6
$\begingroup$

Below is the construction of two non isomorphic groups, $G_1$ and $G_2$ such that $KG_1 \cong KG_2$ for any field $K$.

(My Doubts lie within.)


Consider two groups

$Q_1=\langle x_1,y_1,z_1\ |\ x_1^5=y_1^5=z_1^5=1, [x_1,y_1]=z_1, z_1\ \text{is central}\ \rangle$

$Q_2=\langle x_2,y_2,z_2\ |\ x_2^5=z_2^5=1, y_2^5=z_2, [x_2,y_2]=z_2, z_2\ \text{is central}\ \rangle$.

and let $\langle u_1\rangle$ be cyclic group of order $4$ and $\langle u_2\rangle$ be cyclic group of order $2$.

Now defining actions of $u_i$ on $Q_j$ $(i,j\in \{1,2\})$ as following- $$x_j^{u_i}=x_j,\\y_j^{u_i}=y_j^{24}\\ z_j^{u_i}=z_j^{24}$$

Now w.r.t this action, we can define semidirect products $Q_i \rtimes \langle u_j\rangle$ and hence it makes sense to make groups $$G_1=Q_1 \rtimes \langle u_1\rangle\times Q_2 \rtimes \langle u_2\rangle\\G_2=Q_1 \rtimes \langle u_2\rangle\times Q_2 \rtimes \langle u_1\rangle$$

Now I have to prove that $G_1\ncong G_2$, but their rational algebras are isomorphic.

Now both $G_1$ and $G_2$ has same order i.e. $125\times 125\times 8=125000$.

How should I go about proving $G_1 \ncong G_2$?


What I know, is that by defining $y_i^{x_i}=y_iz_i^{-1}$ and $z_i^{x_i}=y^{-1}z^{-1}y$ we can see that $Q_i= \langle y_i,z_i \rangle \rtimes \langle x_i \rangle$.

Now, To Realize- $G_i'=\langle y_1,z_1\rangle \times \langle y_2,z_2\rangle$ and $(G_i)^{(5)}=\langle z_2\rangle$.

Now, calculating by explicitly writing $[a,b]$ for any two arbitrary elements of $G_i$ is a very tedious process, and while trying that I still could not see how $x_i's$ got eliminated and $[a,b]\in \langle y_1,z_1\rangle \times \langle y_2,z_2\rangle$. Is there a shorter method?

If suppose we prove the "To Realize" part, then If $G_1 \cong G_2$, it must imply $C_{G_1}(z_2) \cong C_{G_2}(z_2)$. Now it is not tough to check that $C_{G_1}(z_2)=Q_1\langle u_1\rangle \times Q_2$ and $C_{G_2}(z_2)=Q_1\langle u_2\rangle \times Q_2 \langle u_1^{2}\rangle$.

Now Sylow $2$-subgroup of $C_{G_1}(z_2)=Q_1\langle u_1\rangle \times Q_2$ and $C_{G_2}(z_2)=Q_1\langle u_2\rangle \times Q_2 \langle u_1^{2}\rangle$ are both of order $4$ but former is cyclic i.e. $\langle u_1 \rangle \times \langle 1 \rangle$ and latter is non-cyclic i.e. $\langle u_2 \rangle \times \langle u_1^2 \rangle$. Thus contradicts $C_{G_1}(z_2) \cong C_{G_2}(z_2)$, hence we prove that $G_1 \ncong G_2$

Now, all my trouble lies in To Realize part. Is there a better way to calculate $G_i'$ and $(G_i)^{(5)}$.


Now, Passman has proved that for any field $K$, group algebras $KG_1 \cong KG_2$. The proof is very tedious and hard, I will be drowning in that soon, but before that I was wondering if there is some field, preferably of characteristic $0$, may be $\Bbb{Q}$ or $\Bbb{C}$, for which two group algebras ($KG_1$ and $KG_2$, for above mentioned groups) are isomorphic, is realized quickly or proved with more ease then picking up the case of arbitrary field.

Thanks in advance for any inputs that you may offer.

P.S.- If something doesn't make sense, it could be a typo as it was a lot of typing. Please lemme know if something is bothering you in it.

$\endgroup$
  • 1
    $\begingroup$ You can use any invariant of a group algebra to show that two group algebras are not isomorphic, in characteristic zero or not. For example, the number of conjugacy classes is an invariant of the algebra, the number of 1-dimensional representations, etc. In general, your question »I was wondering if there is some field\dots» one cannot say much more that this. $\endgroup$ – Mariano Suárez-Álvarez Jun 29 '15 at 3:37
  • $\begingroup$ I'm confused that you first say the groups are non-isomorphic but have isomorphic rational algebras, but then that they have non-isomorphic group algebras over every field. Are you sure this isn't supposed to be an example of non-isomorphic groups such that $KG_1$ is isomorphic to $KG_2$ for every field $K$? $\endgroup$ – Jeremy Rickard Jun 29 '15 at 5:49
  • $\begingroup$ @JeremyRickard Thanks for suggesting that. It was a mistake. Now I have corrected it. Typing so much, stops brain activity.:P. In my Mind I has isomorphism of algebras, but I was using the symbol $\ncong$. It was a very heavy and confusing mistake, I apologize $\endgroup$ – Bhaskar Vashishth Jun 29 '15 at 6:00
  • 2
    $\begingroup$ " I was wondering if there is some field, preferably of characteristic 0, may be ℚ or ℂ, for which two group algebras are isomorphic". Clearly, for any two abelian groups of the same order, their complex group algebras are isomorphic. $\endgroup$ – Ofir Schnabel Jun 29 '15 at 9:55
  • $\begingroup$ @OfirSchnabel I meant some field $K$ for which $KG_1 \cong KG_2$ is realised quickly for above mentioned group. $\endgroup$ – Bhaskar Vashishth Jun 29 '15 at 16:48
2
$\begingroup$

As Mariano noted, you can tell immediately that they are not isomorphic by considering a number of their algebra invariants.

To know that they are isomorphic is usually more complicated. Though note that, as Ofir noted, all abelian groups of the same order over $\mathbb{C}$ (or any field with a primitive root with order the order of the groups in question) have isomorphic group algebras.

The reason for that is that the group algebras can be written as a sum of modules. A necessary and sufficient condition for $kG_1\cong kG_2$ is that they both admit module decompositions which are equivalent as vector spaces: just match up the dimensions and multiplicities of each factor. As an example, over $\mathbb{C}$ the group algebras of $D_8$—the dihedral group of order 8—and $Q$—the quaternion group of order 8—are isomorphic since they have the same number of 1-dimensional and 2-dimensional irreducible representations, and no higher irreducible representations. The high-level way of stating that is to say that the two representation categories have isomorphic Grothendieck rings.

Of course, knowing the $k$-representation theory of the groups in question may not be such a simple thing, but there's no real way of getting around that problem if you want to establish the isomorphism.

There is one (more restrictive) situation for which the question is pretty well-investigated, and that's the situation of isocategorical groups. Two groups are isocategorical when their $\mathbb{C}$-representation categories are monoidally equivalent (importantly, the symmetry structure of the tensor product will not be preserved if the groups are not isomorphic). This is a stronger statement than having isomorphic Grothendieck rings: $Q$ and $D_8$ are not isocategorical. Indeed, the linked paper shows that any group which is isocategorical to a non-isomorphic group necessarily has an abelian normal subgroup of order a power of 4 that can be equipped with a certain isomorphism (the lack of this isomorphism is what stops $Q$ being isocategorical to $D_8$).


For your particular groups, notice that they cannot be isocategorical: they have no abelian normal subgroups a power of 4. They have abelian subroups of order 4, but they're not normal.

So for your two particular groups you're pretty much forced to compute the representation categories (over $\mathbb{C}$, say). Your groups should be relatively easy to compute this for: dealing with the direct products is trivial and the semidirect products should be routine (induce representations of $Q_i$ up). The only potentially hard part is to figure out the representations of $Q_1$ and $Q_2$, but if you've ever tried constructing a character table for a group then that's a good starting point. Note that their commutator subgroups are easy to find, whence their linear character groups are easy to compute.

$\endgroup$
  • $\begingroup$ Please see edit. $\endgroup$ – Bhaskar Vashishth Jun 29 '15 at 16:55
  • 1
    $\begingroup$ Great answer! Small point: note that having the same multiset of character degrees is quite a bit weaker than having isomorphic Grothendieck rings, because the Grothendieck ring also captures the tensor product of irreps. (Consider abelian groups.) So char. table > Grothendieck ring > multiset of char. degrees (where ">" = "is at least as much information as"). For $Q, D_8$ they have the same character tables, but this is much stronger than necessary to have isomorphic group algebras. $\endgroup$ – Joshua Grochow Sep 15 '16 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.