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14h
answered Is there a correspondence between normal subgroups and representations of a (finite) group?
14h
answered Induced coaction on a vector space.
1d
answered Equivalence of definitions of the axiom of induction.
1d
answered $M \neq 0$ but $M^* = 0$.
1d
answered If $G$ is a Finite Group such that $H\le K$ or $K\le H$ for all Subgroups $H,K$ of $G$, then $G$ is Cyclic and of order $p^n$ for some Prime $p$.
Aug
20
answered What shall I learn in order to understand Auslander-Reiten theory and tilting theory?
Aug
7
asked Can MAGMA write Groebner basis elements in terms of the original generators?
Jul
17
awarded  Nice Answer
Apr
30
comment Prove that $\mathrm{Ind}_{\mathbb{I}}^G \cong \mathbb{C}[G]$
They are equal because they are vector spaces of the same dimension with the same action. And you're not restricting to $\mathbb I$, you're restricting to $\{I\}$ which is a subgroup.
Apr
30
answered Matrix Ring of a Semisimple Ring
Apr
30
comment Prove that $\mathrm{Ind}_{\mathbb{I}}^G \cong \mathbb{C}[G]$
Both $\mathbb I \oplus \cdots \oplus \mathbb I$ and $\mathrm{Res}_{\{I\}}\chi$ are vector spaces with the trivial action of the trivial group.
Apr
29
comment Prove that $\mathrm{Ind}_{\mathbb{I}}^G \cong \mathbb{C}[G]$
@1234: I'm not sure what you're asking.
Apr
27
awarded  Nice Answer
Apr
19
revised Writing equations without latex (for labelling, classification, structure analysis and rendering according to mathematical meaning)
edited tags
Apr
18
answered Finitely generated modules and submodules
Apr
15
revised Factorizing expressions
edited tags
Apr
14
comment Corollary to Maschke's Theorem.
@QiaochuYuan: That should be an answer.
Apr
14
comment Does this define a Ring?
For a non-trivial group it's never an integral domain. Consider the function that's just $1$ on all group elements and consider a function whose values when summed over the entire group yield zero.
Apr
10
comment Find an automorphism
That doesn't respond to my comment.
Apr
10
comment Find an automorphism
Why does $\phi(a) = b$? You said $a = b^n$ and $\phi(a^j) = b^{jn}$. Plug in $j = 1$ and we get $\phi(a) = \phi(a^1) = b^{1n} = b^n = a$.