Let $G$ be a finite group. Is $\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G$ free as a $\mathbb{Z}$-module, where $Z$ denotes the centre?
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$\begingroup$ So the only issue is whether there is torsion? $\endgroup$– Geoff RobinsonCommented Dec 6, 2016 at 10:43
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$\begingroup$ Yes. It seems obvious that there is no torsion but I can't prove it. $\endgroup$– M. LiveseyCommented Dec 6, 2016 at 10:45
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$\begingroup$ Isn't $\mathbb{Z}G$ free as a $Z(\mathbb{Z}G)$-module? I would guess that if we take $\mathbf{T}\subseteq G$ to be a transversal set of $G/Z(G)$ it would be obvious that $\mathbf{T}$ spans $\mathbb{Z}(G)$ over $Z(\mathbb{Z}G)$. So your claim should follow from the general fact that a tensor product of free modules over a commutative ring is again free.. $\endgroup$– kneidellCommented Dec 6, 2016 at 11:33
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5$\begingroup$ @kneidell No. For example, if $G$ is dihedral of order $8$ then $Z(\mathbb{Z}G)$ has $\mathbb{Z}$-rank $5$, so $\mathbb{Z}G$ can't be a free $Z(\mathbb{Z}G)$-module. $\endgroup$– Jeremy RickardCommented Dec 6, 2016 at 11:44
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4$\begingroup$ @EhudMeir It is also not necessarily projective. Again, Jeremy's example works as a counterexample. If $\mathbb{Z}G$ were projective as a $Z(\mathbb{Z}G)$-module then $\mathbb{F}_2G$ would be projective as a $Z(\mathbb{F}_2G)$-module. However, since $G$ is a $2$-group $Z(\mathbb{F}_2G)$ is a local ring and so any finite dimensional projective $Z(\mathbb{F}_2G)$-module is a direct sum of copies of $Z(\mathbb{F}_2G)$. Now see Jeremy's comment for a contradiction. $\endgroup$– M. LiveseyCommented Dec 6, 2016 at 12:11
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1 Answer
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There's a distinctly non-zero chance that my calculations are wrong, but I think it has a lot of $2$-torsion for $G=A_5$, the alternating group of degree $5$. I'm afraid the method I've used is a little indirect, and I haven't extracted an explicit torsion element.
Certainly $\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G$ is a finitely generated abelian group, and so it will be free if and only if $\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G,\mathbb{F})$ has the same $\mathbb{F}$-dimension for every field $\mathbb{F}$.
But $$\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G,\mathbb{F})\cong \operatorname{Hom}_{Z(\mathbb{Z}G)}\left(\mathbb{Z}G,\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}G,\mathbb{F})\right),$$ which is isomorphic to $\operatorname{Hom}_{Z(\mathbb{F}G)}(\mathbb{F}G,\mathbb{F}G),$ the endomorphism algebra of the group algebra $\mathbb{F}G$ as a module over its centre.
For $\mathbb{F}=\mathbb{C}$ this is easy to calculate from the irreducible character degrees. It's the sum of the fourth powers of the degrees, which for $G=A_5$ is $1^4+3^4+3^4+4^4+5^4=1044$.
For $\mathbb{F}$ algebraically closed of characteristic two, $\mathbb{F}A_5$ has a simple non-principal block isomorphic to $M_4(\mathbb{F})$ which contributes $4^4=256$. The centre of the principal block has a fairly simple structure $Z=\mathbb{F}[x,y,z]/(x,y,z)^2$, where $x,y$ and $z$ are in the socle of the principal block, each annihilated by all but one of the conjugacy classes of primitive idempotents, and according to my calculation, as a module for this centre, the principal block is the direct sum of one copy of $Z/(y,z)$, four copies of each of $Z/(x,y)$ and $Z/(x,z)$, and $26$ copies of $Z/(x,y,z)$, and the endomorphism algebra is $1258$-dimensional, so that (taking into account the non-principal block) the endomorphism algebra for the whole group algebra as a module for its centre is $1514$-dimensional.
This can't be the best way to do it, but maybe it provides some clues for a more illuminating answer.
By the way, I tried some smaller groups ($A_4$ in characteristic $2$, $D_{2p}$ in characteristic $p$), and for those the dimensions were the same.
answered Dec 7, 2016 at 12:58
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$\begingroup$ You may know (and be using) this already, but I think that when $\mathbb{F}$ is algebraically closed of characteristic $p$ ( which may be $0$), we have ${\rm Hom}_{Z(\mathbb{F} G)}(\mathbb{F}G,\mathbb{F}G) = \oplus_{B}{\rm Hom}_{Z(B)}(B,B) $, where $B$ runs over the blocks of $\mathbb{F}G$. $\endgroup$ Commented Dec 8, 2016 at 10:29
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$\begingroup$ @Jeremy Rickard Thanks Jeremy. That's not the answer I wanted but it seems to be correct! I am acually interested in $$\operatorname{Hom}_{Z(\mathbb{F}G)}\left(\mathbb{F}G,\mathbb{F}G\right)$$ and wanted it to be generated by left and right multiplication by $\mathbb{F}G$. I think you've essentially disproved this. Thanks for stopping me wasting my time! $\endgroup$ Commented Dec 8, 2016 at 10:29
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$\begingroup$ @M.Livesey Are you sure I've essentially disproved that? I assume by "generated by left and right multiplication" you mean that every $Z(\mathbb{F}G)$-endomorphism of $\mathbb{F}G$ is a linear combination of endomorphisms of the form $x\mapsto gxh$, or in other words, that the natural map $\mathbb{F}G^{op}\otimes_{Z(\mathbb{F}G)}\mathbb{F}G\to\operatorname{End}_{Z(\mathbb{F}G)}(\mathbb{F}G)$ is surjective. I haven't thought this through carefully, but doesn't my example just show that both the domain and codomain of that map are larger than you might guess, but by exactly the same amount? $\endgroup$ Commented Dec 8, 2016 at 19:46
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$\begingroup$ @JeremyRickard I think you have disproved it. We know that $\mathbb{F}G^{op}\otimes_{Z(\mathbb{F}G)}\mathbb{F}G$ and $\operatorname{End}_{Z(\mathbb{F}G)}(\mathbb{F}G)$ have the same dimension and so surjectivity is equivalent to injectivity. The torsion elements you've proved exist must act by zero via the map $$\mathbb{Z}G^{op}\otimes_{Z(\mathbb{Z}G)}\mathbb{Z}G\to\operatorname{Hom}_{Z(\mathbb{Z}G)}(\mathbb{Z}G)$$ but they are not zero in $\mathbb{F}G^{op}\otimes_{Z(\mathbb{F}G)}\mathbb{F}G$. Therefore we don't have injectivity. $\endgroup$ Commented Dec 9, 2016 at 14:15
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1$\begingroup$ @M.Livesey Ah! If $\alpha$ is a $Z(\mathbb{F}G)$-module endomorphism of $\mathbb{F}G$ that is generated by left and right mulitplication by $\mathbb{F}G$, then $\alpha(I)\subseteq I$ for any two-sided ideal $I$ of $\mathbb{G}$. But as a $Z(\mathbb{F}A_5)$-module, $\mathbb{F}A_5$ has many isomorphic one-dimensional direct summands some of which are in $\text{rad}^2(\mathbb{F}A_5)$ and some of which are not, so there are endomorphisms that don't preserve $\text{rad}^2(\mathbb{F}A_5)$. $\endgroup$ Commented Dec 11, 2016 at 10:06