If $G$ is an infinite group with finitely many conjugacy classes, what can be said about $G$? (Should $G$ be simple/ solvable/….?) For $n\geq 2$, does there exists a (infinite) group $G$ with exactly $n$ conjugacy classes, which is periodic also? I couldn’t find any information on these questions. One may provide links also for […]

This question already has an answer here: Fields of arbitrary cardinality 5 answers

Let $G$ be an infinite group and $\alpha$ a cardinal number with $\aleph_0\leq \alpha\leq |G|$. Is there a subgroup $H$ of $G$ with $|G:H|=\alpha$ (what about $|H|=\alpha$)?

Is there any finite (resp. infinite) non-abelian group of order $\geq 8$ such that $AB=BA$ for all subsets $A, B$ with $|A|\geq 3$ and $|B|\geq 3$? ($AB=\{ab: a\in A, b\in B\}$)

It is well-known fact that every finite $p$-group $G$ is nilpotent. I am asking to have a counter example when $G$ is infinite $p$-group. Thanks.

An easy argument shows that for any finite group $G$ the cardinal of $Aut(G)$ is less than $(|G|-1)!$. In particular the automorphisms group of a finite group is finite. Basically my question is about the converse statement. If a group $G$ has finite automorphisms group then should $G$ be finite ? The answer to this […]

Let $G$ be an abelian Group. Question is to prove that $T(G)=\{g\in G : |g|<\infty \}$ is a subgroup of G. I tried in following way: let $g_1,g_2\in T(G)$ say, $|g_1|=n_1$ and $|g_2|=n_2$; Now, $(g_1g_2)^{n_1n_2}=g_1^{n_1n_2}g_2^{n_1n_2}$ [This is because G is abelian]. $(g_1g_2)^{n_1n_2}=g_1^{n_1n_2}g_2^{n_1n_2}=(g_1^{n_1})^{n_2}(g_2^{n_2})^{n_1}=e^{n_2}e^{n_1}=e$ Thus, if $g_1,g_2$ have finite order, so is $g_1g_2$.So, $T(G)$ is closed under […]

One of the best ways to get a handle on a group is to recognize it as isomorphic to a set of symmetries of some structure. The dihedral group of order $2n$ is easily recognized as the set of symmetries of a regular $n$-gon, the symmetric group as the permutations of a set, the Klein-four […]

This question already has an answer here: Nonabelian group with all irreducible representations one-dimensional 1 answer

I am working on the following homework problem: Let $\phi$ be an isomorphism from $\mathbb{R}^*$ to $\mathbb{R}^*$ (nonzero reals under multiplication). Show that if $r>0$, then $\phi(r) > 0$. I was trying to prove this by contradiction, but I haven’t been able to find any problems. I know that we must have $\phi(1) = 1$, […]

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