Let $k &gt; 2$ be an integer.  Is there a finite non-abelian group with exactly $k$ conjugacy classes?  The only group with $1$ conjugacy class is the trivial group, and the only group with exactly $2$ conjugacy classes is the group of order $2$, so the restriction $k&gt;2$ is necessary.  (The symmetric group $S_3$ has $3$ conjugacy classes, and is the only such non-abelian group.)  For small values of $k$, the number of non-abelian groups with exactly $k$ conjugacy classes are .  I had originally come to this question for square-free values of $k$, but upon thinking about it, I realised that there was an even more basic form of the question, whose answer I did not know and could not find, as stated above.  I thought that it might be possible to construct a group of the form $G = A\rtimes B$, with $A$ abelian of order $k$, and $B\leq\operatorname{Aut}(A)$ so that $G$ has exactly $k$ conjugacy classes, but I was unable to come up with a general argument.  It does not always work for cyclic $A\cong C_k$.
122686
Given an integer $n\geq 1$, it is known that there are only finitely many finite groups with exactly $n$ conjugacy classes.  Question: For $n\geq 3$, does there exists a finite non-abelian group, with exactly $n$ conjugacy classes?
115635
Let $c(G)$ denote the number of conjugacy classes of a group $G$. I have to show that  $$\lim_{n \to \infty} \inf _{|G|=n}c(G)=\infty.$$ That is, I have to show that $\exists $ a function $f:\mathbb{N} \to \mathbb{N}$ such that if $c(g)\leq n, $ then $|G|\leq f(n).$   I know that if $G$ is a finite group such that any two non-identity elements in $G$ are conjugate, then $G$ is isomorphic to $\mathbb{Z}_2$. The proof is simple, infact:      It can easily be shown that $G$ acts transitively by conjugation on $G \setminus \{e\}$. Then, by  orbit-stabilizer theorem, the claim follows.    I wonder if similar idea could be useful somehow. But I can't think of any to prove the claim above. Please help!
46132
Browsing  I came across this:  The conjugacy class of an element $g\in A_{n}$:  splits if the cycle decomposition of $g\in A_{n}$ comprises cycles of distinct odd length. Note that the fixed points are here treated as cycles of length $1$, so it cannot have more than one fixed point; and does not split if the cycle decomposition of $g$ contains an even cycle or contains two cycles of the same length.   Anybody with a proof?
111044
Suppose $G$ is a non-abelian group and $H,K$ are two abelian subgroups of $G$. Then must $HK$ be an abelian subgroup of $G$?  I know an example, but I am confused. Thus I just want to check that.
37073
Let $G$ be a group where every non-identity element has order 2.     If |G| is finite then $G$ is isomorphic to the direct product $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots  \times \mathbb{Z}_{2}$.  Is the analogous result   $G= \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \ldots $.  true for the case |G| is infinite?
10037
Suppose $n$ is a non negative integer $\geq 4$ and $\sigma\in S_n$ a permutation. Conjugacy classes in $S_n$ are completley determined by the cycle structure of $\sigma$. If we let the alternating group $A_n$.act on $S_n$ by conjugation, the orbits coincide with the conjugacy classes in $S_n$ except for those (even) permutations whose cycle structure is composed of even cycles (which corresponds to cycles of uneven length) of distinct order (fixed points are treated as cycles of length $1$.) See   For instance a $3$-cycle in $A_4$ has cycle structure $[3,1]$ which fits the bill, and a $5$-cycle in $A_5$ or $A_6$ works too, since the cycle structure is $[5]$ and $[5,1]$ respectively. On the other hand, a $3$-cycle in $A_5$ will does not satisfy the criteria as it has cycle structure $[3,1,1]$ with two fixed points.  In this case, the conjugacy class of $\sigma$ inside $S_n$ splits in $2$ orbits of the conjugation action of $A_n$, and if $\tau$ is any odd permutation, for instance $(12)$, then we have $$\mathrm{Conj}_{S_n}(\sigma)=\mathcal{O}(\sigma)\coprod\mathcal{O}(\tau\sigma\tau^{-1})$$  My question is How can we differentiate between the two orbits? Does it have geometric meaning?    I can see that this information can be useful when looking at representations of the alternating groups, since there are as many irreducible representations as conjugacy classes. As for some geometric interpretation, I ask because there is one for the three cycles in $A_4$. If $S_4$ is understood as the group of symmetries of a regular tetrahedron, then $A_4$ is the subgroup of direct symmetries, and a $3$-cycle corresponds to a rotation of angle $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ with axis passing through one of the vertices. Conjugacy classes inside $A_n$ preserve the angle, and that's how we can tell them apart. In this case the computations are so simple we don't really need this picture, but I think it is a nice way to understand the fact $3$-cycles in are not all conjugate inside $A_4$...
25848
Question: If $G$ is a finite group such that the product of its elements (each chosen only once) is always $1$, independent of the ordering in the product, what can we say about $G$?  I was trying to conclude that $G$ must be abelian (I am not sure about this). I proceed as follows: if $|G|\leq 4$, then obviously, $G$ is abelian. Suppose $|G|&gt;4$. Then consider products of all elements in the order $a^{-1}b^{-1}ab cc^{-1}dd^{-1}....$. This, being identity, implies that $a^{-1}b^{-1}ab=1$, i.e. $G$ is abelian.  BUT, the problem is in $a^{-1}b^{-1}ab$, the elements $a$ and $a^{-1}$ should be distinct, and similarly for $b$. Otherwise, the product $a^{-1}b^{-1}ab cc^{-1}dd^{-1}....$ is not in the prescribed form as in question.  Also, note that the cyclic group of order $4$ doesn't satisfy that $abcd=1$ where $\{a,b,c,d\}\cong \mathbb{Z}_4$.
26753
Can you show me a reasonably simple (using only elementary group-theoretic tools) example of infinite group with just 2 conjugacy classes ?
73328
For which $n$ is there a group of order $n^2$ without a subgroup of order $n$.  Such groups can not be nilpotent.   This question is related to .
46258
In Rotman's introduction to the theory of groups, there is an exercise (7.9) where one has to prove that the following facts for a finite group $G$.    If $G$ is abelian and $|G|&gt;2$ then $\text{Aut}(G)$ has even order $\text{Aut}(G)$ is not cyclic when $G$ is not abelian $\text{Aut}(G)$ is never cyclic of odd order >1.    This is all fairly easy, but I was wondering if there even exists a group $G$ with $|G|&gt;2$ which has automorphism group of odd order. Anyone has an example or reference (couldn't find it by simple internet search)?  Clearly $G$ can't be abelian, symmetric, alternating, dihedral, ... so the most obvious counterexamples don't work.  Thanks in advance!
40359
It is an easy exercise to show that all finite groups with at least three elements have at least one non-trivial automorphism; in other words, there are - up to isomorphism - only finitely many finite groups $G$ such that $Aut(G)=1$ (to be exact, just two: $1$ and $C_2$).  Is an analogous statement true for all finite groups? I.e., given a finite group $A$, are there - again up to isomorphism - only finitely many groups $G$ with $Aut(G)\cong A$?  If yes, is there an upper bound on the number of such groups $G$ depending on a property of $A$ (e.g. its order)?  And if not, which groups arise as counterexamples?  And finally, what does the situation look like for infinite groups $G$ with a given finite automorphism group? And what if infinite automorphism groups $A$ are considered?
72407
I need to prove that the center of $A_n$ is trivial for $n \geq 4$.  $Z(A_3) = A_3$, since $A_3 = \mathbb{Z}/3\mathbb{Z}$ is commutative.  One idea is two use "counting" technique. First of, all we count cojugacy classes of even permutations in $S_n$. It's the classes correspodning to the types $[\lambda_1, ..., \lambda_r]$ where $n \equiv r \mod 2$  Now, the conjugacy class in $S_n$ correspoding to the type $[\lambda_1, ..., \lambda_r]$ splits into two equal-sized classes in $A_n$ if $\lambda_1, ..., \lambda_r$ are distinct odd numbers. If they aren't then it is preserved in $A_n$.  Then we can use "the counting formula"(  ) of elements in each "even" class of $S_n$ and divide one by $2$ if needed.  Still, I'm not sure if it can be done. Seems to be a lot of work there. Maybe there are other, he easier ways? Or, maybe, it is the way, in this case, I wouild appreciate any advices.
90920
If $G$ is an abelien group, let $\hat{G}$ be the set of all   homomorphisms of $G$ into the group of nonzero complex numbers under   multiplication. If $\phi_1, \phi_1\in \hat{G}$ define $\phi_1\cdot  \phi_2$ by $(\phi_1\cdot \phi_2)(g)=\phi_1(g)\phi_2(g)$ for all $g\in  G$.      If $G$ is a finite abelian group prove that $o(G)=o(\hat{G})$ and $G$   is isomorphic to $\hat{G}$.   The case when $G$ is finite cyclic is easy because in this case $\hat{G} \cong G$. Let's consider the case when $G$ is abelian group of order $p^n$, $p$ - prime.   By Fundamental theorem on finite abelian groups we have the following: $$G=\langle a_1 \rangle_{p^{n_1}}\times \langle a_2 \rangle_{p^{n_2}} \times \dots \times \langle a_k \rangle_{p^{n_k}},$$  where $n_1\geq n_2\geq \dots \geq n_k&gt;0$ and $n_1+\dots+n_k=n$.   Let $A_i=\langle a_i \rangle_{p^{n_i}}$ and $A_i$ are cyclic groups then $A_i\cong \hat{A_i}$. So we have the following $$G=A_1\times \dots \times A_k\cong \hat{A_1}\times \dots \times \hat{A_k}.$$  Then I was trying to think about $\hat{G}$ and I think that $\hat{G}=\hat{A_1}\times \dots \times \hat{A_k}$ which follows that $G\cong \hat{G}$.   How to prove rigorously that $\hat{G}=\hat{A_1}\times \dots \times \hat{A_k}$? I am not able to prove it.
71935
When people want to classify a group with certain (small) order, they seem to find a normal subgroup $H$ and a subgroup $K$, and then consider how many distinct $H \rtimes_\varphi K$ are there.  My question is: For any group $G$, does it always exist normal subgroup $H$, subgroup $K$ and $\phi : K \rightarrow Aut(H)$ s.t. $G\cong H \rtimes_\varphi K $?  Sorry if this question is too basic...  Thanks the comment and the answers. Then if the $G$ is not simple, can we have can $G\cong H\rtimes_\varphi K$ for proper subgroup $H$, $K$, with $H$ normal?  EDIT:  If we have a group $G$, that is not simple (then $G$ has at least one normal subgroup), are there always a non-trivial normal subgroup $H$ and a proper subgroup $K$ s.t. $G \cong H \rtimes K$. (Note that if $G$ is simple this is clearly not possible).  But the quaternion group and $Z_4$ given in answers have shown that we cannot always have it.
62210
Is there a number $n$ such that there are exactly 1 million abelian groups of order $n$?   Can anyone please explain. I would yes because numbers are infinitive, and so any number n can be expressed as a direct product of cyclic groups of order n.  Can anyone please help  me understand.  Thank you.
117293
What exponent $e$ can guarantee the group to be abelian? Are there any known results except the case $e = 2$? Thanks for help.
62527
Sorry for the last mistaken problem I just posted. Now I know that only having the order being odd square free is not enough for a group to be cyclic.   Here's the complete problem which the main goal is to show that any group of order $n$ satisfies $\gcd (n,\varphi(n))$=1 is cyclic. It asks me to do it in the following way:  (a) If $n &gt; 2$, $n$ is an odd squarefree integer.  (b) Show that there is a $H = G/N$ of prime order.  (c) Show that $G \cong N \times H$ and $G$ is abelian.  (d) $G$ is cyclic.    Any ideas is appreciated.