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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 $\phi(a) = \phi(a^1) = b^{1n} = b^n = a$.
Apr
9
comment Find an automorphism
If $b^n = a$ then the map $\phi(a^j) = b^{jn}$ is the identity map and it doesn't map $a$ to $b$.
Apr
9
comment Find an automorphism
You need to explain what $a$, $b$, and $j$ are, otherwise that's just a random string of letters.
Apr
9
revised Help to understand and complete a proof of a theorem about nilpotent endomorphisms and Jordan-basis
deleted 2 characters in body
Apr
9
answered Find an automorphism
Apr
9
comment Direct limit sheaf.
You can't write them explicitly, they are given by sheafification of a map that's given by a universal property. You'll have to use universal properties to prove that the two compositions in your commutative diagram are identical.
Apr
9
answered $GL_3(\mathbb{F}_2)$ is simple
Apr
9
answered Direct limit sheaf.