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7

Yes you can ! When you have a cube, you can divide that cube into $n^3$ smaller cubes by cutting it into $n$ equal slices along each side. Hence this operation adds $n^3-1$ cubes. $$5772=1+(17^3-1)+(9^3-1)+(5^3-1)+(2^3-1)$$ So take a cube, and in any order : Choose one cube and cut it into $17^3$ cubes Choose one cube and cut it into $9^3$ cubes Choose ...


5

Note that $\langle \zeta\rangle$ is a finite, cyclic group of order $7$, so every non-trivial element has order $7$. In particular $\zeta^3$ is another primitive $7^{th}$ root of $1$, hence is another roots of the irreducible polynomial $\Phi_7(x)={x^7-1\over x-1}$. But then as this is irreducible by Eisenstein's criterion applied to $\Phi(x+1)$, we get that ...


5

The canonical projection $G \to G/K$ composed with your isomorphism does the job.


4

This problem you are intereted in is call the inverse Galois problem. Every finite group is a Galois group, every profinite group is a Galois group. In general, it is quite hard, for a general group $G$, to say if $G$ is a Galois group or not. I would recommand JP Serre's "Topics in Galois theory" for an authorative reference and state of the art (a bit ...


3

Let $r$ and $s$ two lines in the plane through the origin, and denote with the same letters the reflections through them. Then ${\rm ord}(r)={\rm ord}(s)=2$. But the isomorphism type of the group $G=<r,s>$ depends strongly on the angle $\theta$ between $r$ and $s$. If $\theta=\pi/4$ then $G=\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$. If $\theta\notin\Bbb ...


2

That's totally fine. I might think it is more commonly written as $$ \sum_{\substack{ 1 \leq i \leq n \\ 1 \leq j \leq m}} \phi_{v_i}\,\phi_{l_j},$$ or if the ranges of summation are clear, then $$ \sum_{i,\,j} \phi_{v_i}\,\phi_{l_j}.$$


2

Suppose $a\in\operatorname{Rad}(\operatorname{Rad}(I))$. Then $a^m\in\operatorname{Rad}(I)$, for some $m$. Then $(a^m)^n\in I$ for some $n$.


2

Normally, one does not define algebraic sets associated to radical ideals only, but rather defines them for any ideal or even for any set, like $V(S)= \{p \ \colon p(s) = 0 \ \forall s \in S\}$. Then, one shows that every algebraic set is in fact already associated to the algebraic set given by some radical ideal. Since $V(S) = V(\langle S \rangle)$ and ...


2

One should point out the obviousness of the first statement: $\alpha$ is the first map of an injective composition, hence injective. $\gamma$ is the second map of a surjective composition, hence surjective. The second statement can be also seen this way without using the snake lemma: If $\alpha$ is an isomorphism, consider the diagram $$\require{AMScd} ...


2

If $G$ is a group of prime order, then every nontrivial element generates $G$.


2

An idea: let $\;X:=\{h_i\;:\;\;i\in I\}\;$ be a generator set of $\;H\;$, and let $\;\{g_1+H,..,g_k+H\}\;,\;\;g_i\in G\;$ be a free generator set of $\; G/H\;$ == Show that $\;x\in\langle g_1,...,g_k\}\cap\langle X\rangle\implies x=0\;$ == Show that $\;\langle\{ g_1,...,g_k\}\cup\{ X\}\rangle=G\;$


2

What you want to show is that $$ hA = Ah \iff \forall a \in A,\ h a h^{-1} \in A $$ Suppose $hA = Ah$, and let $a \in A$. Since $hA = Ah$, and $ha \in hA$, we have $ha = a'h$ for some $a'\in A$. Then $$ ha = a'h \implies hah^{-1} =a', $$ evidently in $A$. On the other hand, suppose $ h a h^{-1} \in A$ for all $a \in A$. Let $x \in hA$. Then $x = ha$ ...


2

Any permutation representation in $V=K^n$ has an invariant subspace $W$ given by the equation $x_1+x_2+\cdots+x_n=0$ (the vectors with coordinates summing to$~0$; here of course $n=3$). Since $\dim W=n-1$, any complementary subspace to $W$ is spanned by a since vector $v\notin W$. In order for that complementary subspace to be invariant, $v$ must be an ...


1

For $G$ a finite group, and $F$ is a field, the group algebra $FG$ is semisimple if and only if $J(FG) = \{0\}$, where $J(FG)$ is the Jacobson radical off $FG$, which has several equivalent definitions: one is $\{ j \in FG: 1-jx \}$ is nilpotent for all $x \in FG$. We note that if $j \in Z(FG)$ is nilpotent and non-zero, then $jx$ is nilpotent for all $ x ...


1

I believe that this hint helps you to establish that $\mathbb Z[X]$ is countable. For a polynomial $f(X)=a_nX^n+\ldots a_1X+a_0\in\mathbb Z[X]$, $a_n\neq 0$, define number $h(f)$ with: $$h(f)= n+|a_n|+\ldots+|a_1|+|a_0|.$$ The function $h:\mathbb Z[X]\longrightarrow \mathbb N$ measures some kind of a complexity of a polynomial. It is fairly obvious that ...


1

With ordinary integers: Theorem of Gauss that $x^2 + y^2 + z^2$ integrally represents all positive integers except those $$ 4^k (8 n+7). $$ Meanwhile, if $x^2 + y^2 + z^2 \equiv 0 \pmod 4,$ then $x,y,z$ are all even. Put these together, we say that $x^2 + y^2 + z^2$ is anisotropic in $\mathbb Q_2.$ It is also anisotropic in the real numbers, as the sum ...


1

Ther are lots of good answer for this questions and they have very good explanations. But I think you can just apply the subgroup test. Note that $1=(a^{-1})a\in H\not= \emptyset$ Let $g_1=xa$ and $g_2=ya$ be two elements of $H.$ Then $$g_1g_2^{-1}=xaa^{-1}y^{-1}=(xy^{-1}a^{-1})a\in H$$ Hence $H\le G.$


1

Use free groups are projective. Theres a short exact sequence $0\to H\to G \to \mathbb Z^k\to 0$ that splits.


1

Take $Y = \mathbf{F}_{p^{n-2}} = $ a field with $p^{n-2}$ elements and let $X$ be the set of subset of $X$ with exactly $k$ elements. Make act the group $\mathbf{F}_{p^{n-2}}$ on $X$ by translation, and show that an orbit is never reduced to a single element. What is then necessarily the cardinal of an orbit ? Conclude.


1

Your definition of the order of an element is missing one important feature. The order of an element $g$ is the smallest positive integer $n$ such that $g^n = e$. (If no such integer exists, then $g$ has infinite order.) So no element can have order $0$. EDIT: In response to your comment, I would instead use the following argument. Let $d = \gcd(n,m)$ ...


1

If $a_i\in I_i$, $a_i=0+\cdots+a_i+\cdots+0\in I_1+\cdots+I_m$. If $a_i\in I_i$ for each $1\le i\le m$, then $a_1\cdot\ldots\cdot a_m\in I_1\cap\cdots\cap I_m$.


1

Note $J$ this radical of $I$. If $a\in J$ and $b\in A$, $a^n\in I$ for some $n>0$ so that $(ba)^n = b^n a^n \in I$ because $I$ is a ideal, so that $ba\in J$. Now, if $a,b\in J$, whose respective powers $a^n$ and $b^m$ are in $I$, then by commutatativity we have $(a+b)^{n+m} = \sum_{k=0}^{n+m} {n+m \choose k} a^k b^{n+m-k}$ by the well-known binomial ...


1

Well, you can note that $(1+i)$ is maximal because it has a prime norm (hence it is irreducible) and $\Bbb Z[i]$ is a Euclidean ring under the norm (hence PID), so prime ideals are maximal. The image should be a field of some sort. We note that $(1+i)|2$--indeed $(1+i)(1-i)=2$--hence $2\equiv 0\pmod{1+i}$, i.e. the field has characteristic $2$. Indeed, we ...


1

You can also think of this in the following way: Look at the composition $\mathbb Z[i] \to \mathbb Z[i]/(1+i) \cong \mathbb Z[X]/(X^2+1,X+1) = \mathbb Z[X]/(2,X+1) \cong (\mathbb Z/2\mathbb Z[X])/(X+1) \cong \mathbb Z/2\mathbb Z$ The first isomorphism is given by $a+bi \mapsto a+bX$ and the last isomorphism is given by $X \mapsto -1$. Hence the ...


1

A good way to find the minimal polynomial of an element when knowing the Galois group is to compute all the conjuagtes of the element and compute $ \prod_j (X -a_j)$ where $a_j$ are the conjugates. The conjugates in the first case are $\zeta + \zeta^{-1}$, $\zeta^2 + \zeta^{-2}$, and $\zeta^{3} + \zeta^{-3}$. Note the others just repeat, for example, ...



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