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34m
answered How to find the base of Hom(U, V)
1d
comment Number of onto and into group homomorphisms between $\mathbb Z$ and $\mathbb Z$
As a group ${\Bbb Z}=<1>$. Thus any group homomorphism $\phi:{\Bbb Z}\rightarrow{\Bbb Z}$ is determined by $\phi(1)\in\Bbb Z$.
1d
revised Induced subgroup of $\pi_1(S^1)$ by $p_n$
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1d
revised How many times can you fold an a4 paper?
edited tags
2d
comment Some subgroup of $GL_2(\mathbb{Q})$
what are the elements $A\in GL_2$ such that $A^2=I$?
2d
answered Induced subgroup of $\pi_1(S^1)$ by $p_n$
Jan
20
awarded  Taxonomist
Jan
19
comment Galois-theory - a question about the galois group
@Epsilondelta: A Galois automorphism needs to send a root of a polynomial to another root of the same polynomial. Since $\sqrt{2}$ is a root of $x^2-2$, any Galois automorphism either fixes $\sqrt{2}$ or sends it to $-\sqrt{2}$. Since your field is generated by $\sqrt{2}$ and $\sqrt{3}$ you have 4 different ways an automorphism can act on them, hence on the field itself.
Jan
19
answered Find the algebraic set $V(S)$
Jan
19
comment Describe the set of all $z$ such that $Im(z+5)=0$
Adding a real number to a complex number won't change its imaginary part.
Jan
19
comment How do you solve the equation $ (z^2-1)^2 = 4 ? $
You mean $z_1^2=3$ and so on
Jan
19
answered Galois-theory - a question about the galois group
Jan
19
answered Topology on the tensor Bundle $T^{r, s}(M)$?
Jan
18
comment If $x^m=e$ has at most $m$ solutions for any $m\in \mathbb{N}$, then $G$ is cyclic
@Belgi: an element in $A_d$ generates a cyclic group of order $d$, read the definition!
Jan
18
answered Isomorphism of rings (with parameters)
Jan
18
answered Show that if $a$ is an integer, then 3 divides $a^3 - a $
Jan
18
comment If $x^m=e$ has at most $m$ solutions for any $m\in \mathbb{N}$, then $G$ is cyclic
@Belgi: If $A_d=\emptyset$ the inequality is certainly valid,so assume there exists $x\in A_d$. Now consider the cyclic subgroup $<x><G$. It consists of $d$ elements, so it exhausts by hypothesis the elements in $G$ with $z^d=e$. But now $A_d$ is nothing but the set of generators of $<x>$. Thus ...
Jan
13
answered irreducibility in $\mathbb Z/p\mathbb Z[X]$
Jan
13
answered How to find left and right cosets of a subgroup
Jan
13
revised Question regarding a imaginary number problem
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