9

Suppose $|G|=p^n$. Then $G$ has a normal subgroup of order $p^m$ for every $0\le m\le n$.

By induction. It is clearly true for $n=0$. Now suppose $k<n$ and $H_i$ is a normal subgroup of $G$ of order $p^i$ for $i=0,...,k$.

I read in another thread that if we consider $Z(G/H_k)\neq 1$ and the Cauchy's theorem, this will provide us a group of order $p^{k+1}$. I don't see why yet.

By that theorem there exists $gH_k\in Z(G/H_k)$ with order $p$. How can we construct a group of order $p^{k+1}$ from here?

Thank you.

Talexius
  • 2,105
  • 2
    Probably easier (though in the end, the same) to note that every $p$-group has a normal subgroup of order $p$. Now just look at the preimage of such a subgroup in $G/H_k$. – Steve D May 17 '16 at 00:30
  • Just comment to make clear of the induction steps, which are not about $n$ but $m$. We fix $n$ and $|G|=p^n$. We are actually performing (strong) induction on $m$ rather than on $n$. The base case is $m=0$. Then we assume $m\leq n-1$ is correct. At last, we try to prove that $m=n$ is correct by the inductive assumption on $m\leq n-1$. – Sam Wong Dec 16 '22 at 03:46

1 Answers1

14

Lemma 1: If $G$ is a group, every subgroup of $Z(G)$ is normal in $G$.

Proof: If $H\subset Z(G)$ is a subgroup of the center, then $gHg^{-1}=H$ for every $g\in G$ since every $g$ commutes with every element of $H$.

Lemma 2: If $N$ is a normal subgroup of $G$ and $H\subset G/N$ is a normal subgroup of $G/N$, then the preimage in $G$ of $H$ is a normal subgroup of $G$.

Proof: this is just the correspondence theorem.

Okay, now to your situation. You have a subgroup $H_k$ in $G$ of order $p^k$. You have found yourself an element $gH_k\in G/H_k$ of order $p$ that is contained in the center of $G/H_k$. This means that the order-$p$ subgroup it generates in $G/H_k$ is contained in the center of $G/H_k$. By the first lemma, this subgroup is normal in $G/H_k$. By the second lemma, its preimage in $G$ is normal in $G$. And its order is $p$ times the order of $H_k$, which is $p^{k+1}$.

Ben Blum-Smith
  • 21,560
  • How did you get your last line: "And its order is $p$ times the order of $_k$, which is $p^{k+1}$."? – user5826 Jun 27 '19 at 17:43
  • 3
    The group in question is the preimage in $G$ of the order $p$ subgroup of $G/H_k$ generated by $gH_k$. The $p$ elements of this subgroup are the $p$ cosets $gH_k, g^2H_k, \dots, g^{p-1}H_k,H_k$, viewed as elements of the quotient $G/H_k$. The preimage in $G$ is the union of these $p$ cosets, viewed as subsets of $G$, each of which has cardinality $p^k$. So the cardinality of this preimage is $p$ times $p^k$, which is $p^{k+1}$. Does that answer? – Ben Blum-Smith Jun 27 '19 at 19:33
  • Sorry @BenBlum-Smith but why do we know that the subgroup $\mathcal P\le G/H_k$ whose order is $p$ is contained in the center of the quotient group? –  Jan 19 '21 at 16:30
  • 3
    @Alchemist - the needed claim (quoted in the OP, although without proof) is only that there exists a subgroup of order $p$ contained in the center, not that any such subgroup is contained in the center (which would be false). The proof is a famous counting argument. It is enough to prove the center is nontrivial, because a nontrivial subgroup of a $p$-group necessarily contains an element of order $p$, by Cauchy's theorem. To prove the center is nontrivial, decompose $G$ into conjugacy classes. All non-singleton conjugacy classes have length a nontrivial $p$-power by the orbit- (con'td) – Ben Blum-Smith Jan 20 '21 at 18:02
  • 1
    stabilizer theorem, and $G$ itself has nontrivial $p$-power order of course. The number of singleton conjugacy classes must therefore be divisible by $p$ (because the lengths of all conjugacy classes add up to the group order). There is at least one singleton class: the class of the identity. Therefore, there are at least $p$ such classes! Since the center of $G$ is exactly the union of the singleton conjugacy classes, this proves that the center has order at least $p$. QED. – Ben Blum-Smith Jan 20 '21 at 18:06
  • Nice answer (+1). – A learner Feb 24 '21 at 15:42