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3

Suppose we have a map $f:\Bbb Z^n\to\Bbb Z^m$ that is injective. Since $\Bbb Q$ is $\Bbb Z$-flat, we obtain an injection of $\Bbb Q$-vector spaces $1\otimes f:\Bbb Q^n\to \Bbb Q^m$, so that $n\leqslant m$.


2

This is another conjugation problem in disguise: $(2\ 3)(3\ 4)(4\ 5)(4\ 3)(3\ 2) = (2\ 3)[(3\ 4)(4\ 5)(3\ 4)^{-1}](3\ 2)$ $= (2\ 3)(3\ 5)(2\ 3)^{-1}$ (since $(3\ 4)$ takes $4 \to 3$ and fixes $5$) $= (2\ 5)$ (since $(2\ 3)$ takes $3 \to 2$ and fixes $5$).


2

Note that 1) can be reformulated in the following way: Given a finite field $F$ and a finite dimensional vector space $V$ with a subspace $V_0 \subseteq V$ of dimension $m$, how many subspaces of $V$ of dimension $k$ contain $V_0$? We are thus looking for the subspaces of $V/V_0$ of dimension $k-m$. Given that $V$ has dimension $n$, the question reduces to ...


2

Denote by $A_g$ the element of $P$ that contains $g \in G$. The proof mostly uses the facts that the operation is well-defined and $A_g=A_1 \iff g \in A_1$ Let's check that $A_1$, the element of $P$ containing the identity, is a subgroup: Identity: It's given that $1 \in A_1$ Closure: Suppose $a, b \in A_1$. Then $A_a = A_1$ and $A_b=A_1$. By the ...


1

If you were asking about matrices which commute with the action of $GL_n$ instead of $U(n)$, the answer would be Schur-Weyl Duality: the space of intertwiners is spanned by the matrices which permute the different tensor factors of $\mathbb{C}^n$. I believe that the answer should be the same also if you consider just the unitary matrices.


1

Let $A\in M_n(\pi^iR/\pi^{i+1}R)$. Then $A^2=0$ since $ab=0$ for $a,b\in\pi^iR/\pi^{i+1}R$. Since $p(R/\pi R)=0$ it follows $pR\subseteq\pi R$ hence $p\pi^iR\subseteq\pi^{i+1}R$, that is, $p(\pi^iR/\pi^{i+1}R)=0$. Therefore $pA=0$. We then have $(I+A)^p=\sum_{i=0}^p{p\choose i}A^i=I+pA=I$.


1

Let's go through the product(s) $$ (2 3) (3 4) (4 5) (4 3) (3 2) $$ step by step. Start with $2$, by writing $(2$, and see how the elements map out: $2\mapsto 3\mapsto 4\mapsto 5\mapsto 5\mapsto 5\qquad:\qquad(2\; 5$ $5\mapsto 5\mapsto 5\mapsto 4\mapsto 3\mapsto 2\qquad:\qquad(2\; 5)(3$ $3\mapsto 2\mapsto 2\mapsto 2\mapsto 2\mapsto 3\qquad:\qquad(2\; ...


1

As per the logic you mentioned above you have: $(2\ 3\ 5\ 9\ 17\ 33\ 14\ 27)$ Then the next cycle would be $(4\ 7\ 13\ 25\ 49\ 46\ 40\ 28)$ then $(6\ 11\ 21\ 41\ 30\ 8\ 15\ 29)$ and $(10\ 19\ 37\ 22\ 43\ 34\ 16\ 31)$ and $(12\ 23\ 45\ 38\ 24\ 47\ 42\ 32)$ and $(20\ 39\ 26\ 51\ 50\ 48\ 44\ 36)$ then oddly the pair $(18\ 35)$ and of course $(1)(52)$. ...


1

It is not quite that simple, these are the disjoint circles: (2 3 4 9 17 33 14 27 2) (4 7 13 25 49 46 40 28) (6 11 21 41 30 8 15 29) (10 19 37 22 43 34 16 31) (12 23 45 38 24 47 42 32) (20 39 26 51 50 48 44 36) (18 35) I don't know of any method other then writing it out. Maybe something including modulus, would be interesting to see.


1

Your professor is saying the following. The left-action of $G$ on itself gives a "permutation representation,'' that is, an embedding $G \hookrightarrow S_{2k}$. If you are not familiar with the phrase "action" in the previous sentence, all I am doing is copying the proof of Cayley's theorem that every group is a subgroup of the symmetric group; it ...



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