G)}$$ with $|H_{i+1}/H_{i}|=p$. By the isomorphism theorems, there exist subgroups $\widehat{H_i}$ of $G$, with $\widehat{H_i}/(H\cap Z(G)) = H_i$, and we have $\widehat{H_i}\triangleleft \widehat{H_{i+1}}$, $[\widehat{H_{i+1}}:\widehat{H_i}] = [H_{i+1}:H_i]=p$, giving the desired chain $$H = \widehat{H_0}\triangleleft \widehat{H_1} \triangleleft\cdots\triangleleft \widehat{H_n}=G.$$

If $Z(G)\cap H=\{1\}$, then let $g\in Z(G)$ be an element of order $p$ (always possible, since $Z(G)\neq\{1\}$). Let $H'=\langle H,g\rangle$. Then $H\triangleleft H'$, and $|H'/H| = p$. Now we have $H'\cap Z(G)\neq\{1\}$, so we can apply the previous case to $H'$ to obtain a chain: $$H'=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n = G,$$ with $|H_{i+1}/H_i|=p$. Now we have: $$H = H_{-1}\triangleleft H_0\triangleleft\cdots\triangleleft H_n=G$$ as a chain with the properties we want.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.