Let $G$ be a finite group. If $L$ is a finite free $\mathbf{Z}$-module with an action of $G$, say $L$ is induced if it's isomorphic as a $G$-module to $Ind_H^G(\mathbf{Z})$ with $H$ a subgroup of $G$ and $H$ acting trivially on $\mathbf{Z}$. And, for want of better terminology, let's say $L$ is a sum-of-induceds if it's isomorphic to a direct sum of induced modules in the sense above. [EDIT: Ben Webster points out that "permutation representation" is a rather better name for this notion! It's just the $\mathbf{Z}$-module coming from the action of $G$ on a finite set.]

The question: Is a finite free $\mathbf{Z}$-module which is an extension of one sum-of-induceds by another, also a sum-of-induceds?

Over $\mathbf{Q}$ this is trivial because every short exact sequence splits. But this is not true over $\mathbf{Z}$. For example there are two actions of $\mathbf{Z}/2\mathbf{Z}$ on $\mathbf{Z}^2$ which become the group ring over $\mathbf{Q}$: one has the non-trivial element acting as $(1,0;0,-1)$ and the other has it acting as $(0,1;1,0)$. The latter is a non-split extension of the trivial 1-d representation by the non-trivial one (and also a non-split extension of the non-trivial one by the trivial one). Note that this non-split extension is induced.

There are mod $p$ extensions that don't split either. For example over $\mathbf{Z}/p\mathbf{Z}$ there is a non-trivial extension of the trivial representation by itself. But this extension does not lift to an extension of the trivial $\mathbf{Z}$-module by itself.

Why am I interested? For those that know what a $z$-extension of a connected reductive group over a number field is, my "real" question is: is a $z$-extension of a $z$-extension still a $z$-extension? I've checked the geometric issues here but the arithmetic one above is the one I haven't resolved. $G$ is a Galois group and the $\mathbf{Z}$-modules are the character groups of the central tori in question. If I've understood things correctly, a $z$-extension of a $z$-extension is a $z$-extension iff the question I ask above has a positive answer.

Note finally that applying the long exact sequence of cohomology, and using the fact that induced representations have no cohomology by Shapiro's Lemma, we see that the extension I'm interested in also has no cohomology (and furthermore its restriction to any subgroup has no cohomology either). Is this enough to show it's induced?

  • 2
    $\begingroup$ I think "sum of induced" just means "permutation representation." Am I missing something? $\endgroup$
    – Ben Webster
    Jan 10, 2010 at 18:41
  • $\begingroup$ I think you're right. $\endgroup$ Jan 10, 2010 at 19:17

1 Answer 1


In my interpretation of your description, a sum-of-induceds is just the permutation representation of some (not necessarily connected) G-set. The representation $\mathrm{Hom}(V_1,V_2)$ for two permutation representations is itself permutation: it's the permutation action on product of the G-sets.

So, $\mathrm{Ext}^1(V_1,V_2)=H^1(G,\mathrm{Hom}(V_1,V_2))$ which you claim vanishes. So it sounds to me like every extension between permutation representations splits. [EDIT: I'm no longer nervous] I'm still a little nervous about this cohomology vanishing, but I don't know integral representation theory so well, so it's hard for me to say.

  • $\begingroup$ I don't think I'm claiming that the ext group vanishes. I'm just saying that every element of the ext group gives rise to a sum-of-induceds representation, not that it's the sum of the two sum-of-induceds we're starting with. $\endgroup$ Jan 10, 2010 at 19:12
  • $\begingroup$ Oh, but stop, the ext group clearly vanishes by Shapiro. So done! Thanks! $\endgroup$ Jan 10, 2010 at 19:20
  • 6
    $\begingroup$ For any finite group G, the first cohomology group H^1(G,Z) of the group G with constant integral coefficients Z vanishes, since there are no nonzero group homomorphisms G->Z. By the Shapiro Lemma, it follows that the first cohomology of G with coefficients in any integral permutational representation vanishes, too. Hence your argument is indeed correct and any extension of integral permutational representations splits. $\endgroup$ Jan 10, 2010 at 19:49

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