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bio website fkraiem.org
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visits member for 11 months
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1d
comment If $H \leq G, \exists g \in G$ such that $HgHg^{-1} = G$, then $H = G$
$g = h_1^{-1}h_2^{-1}$, surely?
Nov
15
comment Why do we consider that $p$ & $q$ are co-primes when proving square root of a prime number is irrational?
Because it makes the proof easier. Or are you asking why we can limit our attention to the case where $p$ and $q$ are coprime?
Nov
15
comment Subgroups and corresponding subfields of galois group of $x^6 + x^3 + 1$
It is important to say that $\omega$ is a primitive $9$th root of unity, because if you just say that it is a $9$th root of unity, you can get different fields depending on which precise root you take (maybe this is what you meant by a "complex" root, but you should avoid this terminology: $1$ is just as well an element of $\mathbf{C}$ as $e^{2i\pi/9}$).
Nov
9
comment Why is the group G a normal subgroup of itself?
Be careful also that contrary to what you seem to think, even having "duplicates" does not imply that you do not "generate the entire group", unless the group is assumed to be finite. In other words, a non-injective map from an infinite set to itself can be surjective.
Oct
13
answered Suppose that $K$ is a field and that $f$ and $g$ are relatively prime in $K[x]$. Show that $f - Yg$ is irreducible in $K(y)[x]$.
Oct
9
revised Dihedral groups
Changed tags and added TeX.
Oct
9
suggested suggested edit on Dihedral groups
Oct
8
comment If $R$ is a ring and $A$ is a maximal ideal of $R$ then $R/A$ is a field
A very important property to keep in mind (which might be introduced in your class later, but it doesn't hurt to start thinking about it now) is that there is a bijection between the ideals of $R$ which contain $A$ and the ideals of $R/A$. (The result then follows trivially: if there is no proper ideal of $R$ which contains $A$, then $R/A$ has no proper non-trivial ideal, which is the definition of a field.)
Oct
7
comment Find a relationship between $f(f^{-1}(f(A)))$ and $f(A)$. Prove it in the general.
The standard way to prove that $A$ contains $B$ (or that $B$ is a subset of $A$) is to prove that if $x \in B$, then $x \in A$. Here, we show that if $x \in A$, then $x \in f^{-1}(f(A))$.
Oct
7
comment Find a relationship between $f(f^{-1}(f(A)))$ and $f(A)$. Prove it in the general.
Yes. And of course if $A$ contains $B$, then $f(A)$ contains $f(B)$ and you are done.
Oct
7
comment Find a relationship between $f(f^{-1}(f(A)))$ and $f(A)$. Prove it in the general.
Let $x \in A$, then $f(x) \in f(A)$. But $f^{-1}(f(A))$ is the set of all $x$ such that $f(x) \in f(A)$.
Oct
7
comment Group with exactly 2 elements of order 10.
Well, I added a direct quote.
Oct
7
revised Group with exactly 2 elements of order 10.
added 87 characters in body
Oct
7
comment Group with exactly 2 elements of order 10.
There is only one question, which is "but can I apply that idea to non cyclic groups such as dihedrals or symmetric groups?". I answered that.
Oct
7
answered Group with exactly 2 elements of order 10.
Oct
7
comment Find a relationship between $f(f^{-1}(f(A)))$ and $f(A)$. Prove it in the general.
Try to show first that $A \subseteq f^{-1}(f(A))$.
Oct
7
answered $\mathbb{Q}[i]$ is the smallest subfield of $\mathbb{C}$ that contains the ring $\mathbb Z[i]$
Oct
5
awarded  Explainer
Oct
5
comment the factor ring $\mathbb Z[i]/\langle3-i\rangle$
As far as I know, those two notations mean the same thing (the quotient of $\mathbf{Z}[i]$ by the principal ideal generated by $3-i$).
Oct
5
awarded  Citizen Patrol