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10

First proof Suppose $\mathbb{Z}^n\cong\mathbb{Z}^m$, with $m\ne n$. Then you can find a set of $m$ free generators in $\mathbb{Z}^n$, say $\{f_1,f_2,\dots,f_m\}$. Then, for some integers $a_{ij}$, $b_{ij}$, you have \begin{align} f_j&=\sum_{i=1}^n a_{ij} e_i &&j=1,\dots, m\\ e_i&=\sum_{j=1}^m b_{ij} f_j &&i=1,\dots, n \end{align} ...


7

$K(\alpha, \beta)$ is Galois over $K$ and the corresponding Galois group is a subgroup of $\text{Gal}(K(\alpha)/K)\times \text{Gal}(K(\beta)/K)$. So $K(\alpha, \beta)/K$ is abelian. Thus, any intermediate Galois extension must be abelian since the corresponding Galois group would be a quotient of $\text{Gal}(K(\alpha,\beta)/K)$


6

$\mathbb{Z}^n \cong \mathbb{Z}^m$ implies $(\mathbb{Z}/2)^n \cong \mathbb{Z}^n / 2 \mathbb{Z}^n \cong \mathbb{Z}^m / 2 \mathbb{Z}^m \cong (\mathbb{Z}/2)^m$. By comparing the number of elements, we get $2^n=2^m$, i.e. $n=m$. (No linear algebra is necessary here!)


3

Whenever you have a condition $g^2=e$ in a group, it's equivalent to $g=g^{-1}$ (multiply both sides by $g^{-1}$). In this case, it applies to every element of the group, so you can add or remove inverses from any expression freely. So the proof is simply $$ab=(ab)^{-1}=b^{-1}a^{-1}=ba.$$


3

This is known as Smith normal form, and you can find lots of references on Google books. In particular, this textbook has exhaustive coverage of Smith normal form and all of its consequences.


3

You can simplify things a lot by noting that $|a|$ divides $K$ iff $a^K=1$. Using this, verifying that $H$ is a subgroup is easy. If you can use homomorphisms, then it is easier because $H=\ker\phi$, where $\phi(x)=x^K$. Since $G$ is Abelian, $\phi$ is a homomorphism.


2

A way that doesn't use group actions (which are awesome, by the way, so try to use them when you can): We want to know the order of the centralizer of $g,\ |C(g)|$. Now $x \in C(g)$ means that $xgx^{-1} = g$. Since $x(1\ 2\ 3)x^{-1} = (x(1)\ x(2)\ x(3))$ we can immediately see two things: $x$ must map $\{1,2,3\} \to \{1,2,3\}$ and thus $\{4,5\} \to ...


2

Let $G$ acts on itself by conjugation. Consider the orbit of $g$, it's the conjugacy class of $g$ in $S_{5}$, so the set of $3$ cycles. Now, $H$ is the stabilizer of $g$. If $x \in H$, then $xg=gx$ so $xgx^{-1}=g$. And if $x \in Stab(g)$, then $xgx^{-1}=g$ so $xg=gx$ so $x \in H$. Now the orbit stabilizer theorem tells us that $|G|=|Orb(g)||Stab(g)|$, so ...


2

use the fundamental theorem of finitely generated Abelian groups: The invarint factor version. G is nothing but a k finite direct sum of Z_n's ordered in an ascending order of divisibility of the n's. Then certainly there is a summant Z_n. Now take a k-tuple with all entries 0 except one entry corresponding to Z_n in which it is 1.


2

The statement you want to prove is false in general. Let $k=1$ for example, and let $$G = \langle x,y,z \mid x^3 = y^3 = z^3 = 1, yz = zyx, xy = yx, xz = zx\rangle$$ be the group found in this answer. Then $G$ is not abelian, but $G$ has exponent three. It thus holds that $(ab)^3 = e = a^3 b^3$ for all $a,b \in G$, and moreover it obviously holds that ...


1

Doesn't the first nontrivial case, with $k=2$, $m_1=2$, and $m_2=3$, give a counterexample? EDIT: User64726 has pointed out that this counterexample doesn't work if one allows $1$ to be among the $d_i$'s. Under that convention, the problem becomes whether we can always take the number of nontrivial $d$'s to be $\leq k$, where $k$ is the number of $m$'s; if ...


1

@Egreg's answer above is very complete and concise, but I thought I'd add another answer just for the heck of it. Suppose toward a contradiction that there exist distinct natural numbers $m,n\in\Bbb{N}$ with $\Bbb{Z}^n=\Bbb{Z}^m$. Let $m\in\Bbb{N}$ the least natural number for which there exists $n\in\Bbb{N}$ with $m\neq n$ such that ...


1

Every group homomorphism $\phi: \mathbb Z^n \to \mathbb Z^m$ extends (uniquely) to a linear transformation $T: \mathbb Q^n \to \mathbb Q^m$ of vector spaces over $\mathbb Q$. Moreover, $\phi$ is injective iff $T$ is injective. But $T$ injective implies $n \le m$. Applying this to both $\phi$ and $\phi^{-1}$, we conclude that $n \le m \le n$.


1

You want to prove that the map $$ x \mapsto (\chi \mapsto \chi(x)) $$ is injective. I assume that you have already proved that this is a homomorphism. So all you need to show is that the kernel of this map is trivial. That is, you want to show that $$ \ker (x \mapsto (\chi \mapsto \chi(x))) = \{e\} $$ where $e$ is the identity in $G$. Now the kernel is ...


1

Given that $(a\cdot b)^{n+2}=a^{n+2}\cdot b^{n+2}$ we can multiply by $a^{-1}$ and $b^{-1}$ on the left and right to get $$(b\cdot a)^{n+1}=a^{n+1}\cdot b^{n+1}=(a\cdot b)^{n+1},$$ the latter equality holding by assumption. Repeating this trick we get $$(b\cdot a)^n=a^n\cdot b^n=(a\cdot b)^n,$$ and dividing these two left hand sides and right hand sides of ...


1

All you have done is correct. However it seems you made some claims which at first sight are correct, but in fact they could be based on some confusion. One of them is the following: when $G=\mathbb Z/p^{\alpha}\mathbb Z\times\mathbb Z/p^{\beta}\mathbb Z$ you said that $\mathbb Z/p\mathbb Z\subset G$. This is right up to isomorphism: we identify $\mathbb ...


1

The map $f:G \to G$ given by \begin{align*} f(g_i) &= \begin{cases} g_i & \text{if}\;\; i \leq r; \\ 0 & \text{if}\;\; i > r \end{cases} \end{align*} has kernel $\langle{g_{r+1}, \dots, g_k\rangle} = H$ (since $G$ is free). Hence $G/H = f(G) = \langle g_1, \dots, g_r\rangle\subset G$.


1

You can take $v_1=(1,4)$, $k_1=1$, $v_2=(0,1)$, $k_2=7$.


1

Another way to see that $|zw| = |z||w|$: Let $z = a+ib$, and $w = c + id$. Then: $|zw| = |(a+ib)(c+id)| = |(ac-bd) + i(ad+bc)| = \sqrt{(ac-bd)^2 + (ad+bc)^2}$ $= \sqrt{a^2c^2 - 2abcd + b^2d^2 + a^2d^2 + 2abcd + b^2c^2}$ $= \sqrt{a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2} = \sqrt{a^2(c^2 + d^2) + b^2(c^2 + d^2)}$ $= \sqrt{(a^2 + b^2)(c^2 + d^2)} = ...


1

$\mathbb C^{\times}$ is an Abelian group. The map $z \mapsto |z|$ is a group homomorphism $\mathbb C^{\times} \to \mathbb R^{\times}$. The circle is the kernel of this map and so is a subgroup of $\mathbb C^{\times}$. The crucial point is that $|zw|=|z|\ |w|$ and this can be proved as follows: $$|zw|^2=(zw)(\overline{zw})=z w \bar z \bar w=z \bar z w \bar ...


1

Hint: if $z\in G$, there exists $\varphi\in(-\pi,\pi]$ with $z=e^{i\varphi}$. (Of course one can choose $\varphi \in [0,2\pi)$ or $\varphi \in I$ for any half closed intervall $I$ of length $2\pi$).



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