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13h
awarded  Nice Question
23h
comment $ G $ is soluble, then every properties of $ G $ is inherited by $ G/N $ ?
Well, what does "property" mean here? Cardinality is not inherited, say!
2d
answered Let $q$ be a prime integer. Show that for each $x∈GF(q)$ there exist elements $r$ and $s$ in $GF(q)$ satisfying $x=r^2+s^2$.
2d
comment Are all integers fractions?
Is that a rational? Yes it is!
2d
comment Show that a one dimensional $\mathfrak {g}\!-\!\operatorname{module}$ is irreducible
A nonzero submodule has either dimension $1$ or $0$. In any case, it is either the trivial submodule or the whole module, which is what you want. Note the fact a one dimensional representation (of whatever structure you're dealing with) is irreducible is more or less independent of what structure you're considering. The important fact is that subrepresentations are also subspaces, are you can control subspaces by their dimension.
Aug
24
answered An element of $GL_n(\mathbb F_p)$ cannot have order $p^2$ if $n < p$
Aug
24
comment An element of $GL_n(\mathbb F_p)$ cannot have order $p^2$ if $n < p$
The eigenvalues of $M$ are the roots of $\chi_M$. Is that what's troubling you? Since in this case $\chi_M$ has only the root $1$, that's it. And you don't need to consider any algebraic closure, since $\chi_M$ and already splits completely!
Aug
24
revised Inverse sum representation of sine
added 19 characters in body
Aug
24
comment Inverse sum representation of sine
@GEdgar Yes, I agree one should combine the symmetric terms.
Aug
24
revised Inverse sum representation of sine
added 1 character in body
Aug
24
answered Inverse sum representation of sine
Aug
24
comment Combinatorial interpretation of an alternating binomial sum
Could you provide some insight as to where the definition of good word comes from?
Aug
22
answered Dimension of totally isotropic subspaces for a given quadratic form
Aug
22
comment Does $f\otimes_A 1_{A/m}:M\otimes A/m\to N\otimes A/m$ injective for all maximal $m$ imply $f$ is an isomorphism?
@user114539 Yes, thats the idea.
Aug
21
comment Is this ring a PID?
What have you tried?
Aug
20
comment A group of order 2p (p prime) and other conditions - prove abelian.
@Ilya.K. If $G/Z(G)$ is cyclic then it is the trivial group $1$; that is, $G$ is abelian. You can always try to prove this, or search it here.
Aug
20
comment I have no idea what “smooth structure” is
What is your question? If you don't know what something is you can always read about it in a book. Now, if you don't understand what you read or you have a doubt, say, you can always make a precise question here!
Aug
20
answered A group of order 2p (p prime) and other conditions - prove abelian.
Aug
20
comment Does $f\otimes_A 1_{A/m}:M\otimes A/m\to N\otimes A/m$ injective for all maximal $m$ imply $f$ is an isomorphism?
@user26857 Sure, just thought it was something worth mentioning.
Aug
19
comment Proof: A matrix with $m$ rows and $n$ colums has $nm$ entries.
What is your definition of matrix?