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6h
comment What's a group whose group of automorphisms is non-abelian?
@PyRulez I think you meant "the automorphism group contains the permutation group as a subgroup." Also, you should fix the last paragraph too (there are homomorphisms of the free group that don't come from functions on the basis elements)
7h
comment What's a group whose group of automorphisms is non-abelian?
Pate's right. Still, this is a great approach which answers the question, since the automorphism group of the free group will contain a copy of the symmetric group (and the homomorphisms of the free group on $\mathbb{Z}$ will contain the collection of functions on $\mathbb{Z}$).
7h
awarded  Nice Answer
1d
awarded  abstract-algebra
1d
comment What's a group whose group of automorphisms is non-abelian?
This is my favorite answer so far.
1d
answered What's a group whose group of automorphisms is non-abelian?
Jun
28
comment How to define a group ring when the group is infinte?
That just follows from the definition. View $M$ in multiplicative notation as the free abelian group generated by $x_1,\dots,x_n$. Then $\mathbb{C}[M]$ is the $\mathbb{C}$-vector space with basis $M$ and a certain multiplication (which I won't write, but you know what it is). Then just see that the multiplication is the same as it is in the space of Laurent polynomials.
Jun
28
revised How to define a group ring when the group is infinte?
edited body
Jun
28
answered How to define a group ring when the group is infinte?
Jun
24
comment A question about discrete topological space
Well, why don't you start be recalling the definition of locally compact?
Jun
22
answered page 4 of Milnor's book on Morse Theory
May
17
awarded  Yearling
May
16
comment Show that the radius of convergence of a sum of series is at least as big as minimum of radii of these series.
Take a point within radius of convergence of both series. Use finite sums to show the sum of the series converges at that point. Then use basic facts about radius of convergence of power series.
Apr
23
comment Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function and $C<1$ such that $|f'(x)|\le C$ for all $x$
Are you asking a different question in the title and the body?
Apr
23
revised Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function and $C<1$ such that $|f'(x)|\le C$ for all $x$
edited title
Apr
10
comment All prime ideals are maximal - Counterexample
Can you think of an integral domain which is not a field?
Apr
8
comment Notation help abstract algebra
Yes. Please stop using all caps though.
Apr
8
comment Notation help abstract algebra
Note that $R$ needs to be an integral domain for this to make sense.
Mar
30
comment The theorem on formal functions
Taking inverse limits is a way of looking at all infinitesimal neighborhoods at once, so to me it seems like a natural thing to do, not something that muddy's the comparison.
Mar
27
comment Give an example to show that there need not exist $x\in X$ such that $\|x\|=1$ and $Lx=1$.
Ok, I fixed the answer.