Vtyjkyuk

Let $G$ be a group of order $p^n$. I have to show that there are normal subgroups of $G$ of orders $1,p,...,p^n-1,p^n$.

My attempt :

The trivial subgroup is a normal subgroup of order $1$. Assume $G$ has a normal subgroup $N$ of order $p^k,k<n$. To show: $G$ has a normal subgroup of order $p^k+1$.

Let $gamma : G rightarrow G/N$ be the natural homomorphism. The kernel is $N$, whose size is $p^k$. So $gamma$ is $p^k-textto-1$. Since $k<n$, $G/N$ is also a $p$-group. Hence, $G/N$ has a non-trivial centre. $Z(G/N)$ is then also a $p$-group. So by Cauchy's theorem, it has an element $a$ of order $p$. The subgroup $langle arangle$ generated by this element is a subgroup of $Z(G/N)$. Therefore, it is normal in $G/N$. Then the pullback of $langle arangle$ is normal in $G$. Since the order of $langle arangle$ is $p$, the order of $gamma^-1langle arangle$ is $p^k×p=p^k+1$. Thus we have found a normal subgroup of $G$ of order $p^k+1$. This completes my proof.

Your answer is partially correct: you have made use of non-triviality of center in $p$-groups after passing to $G/N$, but not for $G$. In other words, you took $N$ a normal subgroup of order $p^k$ and $Grightarrow G/N$ the natural homomorphism; but how do you know that $Nneq 1$? IF $N=1$ then the homomorphism is no longer helpful to carry out induction.

In short, the existence of normal subgroups of every order follow from following two facts:

• For a $p$-group $Gneq 1$, the center is non-trivial.




• If $Ntrianglelefteq G$ and $Grightarrow G/N$ is natural homomorphism then there is a (nice) bijection between normal subgroups of $G/N$ and the normal subgroups of $G$ which contain $N$.