1
$\begingroup$

I have a question about the Maschke's theorem in the group representation.

I know that Maschke's theorem says that "Every representation of a finite group having positive degree is completely reducible, i.e. it has a direct sum of irreducible G-modules.".

I'm very confused when a G-module is different to the given G-module.

More precisely, if G is a finite group and V is a G-module with $\dim V>0$, then V is a direct sum of irreducible G-submodules $V_i$. Now, if W is another G-module, then by the Maschke's theorem, W also has a decomposition of irreducible G-submodules. I wonder if the irreducible G-submodules of W are precisely $V_i$. That is, I think the irreducible factors does not depend on G-module, and depend only on the given group G.

Is it right? I need your help. Thanks.

$\endgroup$
14
  • $\begingroup$ No, this is clearly not true, as can be checked by a few small examples. What is true is that the possible $V_i$ are determined by $G$ (up to isomorphism of $G$-modules), as these are just the irreducible $G$-modules. $\endgroup$ – Tobias Kildetoft Jan 29 '16 at 12:22
  • $\begingroup$ Then, if V is isomorphic to W as G-modules, then is it right? $\endgroup$ – JeongHobin Jan 29 '16 at 12:29
  • $\begingroup$ Yes, clearly if the modules are isomorphic then they have the same summands. $\endgroup$ – Tobias Kildetoft Jan 29 '16 at 12:29
  • $\begingroup$ Is the reason Maschke's theorem?? $\endgroup$ – JeongHobin Jan 29 '16 at 12:30
  • $\begingroup$ I am not sure what you are really asking here. Maschke's theorem says that the modules are completely reducible. That the indecomposable summands of isomorphic modules are the same is not related to that theorem. $\endgroup$ – Tobias Kildetoft Jan 29 '16 at 12:34
0
$\begingroup$

You haven't mentioned what field you are working over, but I imagine it's probably something safe like $\Bbb C$. Really you ought to say that the irreducible factors depend on both $G$ and the field $F$.

Obviously the irreducible submodules of two given modules may be totally unrelated. Just consider two nonisomorphic simple modules.

What is true is that all the simple modules will appear in the regular representation $F[G]$.

$\endgroup$
4
  • $\begingroup$ Yeah, I mean $F=\mathbb{C}$. I want to know that "all the simple modules will appear in the regular representation $F[G]$". Are the simple modules G-submodules of $F[G]$? $\endgroup$ – JeongHobin Jan 29 '16 at 14:08
  • $\begingroup$ @JeongHobin I haven't really thought about $G$-submodules, but my understanding is that they are equivalent to $F[G]$ submodules of $F[G]$, if you extend the $G$ action linearly. $\endgroup$ – rschwieb Jan 29 '16 at 14:32
  • $\begingroup$ I'm sorry but I don't understand your comment :( $\endgroup$ – JeongHobin Jan 30 '16 at 2:18
  • $\begingroup$ An action $G\times V\to V$ should correspond with a module action $F[G]\times V\to V$ $\endgroup$ – rschwieb Jan 30 '16 at 3:25

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.