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I'm trying show that if a group $G$ has order $p^n$ then there exists a surjective homomorphism from $G$ to $\mathbb{Z}/\mathbb{Z}p$. I know from Cauchy's Theorem there is a cyclic subgroup of order $p$ and that will be isomorphic to $\mathbb{Z}/\mathbb{Z}p$, but what to do about the rest of the group, in order to ensure my mapping is a surjection, I don't know.

Addem
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    There is a normal subgroup of order $p^{n-1}$ by Sylow – Randall Dec 05 '17 at 03:54
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    Use the fact that $G$ has nontrivial center and induct on $n$. – anomaly Dec 05 '17 at 03:59
  • @Randall: I don't follow. The Sylow theorems only tell you that there is a subgroup of order $p^n$, which in this case is all of $G$. – Qiaochu Yuan Dec 05 '17 at 04:01
  • @QiaochuYuan I can't recall the precise argument, but I'm sure this is an exercise in Hungerford. I remember proving it as a graduate student. Horrible evidence, I know. If my memory fails and I am wrong, I will retract. – Randall Dec 05 '17 at 04:04
  • Ah, of course, the argument is posted at this very site: https://math.stackexchange.com/questions/549635/a-p-group-of-order-pn-has-a-normal-subgroup-of-order-pk-for-each-0-le-k – Randall Dec 05 '17 at 04:06
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    The statement is definitely true but I don't think the Sylow theorems are relevant. – Qiaochu Yuan Dec 05 '17 at 04:11
  • It may be possible to prove it without ST, but it's a stretch to say they're not relevant: it depends on the version of ST at your disposal. The version of Sylow theory that I learned from Fraleigh (that $G$ contains subgroups of order $p^k$ for all such divisors, not just the maximal one) is definitely relevant. This version proves the normality of any subgroup of order $p^k$ in a subgroup of order $p^{k+1}$. – Randall Dec 05 '17 at 04:24
  • Any group of order $p^n$ contains subgroups of order $p^j$ for all $0 \leq j \leq n$. This is a fairly easy induction argument which exploits the fact that the center is nontrivial. You don't really need the Sylow existence theorem (i.s. that there exists a subgroup of order $p^n$ when this is the largest power of $p$ dividing $|G|$) for this, but some authors do include the smaller powers of $p$ in their statement of that theorem. –  Dec 05 '17 at 04:35
  • I would always advocate for a simple induction over Sylow theorems, for sure. – Randall Dec 05 '17 at 04:36
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    P.S. The key result you use here is that there is a normal subgroup of each possible order (hence a homomorphism with that subgroup as a kernel). This is part of the induction argument as well, but I don't think it's usually included even in the more general statement of the Sylow existence theorem. (I could be wrong.) –  Dec 05 '17 at 04:42
  • @anomaly I've been staring at it for hours, maybe I'm just tired, but not seeing how to do the induction. $G$ has a non-trivial center and its size is a $p$-power, but I don't see what to do with it. The class equation? Mod out by it? Since I don't know how big the center is, I'm not guaranteed that the center is the kernel of the surjective homomorphism. – Addem Dec 05 '17 at 05:18
  • @Randall Since I don't think I have the theorem that a $p$-group has normal subgroups of every $p^k, k\leq n$ then proving it in order to solve the problem seems like it'd be too much. – Addem Dec 05 '17 at 05:20
  • See the answer below. – anomaly Dec 05 '17 at 05:28
  • @QiaochuYuan Some authors include the part about subgroups of all prime power orders as part of the Sylow theorems (I don't like this myself, as it is really a statement about subgroups of $p$-groups, while the Sylow theorems are specifically useful for non- $p$-groups). – Tobias Kildetoft Dec 05 '17 at 08:29

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Assume the result holds for all $p$-groups $G'$ with $|G'| < |G|$. The quotient $G/Z(G)$ then admits a surjection $G/Z(G) \to \mathbb{Z}_p$, and the composition $G \to G/Z(G) \to \mathbb{Z}_p$ is also surjective.

anomaly
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