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I've been asked some questions by a friend interested in crystallography, and the following questions (I'm not an expert) came spontaneous to me:

1) Are there two finite subgroups $P,P'\subset\mathrm{GL}(n,\mathbb{Z})$ that are abstractly isomorphic but not conjugate in $\mathrm{GL}(n,\mathbb{Z})$?


2) Are there two finite subgroups $P,P'\subset\mathrm{GL}(n,\mathbb{Z})$ that are abstractly isomorphic but not conjugate in $\mathrm{GL}(n,\mathbb{R})$?


3) What if we don't assume $P$ and $P'$ to be finite? (ok, this has nothing to do with crystallography)

(They may well be classical and well known results, hence the tag "reference request")

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Maybe you find some useful information (like the answer for 1) in the (answers to the) related questions… and – j.p. Mar 19 '12 at 11:47
up vote 14 down vote accepted

The answer to all three questions is yes and certainly is classical.

One simple example is the following:

Let $C_2$ act faithfully on the set $\{1,2,3,4\}$ in two ways. In the first the non-trivial element of $C_2$ swaps 1,2 and also swaps 3,4. In the second the non-trivial element swaps 1,2 and fixes 3,4.

Each action defines a representation of $C_2$ on $\mathbb{Z}^4$ via permuation matrices. In one case the trace of the non-trivial permutation matrix is $0$ in the other $2$ so the images cannot be conjugate in $GL_4(\mathbb{Z})$ or $GL_4(\mathbb{R})$ however they both generate a subgroup $C_2$ in the former.

It is fairly clear this idea generalises to any isomorphism class of finite groups.

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Since no-one has explicitly mentioned character theory and representation theory, I will say a few words. If we have a finite group $G$, and two (faithful, ie with trivial kernel) representations $\sigma, \tau : G \to {\rm GL}(n,F)$ where $F$ is a field of characteristic zero, asking whether $G\sigma$ and $G\tau$ are conjugate in ${\rm G}(n,F)$ is the same as asking whether the representations are equivalent over $F.$ An obvious necessary condition is that $g \sigma$ and $g\tau$ have the same trace for all $g \in G,$ but character theory (and some Schur index theory, etc.) tell us that this condition is sufficient if the representation is irreducible, and then, with some work, in general (the field $F$ need not be algebraically closed for this conclusion. The (well-known) point is is that in the irreducible case, if the two representations can be intertwined over a n extension of $F,$ they already can be intertwined over $F.$ The question for integral representations is more difficult. It may be that representations $\sigma, \tau : G \to {\rm GL}(n,\mathbb{Z})$ are equivalent after extending the ground ring to $\mathbb{Q},$ but are not equivalent as representations over $\mathbb{Z}.$ A relevant theorem putting some control on the situation is one of Jordan-Zassenhaus. An example is that ${\rm GL}(2,\mathbb{Z})$ has two subgroups isomorphic to the dihedral group $D$ with $8$ elements which are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but are conjugate as subgroups of ${\rm GL}(2,\mathbb{Q}).$ The group $D$ had two normal Klein $4$-subgroups $U$ and $V$. If we induce a non-trivial $1$-dimensional representation of $U$ to $D$ we get a subgroup $E$ of ${\rm GL}(2,\mathbb{Z})$ isomorphic to $D,$ and we can to the same for $V$ to get another subgroup $E^{\prime}.$ The subgroups $E,E^{\prime}$ are not conjugate within ${\rm GL}(2,\mathbb{Z}),$ but the representations afford the same character, so are equivalent as rational representations. (The technical reason is that if we pass to an appropriate local ring with residue field of characteristic $2,$ we have two indecomposable modules which have non-conjugate vertices, so are not isomorphic).

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The answers are "yes" to all three questions. Consider two matrices $a=diag(-1,-1,-1)$ and $b=diag(1,1,-1)$. The subgroups of $GL_3(\mathbb{R})$ generated by these matrices are isomorphic (of order 2) but not conjugate because there are different multiplicities of eigenvalues.

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You mean "yes" :). – Martin Brandenburg Mar 19 '12 at 11:50
Doesn't that mean that the answer is 'yes'? – HJRW Mar 19 '12 at 11:50

If two homomorphisms $h,h':G\to GL_n(\mathbb Q)$ are conjugate in $GL_n(\mathbb R)$ then they are conjugate in $GL_n(\mathbb Q)$. To see this, consider the set of all rational matrices $X$ such that $Xh(g)=h'(g)X$ for all $g$. This is a rational vector space, and a basis for it is a real basis for the corresponding space of real matrices. It follows easily that the existence of such a real matrix with non-zero determinant implies the existence of such a rational matrix with non-zero determinant. This argument does not require $G$ to be finite (or, for that matter, to be a group).

But you can easily find two non-conjugate elements of order $2$ in $GL_2(\mathbb Z)$ that are conjugate in $GL_2(\mathbb Q)$.

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