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7 votes
Accepted

$| \ker{\phi} |<\infty \iff [G: \phi(G)]<\infty$ for group homomorphism $\phi$.

In general, the answer is no. Consider $G=\Bbb{Q}/\Bbb{Z}$. It is known that $\Bbb{Q}/\Bbb{Z} = \bigoplus_p \Bbb{Z}_{p^\infty}$, where $\Bbb{Z}_{p^\infty}$ is the abelian group generated by elements $...
SomeCallMeTim's user avatar
6 votes
Accepted

If $G$ is abelian, can the doubling map $s: G \to G$ be a group isomorphism?

Sure. Consider the integers modulo $3$ under addition. Or, if you want an infinite example, you can just take an infinite direct sum of that group with itself.
Arthur's user avatar
  • 201k
5 votes
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Do sets of commuting permutations with no fixed points generate Abelian groups with no fixed points?

It is difficult for permutation actions to have the property that no non-identity element has fixed points. This property is called being free, and for a group $G$ acting on a set $X$ it is equivalent ...
Qiaochu Yuan's user avatar
4 votes
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Is a topological abelian group with no open subgroup connected?

If you add a compactness condition, then the result is true. Consider a topological group $G$ with no proper open subgroup (I am not assuming $G$ to be abelian). The connected component of the ...
Fançois Gatine's user avatar
4 votes
Accepted

If $G/Z(G)$ is isomorphic to a subgroup of $\mathbb Q$ then $G$ is abelian.

Hint: assume for contradiction that some $a, b \in G$ don't commute, and consider $H = \left< a, b \right> \leqslant G$. PS It is also not hard to unpack the reasoning so that it does not use ...
Adayah's user avatar
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3 votes
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Classifying maps of finitely generated abelian groups up to automorphism

This is a very natural question, but the answer is that such a classification is almost certainly hopeless. In this answer of mine to a related MathOverflow question, I give a reference to the paper ...
Jeremy Rickard's user avatar
3 votes

If $G$ is abelian, can the doubling map $s: G \to G$ be a group isomorphism?

Suppose that $G$ is a finite group. Then the map $f:G\rightarrow G$; $g\mapsto g^2$ is a homomorphism if and only if $G$ is abelian (so we can think of $g^2$ as $g+g$). Then $f$ is an isomorphism if ...
Dietrich Burde's user avatar
2 votes
Accepted

Let $a$ be the reflection of the plane $\mathbb{R}^2$ over the bisector of the odd quadrants

For part (a), note that $G = \langle a, b \rangle$ indicates the subgroup generated by $a$ and $b$ (in which larger group?). So, in general, this can be rather large, since any word we write down in ...
Sammy Black's user avatar
  • 26.2k
2 votes
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Example for application of theorem: $G = \langle C_G(a) \mid a \in Q \setminus \{1\}\rangle$

Here is a possible example. Take $p=3, q=2$. Consider the group $$ \left\{ \begin{pmatrix} \epsilon_1 & a & b\\ 0& \epsilon_2 &c\\ 0 & 0 & \epsilon_3 \end{pmatrix} : \epsilon_1,...
ancient mathematician's user avatar
1 vote
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Locally compact group whose unitary irreducible reps are one dimensional

This is always true, no assumptions concerning direct integral decomposition are needed. More precisely: if $G$ is a locally compact, Hausdorff topological group and all irreducible representations of ...
Mogget's user avatar
  • 641

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