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I'm trying to prove Schur-Zassenhauss Theorem without using Cohomology Theory or Representation Theory.

I started, as usual, by negating the thesis and supposing there exists a counterexample to SZ, then taking the minimum of those.

After many attempts, I managed to reduce the proof to this:

Given a finite group $ | G | = p^n \cdot m $, with $p \nmid m $. Let $N$ be a normal subgroup with the following properties:

  • $N \simeq \left ( \mathbb{Z}/p\mathbb{Z}\right)^n $
  • $N$ is included in all the normal subgroup
  • $N = \Phi(G)$, as the frattini subgroup.
  • $Z_G\left( N \right) \geq \langle K | K \triangleleft G \rangle $
  • $\forall \space K \triangleleft G $, $\Phi(K) = \left\{ e\right\}$

Moreover we have that for every group with cardinality inferior to the one of G, SZ holds. The goal is to prove that G doesn't exist, to this purpose the idea I had is to find a complement of a maximal subgroup, which doesn't seem to lead anywhere.

I wasn't able to go much further than this. Any suggestion or solution is appreciated.

Alessandro Avellino
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    Keith Conrad has tried this already, see here, steps $1-4$ on page $2-3$, after Lemma $4$. There is always a problem in the end, i.e., step $4$, when not using representations or cohomology. I feel that one should use this, why not. – Dietrich Burde Oct 20 '25 at 19:02
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    In any case there is no need to study cohomology theory in order to understand the proof, which is really just an averaging argument over group elements. – Derek Holt Oct 20 '25 at 19:08
  • @DerekHolt the point is not understanding it, I'm explicitly trying to construct a proof that avoids cohomology, even though I've already understood the standard cohomological proof. – Alessandro Avellino Oct 21 '25 at 12:29
  • I am not sure exactly what you mean by "avoiding cohomology". – Derek Holt Oct 21 '25 at 14:07
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    @DerekHolt I'm sorry for not being clear. As in the PDF sent by Dietrich Burde, the standard proof reduces to the abelian case. From there, cohomology theory immediately settles the result. I'm trying to avoid that approach — in particular, I'm looking for conditions on G and N that lead to a contradiction directly. Another proof I know uses the action on representatives of the quotient, and yet another uses representation theory. My goal is to find a proof relying only on more elementary tools. – Alessandro Avellino Oct 22 '25 at 20:36

1 Answers1

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Since I still do not understand what you mean by not using cohomology theory and using only more elementary tools, here is a proof, which is more or less the same as the one in Isaacs' book on Group Theory.

Assume $N \unlhd G$ with $N$ abelian of order $p^n$ and $|G/N|=m$ where $p$ does not divide $m$. We claim that $N$ has a complement in $G$.

Let $\{t_i :1 \ \le i \le m\}$ be a set of coset representatives of $N$ in $G$. Note that for $g \in G$ and $1 \le i \le m$ we can write $t_ig = t_{i^g}n(g)_i$, where $i \mapsto i^g$ is a permutation of $\{1,\ldots,m\}$ and $n(g)_i \in N$.

Define $\phi:G \to N$ by $\phi(g) = \prod_{i=1}^m n(g)_i$.

Then, writing $x^y$ for $y^{-1}xy$, we have $$t_igh = t_{i^g}n(g)_ih = t_{i^g}h n(g)_i^h = t_{i^{gh}} n(h)_{i^g} n(g)_i^h$$ so, since $N$ is abelian and $i \mapsto i^g$ is a permutation of $\{1,\ldots,m\}$, we have $$\phi(gh) = \prod_{i=1}^m n(h)_{i^g} n(g)_i^h = \phi(g)^h \phi(h).$$ (So $\phi:G \to N$ is a crossed homomorphism or derivation .)

Note that for $n \in N$ we have $i^n=i$ and $n(i)=n$ for all $i$, so $\phi(n)=n^m$, and since $p$ does not divide $m$, the restriction of $\phi$ to $N$ is a group isomorphism.

Let $C = \{g \in G : \phi(g) = 1_N \}$. Then $\phi(gh)=\phi(g)^h \phi(h)$ implies that $C$ is closed under the group operation and hence, since $G$ is finite, provided that $C$ is non-empty, it will follow that $C$ is a subgroup of $G$.

Now, for $g \in G$ and $n \in N$, we have $\phi(ng) = \phi(n)^g \phi(g)$ so, since $\phi$ induces a bijection on $N$, for a fixed $g$ and $n$ ranging over $N$, $\phi(ng)$ takes each value in $N$ exactly once.

In particular, for each $g \in G$, there is a unique $n \in N$ with $ng \in C$, so $C$ contains a unique representative from each coset of $N$ in $G$, and hence $C$ is the required complement of $N$ in $G$.

I would be very surprised if there was a proof that was more elementary than that and, although the proof could be worded using cohomolical terminology ($\phi$ is a $1$-cocycle), I have not made any explicit reference to cohomology.

It is interesting to note that the definition of $\phi$ is similar to that of the transfer homomorphism $G \to N$, but for the transfer you would write $t_i g = n(g)_i t_{i^g}$ rather than $t_i g = t_{i^g} n(g)_i$.

Derek Holt
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