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1d
revised Nilpotent ideal and ring homomorphism
added 110 characters in body
1d
answered Nilpotent ideal and ring homomorphism
Sep
2
comment Is any Direct Summand of a Free Module over a PID Also Free?
Any submodule of a free module over a PID is free, regardless of whether the free module is finitely generated or not. The proof is not trivial, you can find one here.
Aug
30
revised Why is the arrow ideal $R_Q$ of a finite, connected, acyclic quiver $Q$ equal to the Jacobson radical?
edited body
Aug
30
answered Why is the arrow ideal $R_Q$ of a finite, connected, acyclic quiver $Q$ equal to the Jacobson radical?
Aug
28
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Aug
25
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Aug
25
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Aug
24
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Aug
24
comment $G$ has order $p^a$, then the center of $G$ counts more than the identity
@GuerlandoOCs Yes, that's better. There's no need to introduce $|C|$ in the last equation, just leave it as $\sum |C_k|$, this whole sum will vanish mod $p$ since you've shown in the previous line that each summand $|C_k|$ divides $|G|$.
Aug
24
comment $G$ has order $p^a$, then the center of $G$ counts more than the identity
@GuerlandoOCs It's more or less right, you should have cardinality signs around the $g_k$ and $C_k$, and $\sum g_k$ is $Center(G)$ should be $\sum |g_k|=|Z(G)|$, you need to add numbers not group elements. Lastly, it should be $|C_k|=|G|/|N_G(x)|$ for any particular $x\in C_k$, this applies to each individual conjugacy class. It doesn't make sense to sum up conjugacy classes, they're not numbers.
Aug
23
comment $G$ has order $p^a$, then the center of $G$ counts more than the identity
@GuerlandoOCs If you have that result, it says the number of conjugates of $x$ (that is, the cardinality of its conjugacy class) is the index of $N_G(x)$, which means the cardinality of the conjugacy class $C$ of $x$ is $|C|=|G|/|N_G(x)|$. This now gives $|C||N_G(x)|=|G|$, which is what is needed to prove $|C|$ divides $|G|$. So you can jump over any uses of the Orbit-Stabilizer above, although that is the more general phenomenon.
Aug
23
comment $G$ has order $p^a$, then the center of $G$ counts more than the identity
@GuerlandoOCs For the second comment, the normalizer and centralizer are the same in the case of a single element: $$N_G(x)=\{g\in G:gxg^{-1}\subseteq\{x\}\}=\{g\in G:gxg^{-1}=x\}=C_G(x).$$ So that exercise seems like it's asking you to prove a special case of the orbit stabilizer theorem with the conjugation action.
Aug
23
comment $G$ has order $p^a$, then the center of $G$ counts more than the identity
@GuerlandoOCs By definition, if $C$ is a conjugacy class, it consists of the class of elements which are all conjugate to each other. So if $x\in C$, $x$ is conjugate precisely to the other elements in $C$, so $C$ is the conjugacy class of $x$. Since the orbit of $x$ is also its conjugacy class (see third paragraph), $Gx$ and $C$ must be the same.
Aug
23
comment $G$ has order $p^a$, then the center of $G$ counts more than the identity
@GuerlandoOCs Those aren't just any singletons, those are the singleton conjugacy classes, which are precisely the elements of $Z(G)$ (see the second paragraph of the answer). So adding them up is the same as counting the elements of $Z(G)$.
Aug
23
comment $G$ has order $p^a$, then the center of $G$ counts more than the identity
I should note the equation $(\ast)$ is the famous class equation, which is the way I've usually seen this result derived.
Aug
23
answered $G$ has order $p^a$, then the center of $G$ counts more than the identity
Aug
23
comment Set of all inner automorphisms is a normal subgroup
@GuerlandoOCs No, $g$ is an automorphism of $G$, not an element of $G$, so $$g\varphi_ag^{-1}(x)=g\varphi_a(g^{-1}(x))=g(a^{-1}g^{-1}(x)a)=g(a^{-1})g(g^{-1‌​}(x))g(a)=g(a)^{-1}xg(a)=\varphi_{g(a)}(x).$$
Aug
22
comment Set of all inner automorphisms is a normal subgroup
For clarity, instead of $f$, let $\varphi_a$ denote the inner automorphism $\varphi_a(x)=a^{-1}xa$. Then show $g\varphi_a g^{-1}=\varphi_{g(a)}$, hence $g\varphi_a g^{-1}$ is inner.
Aug
10
comment This is an ISI B.Math Entrance Test Problem
The line passing through $(a,b)$ and $(-1,-2)$ will intersect the circle in two points, these will be the nearest and farthest points from $(a,b)$.