# Tag Info

## Hot answers tagged abelian-groups

6

Sure. Fix a Hamel basis $B$ for $\Bbb R$ over $\Bbb Q$, fix some $r\in B$ and map $B\setminus\{r\}$ to $0$, and $r$ to $1$. Then you're done. If you want to avoid the axiom of choice, you can't. It is consistent that every homomorphism from $\Bbb R$ to $\Bbb Q$ is continuous, and therefore its image is connected. But this means that every homomorphism is ...

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Here is a proof that appears in Weil's gem Number theory for beginners and in several other books. Let $m$ be the order of $A$ and let $\psi(n)$ be the number of elements of order $n$ in $A$. Then, by Lagrange's theorem, $m=\sum_{d\mid m} \psi(d)$ because every element has an order that is a divisor of $m$. If $\psi(d)>0$, then there is an element of ...

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I believe that Propositions 1.10 and 1.16 of this paper of mine give (a generalization of) what you are looking for. The notation may take some getting used to. Indeed all the key ingredients of the discussion above appear in this somewhat abstracted context...and they comprise the proof.

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If $X$ has at least $3$ elements, then the group of all invertible maps on $X$ contains $S_3$, which is not abelian. Concretely, let $x_1, x_2, x_3$ be three distinct elements of $X$. Then the map that swaps $x_1$ and $x_2$ does not commute with the map that swaps $x_1$ and $x_3$.

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1. When $\varphi$ is the identity, the semi-direct product coincides with the direct product. 2. It generalizes the notion of direct product. 3. \begin{eqnarray*} (a,b)+(\varphi(-b)(-a),-b)&=&(a+\varphi(b)\varphi(-b)(-a),0)\\ &=&(a+\varphi(b-b)(-a),0)\\ &=&(a+\varphi(0)(-a),0)\\ &=&(a-a,0)\\ &=&(0,0) ...

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Yours is one possible way to show this. One can also directly use the universal property of the free abelian group: Given an abelian group $G$ let $A(G)$ be the free abelian group on the underlying set of $G$. Then by the universal property of the free abelian group there exists a unique group homomorphism $\varphi \colon A(G) \to G$ with $\varphi(g) = g$ ...

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The last step of your proof follows from the fact that for groups $G_1$ and $G_2$ and normal subgroups $N_1 \subseteq G_1$ and $N_2 \subseteq G_2$ we have that $N_1 \times N_2 \subseteq G_1 \times G_2$ is a normal subgroup with $(G_1 \times G_2)/(N_1 \times N_2) \cong (G_1 / N_1) \times (G_2 / N_2)$. To see the isomorphism notice that the group epimorphism ...

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In fact, $(P)$ is wrong: there is only one normal subgroup $H$ of order $4$ in $S_4$, and $S_4/H\simeq S_3$. See Show that $S_4/K \cong S_3$.

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You are absolutely correct about (Q). As to (P), you should know there is a unique normal subgroup $V$ of order $4$ in $S_{4}$. This is a transitive subgroup, so $S_{4} = V T$, where $T$ is the stabilizer of $4$, say. But then $T \cong S_{3}$.

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Hint: If $S$ is a nontrivial subgroup of $\Bbb Q$ there exists an isomorphism $\phi$ of $\Bbb Q$ such that $\Bbb Z\subset \phi(S)$. Hence the answer for "subgroup of $\Bbb Q$" must be the same as for "subgroup of $\Bbb Q$ containing $\Bbb Z$".

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Thanks to @DanielFischer comment I was able to get the answer by myself. Considering $i(1)=(5,0)$ then the $A= \mathbb{Z} \oplus \mathbb{Z}$ is another valid option. These are the two only valid options. Let's see the intuition behind this. $A = LT(A) \oplus T(A)$ where $T(A)$ is the torsion part of $A$. We can racionalize the sequence and taking into ...

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