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To be precise, let $G$ be a group of order $p^k$, where $p$ is a prime number and $k$ is a positive integer. We wish to prove, for all $n = 1,...,k$, that $G$ has a normal subgroup $H$ of order $p^n$.

The strategy of a proof from Dummit and Foote is to first to show that the center of a $p$-group must be nontrivial, using the class equation. Then we can get the desired result by inducting on the order of the group. Using the above result, we can find a normal subgroup of order $p$ in $G$, take the quotient, use the inductive hypothesis and bring it back to $G$ by the 4th isomorphism theorem to complete the proof.

However, I was wondering if there was a way to show this without using the class equation? Dummit and Foote prove a more general statement than I am asking about, so there might be a simpler way for this special case.

user1729
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Ben
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  • If you can show that a $p$-group must have a subgroup of index $p$, then it would follow from the theorem that if the index of a subgroup is the smallest prime dividing the order of the group, then the subgroup is normal. – Arturo Magidin Aug 03 '21 at 02:49
  • How could you use this to find subgroups of order $p,...,p^{k-2}$ that are normal in $G$? – Ben Aug 03 '21 at 02:53
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    I was thinking induction, but on further thought it would take more than that (since those groups are not characteristic)... it seems hard to do without somehow using that centers are nontrivial, or essentially proving the class equation in disguise. – Arturo Magidin Aug 03 '21 at 03:09
  • Related: https://math.stackexchange.com/a/4194684/943729 –  Aug 03 '21 at 04:25
  • If you could show that there is a normal subgroup of order $p$, then you could follow the usual proof from there. – verret Aug 03 '21 at 05:47
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    But I guess it's easy to see a normal subgroup of order $p$ should be central (because the automorphism group has order coprime to $p$) so this doesn't help much. Is there a reason you want an alternate proof? It's not like the class equation one is very difficult... – verret Aug 03 '21 at 07:47
  • "However, I was wondering if there was a simpler way to show this". I don't think so. Of course, it depends on what "simpler" really means. The standard proof also has many advantages, and the class equation is important in itself. I think it is not useful to avoid such topics in this context. – Dietrich Burde Aug 03 '21 at 08:19
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    I agree with @DietrichBurde's comment that trying to prove this without the class equation isn't necessarily a good idea. However, the question is still a valid question and not a duplicate of the suggested one, which does use the class equation, so I reopened it. (I also edited out the word "simpler", because leaving it in makes the question subjective.) – user1729 Aug 03 '21 at 14:34
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    You could prove that the number of subgroups of order $p$ is congruent to $1$ mod $p$, and then use the conjugation action of the group to show that there must be a normal subgroup, but that's not exactly simpler! – Derek Holt Aug 03 '21 at 15:34

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