On local group rings

Question

Let $G$ be a finite group and $k$ a field of characteristic $p$. Suppose that the ring $k[G]$ is local. How this implies that G is a p-group?

In fact, there are different proofs as in the answers below. But, I'm confused with the hint that rised in the exercise ( page 54) given in ( library.msri.org/books/Book51/files/02iyengar.pdf).

Since all maximal ideals are prime, the nilradical $nil(k[G])$ is contained in the Jacobson radical $J(k[G])$. Since $k[G]$ is a finite dimensional $k$-algebra, it is a left-Artinian ring. So all prime ideals are maximal and then $nil(k[G])=J(k[G])$. However, I'm not sure how to use this hint to prove that $G$ is a $p$-group.

I would appreciate any help. Thank you in advance.

Answer 1

It seems a more elementary approach will work. We use two standard observations for any finite subgroup $H$ of $G$:

1. When the order $n$ of $H$ is invertible in $k$, then $e=n^{-1}\sum_{h\in H}h$ is an idempotent; and
2. $he=e$ for any $h\in H$

Now pick a nonidentity element $x'$ of $G$ and consider $H'=\langle x'\rangle$. If $|H'|$ is not a power of $p$, then we can find a nonidentity element of $H'$ with order coprime to $p$. Call this element $x$ and the subgroup it generates $H$. We've guaranteed the order $n$ of $x$ (and $H$) is invertible in $k$ to satisfy 2 above.

Considering that $R[G]$ is local, it only has trivial idempotents, and the idempotent from 1 above is nonzero, so $e=1$. But multiplying both sides of $e=1$ with $x$, with 2 above we have $x=xe=e=1$. This is a contradiction, however, since $1_G$ and $x$ are linearly independent elements in $k[G]$.

Thus the only possible order for an element in $G$ is a power of $p$.

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You may find it interesting to know that it is unnecessary to assume $G$ is finite.

Theorem 1 of

Renault, G. “Sur les anneaux de groupes.” CR Acad. Sci. Paris Sr. AB 273 (1971): A84-A87.

shows that $G$ is a $p$-group for some $p$ when $R[G]$ is local. You can find an English translation here.

Answer 2

Here's an elementary proof. Since $kG$ is local, there is a unique simple $kG$-module, which is necessarily the trivial module. Thus $kG$ is a polytrivial module, i.e., a module all of whose composition factors are trivial.

Now take any composition series for $kG$. By choosing an obvious basis such that the $i$th element generates the $i$th submodule of this series, the elements of $G$ are expressible as triangular matrices. Furthermore, since the composition factors are trivial, the diagonal entries of these matrices are $1$.

Finally, the set of upper unitriangular matrices in $\mathrm{GL}_n(q)$ ($q$ a power of $p$) is a $p$-group, and the same holds for $\mathrm{GL}_n(k)$, and we are done.

Answer 3

Here is a completely different proof. It uses the fact that the Jacobson radical is nilpotent and maximal, and the residue ring is a division ring.

Let $G$ be a finite group, let $k$ have characteristic $p$, and suppose that $kG$ is local. Thus $I=J(kG)$ is nilpotent and maximal, and $kG/I$ has no zero divisors.

Let $q\neq p$ be a prime and suppose that $x\in G$ has order $q$. Then $x^q-1=0$ in $kG$. Since $kG/I$ is a division ring, and $x^q-1=(x-1)(x^{q-1}+\cdots+x+1)$, one of those factors is zero in $kG/I$, i.e., lies in $I$. If $x-1$ lies in $I$ then (as $I$ is nilpotent) some sufficiently large power $(x-1)^{p^i}=x^{p^i}-1$ is zero, which is obviously false. Thus the second term lies in $I$.

Raise $\alpha=x^{q-1}+x^{q-2}+\cdots+x+1$ to the $p$th power: $$(x^{q-1}+x^{q-2}+\cdots+x+1)^p=x^{p(q-1)}+x^{p(q-2)}+\cdots+x^p+1.$$ Since $q$ is a prime, the set $\{pi\mid 1\leq i\leq q-1\}$ is equal to $\{i\mid 1\leq i\leq q-1\}$ modulo $q$. Thus $\alpha^p=\alpha$, so clearly this is not nilpotent either. This is a contradiction.

Answer 4

We prove the statement by contradiction. Let $x \in G$ whose order is not a power of $p$. Let $H=\langle x \rangle$. Then $H$ has an irreducible nontrivial representation $V$ over $k$. (Note: this is wrong if the order of $x$ is a power of $p$, because then the only irreducible representation is the trivial one.) Consider the co-induced representation $W:=\mathrm{Hom}_{k[H]}(k[G],V)$. This is a representation of $G$. We compute its invariants. Because we used the co-induced representation (and not the induced one) we have without any assumptions regardings the finiteness of $[G:H]$ via Shapiro's lemma [note: or just the Tensor-Hom adjunction, see below]

$$W^G=H^0(G,W)=H^0(H,V)=V^H=0$$

So we have found a nontrivial representation with trivial invariants.

But if $k[G]$ is local, then the only left maximal ideal is the augmentation ideal, so every irreducible representation is trivial, because every such representation is isomorphic $k[G]/\mathfrak{m}$ for a left maximal ideal $\mathfrak{m}$.

But every nontrivial representation has a irreducible subrepresentation, so if $k[G]$ is local, then $W^G \neq 0$ for all nonzero $k[G]$-modules $W$.

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If you don't want to cite a result from group cohomology, here's an elementary proof of the equality $$\mathrm{Hom}_{k[H]}(k[G],V)^G=V^H$$ for any $k[H]$-module $V$. Let $k$ denote the trivial $k[H]$-module or $k[G]$-module, respectively. Then we have $$\mathrm{Hom}_{k[H]}(k[G],V)^G=\mathrm{Hom}_{k[G]}(k,\mathrm{Hom}_{k[H]}(k[G],V))=\mathrm{Hom}_{k[H]}(k \otimes_{k[G]} k[G],V)$$ $$=\mathrm{Hom}_{k[H]}(k,V)=V^H$$