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5

Here are the 101* finite nonabelian groups with 16 conjugacy classes whose order is less than 2000. (*I don't remember if there are any exceptional orders left out of the census.) The names are those given by GAP's structure description, and have not been cleaned up for this answer since there are so many groups. SmallGroup(40,1) = C5 : C8 SmallGroup(40,5) ...


4

Here are the 18 isomorphism classes of finite groups with 8 conjugacy classes: SmallGroup( 20, 1) = $\operatorname{AGL}(1,5)$ SmallGroup( 20, 4) = $D_{10}$ SmallGroup( 24, 13) = $C_2 \times A_4$ SmallGroup( 26, 1) = $D_{13}$ SmallGroup( 48, 3) = $C_3 \ltimes (C_4 \times C_4)$ SmallGroup( 48, 28) = $\operatorname{SL}(2,3) \mathsf{Y} C_4$ SmallGroup( 48, ...


4

As far as I have always understood this, it is completely possible and common for $N$ to not be contained in $U$ at all. So, it would not make sense to mod out by $N$. But, taking $UN$ first, will be the smallest subgroup that contains both $U$ and $N$, which then gives us what we need to make sense of the quotient. Moreover, the second Isomorphism ...


4

This is not a complete answer, but I am starting to think that the answer might be yes, $G \times G$ is $2$-generated. (Added later: I believe that it is now a complete answer, and that the answer is indeed yes.) I can prove it under the following additional assumption: There exist $a \in G$, $b \in G'$ with $\langle a,b \rangle = G$. Assuming that is ...


4

The comments suggest that you mean only centralizers of involutions. Even in that case, no finite group $G$ can have only the two involutionn centralizers you suggest, so Q2 seems to have a negative (or empty) answer. One involution centralizer must contain a Sylow $2$-subgroup. Hence the Sylow $2$-subgroup of $G$ must have order $16$. But the elementary ...


3

Yes, any (abelian or non abelian) group of order $7$ is cyclic, and this is true for any group of prime order. Now, an abelian group of order $25$ can be only $C_{25}$ or $C_5\times C_5$, so that your group $G$ can be $G=C_{25}\times C_{7}$ or $G=C_{5}\times C_{5}\times C_{7}$. Now you have to compute the number of elements of order $5$ in these two groups, ...


2

Let $G$ be a counterexample of minimal order. Since $G$ is a $p$-group, its center is non-trivial. Therefore $\overline{G}:=G/Z(G)$ has smaller order than $G$ and also has a unique maximal subgroup $\overline{H}$, and so by the minimal choice of $G$, $\overline{G}$ must be cyclic. Now there exists a very popular exercise stating that under these conditions, ...


2

Denote set of left cosets by $G/H$ and let $S\left(G/H\right)$ denote the symmetric group of bijections on it. Then $f:G\rightarrow S\left(G/H\right)$ defined by $g\mapsto\lambda_{g}$, where $\lambda_{g}\in S\left(G/H\right)$ on its turn is defined by $xH\mapsto gxH$, is a grouphomomophism with kernel $N:=\cap_{x\in G}xHx^{-1}$. Here $S\left(G/H\right)$ ...


2

This is really straightforward calculation in GAP: gap> G:=SmallGroup(64,138); <pc group of size 64 with 6 generators> gap> StructureDescription(G); "(((C4 x C2) : C2) : C2) : C2" gap> cc:=ConjugacyClasses(G); [ <identity> of ...^G, f1^G, f2^G, f3^G, f4^G, f5^G, f6^G, f1*f2^G, f1*f3^G, f1*f6^G, f2*f3^G, f2*f5^G, f3*f4^G, f4*f5^G, ...


2

The space consists of just nine points $(0,0)$, $(0,1)$, $(0,2)$, $(1,0)$, $(1,1)$, $(1,2)$, $(2,0)$, $(2,1)$, $(2,2)$ which can be naturally put on a $3\times3$ grid. The pictures show arrows which connect each point to the point it is mapped to (the latter being where the arrow points to), with the arrows for points mapping to itself omitted (thus the ...


2

Suppose $G$ is a noneabelian group of order $6$. By Cauchy's theorem, there exists an element of order $2$, call it $b$, and an element of order $3$, call it $a$. The subgroup $H=\langle a\rangle$ is normal in $G$, since it has index $2$. Since $\langle a \rangle\cap\langle b\rangle =1$, $\langle a,b\rangle =G$. We have $bab^{-1}=a^i$, it cannot be $i=1$ ...


2

It's pretty clear that every pair of distinct elements with order 2 will generate such a subgroup. Seek all of those pairs! If I didn't miscount, I think I see 7 elements of order 2. The subgroup will contain three elements of order 2. Experiment and see what combinations are distinct. If you try all $_7C_2$ pairs, you should find that everything is ...


2

How about a mix of DonAntonio's two answers? He writes: Anyway, the non-abelian group of order 21 can be given as $G=\langle\,x,y\;\;;\;\;x^3=y^7=1\;,\;xy=y^2x\,\rangle\cong\langle\,(235)(476)\;,\;(1234567)\,\rangle\leq S_7$ Give both the presentation with generators and relations, and its realization with x and y in $S_7$. It's easier to calculate with ...


1

The orbit-stabilizer theorem (easy to prove!) says that whenever you have a group acting on a set $X$, for any $x\in X$, the size of the orbit $Gx$ is $|G|/|\text{Stab}_G(x)|$. This means that since $|G| = p^k$, that every orbit will have size a power of p (possibly 0!). Since the orbits partition $X$, and since $p\nmid|X|$, this means that it's impossible ...


1

For example, say you have a finite Abelian group which is a product of cyclic groups of orders $q_1,q_2,\ldots,q_n$. Then let $N = \sum_j q_j$ and consider the permutation group $S_N$. For each $q_j$, choose a subset of $\{1,2,\ldots,N\}$ of size $q_j$ where all the subsets are disjoint. Then consider the subgroup generated by a set of cycles, where you have ...


1

You ask for a simple example, so here is one: Consider the group $\mathbb{Z} / n\mathbb{Z}$. This is a finite abelien group under addition of order $n$. Let $$ \phi: \mathbb{Z} / n\mathbb{Z} \longrightarrow S_n $$ given by $$ \phi([1]) = (1\; 2\; 3\; \dots \; n ). $$ So $$ \phi([m]) = (1\; 2\; 3\; \dots \; n )^m. $$ Then $\phi$ is a injective. Note that ...


1

Factor $|G|$ into primes as $\prod_{i=1}^n p_i^{k_i}$ where the $p_i$ are distinct. Proceed by induction on $n$: the base case is done assuming you know prime-order groups are cyclic. Then for the induction step factor out all instances take a generator $g_1$ of a subgroup of order $p_1^{k_1}$ and $g_2$ of a subgroup of order $|G|/p_1^{k_1}$. (By induction ...


1

Since $f$ is irreducible, $G$ must act transitively on the roots. Thus the stabilizer of a point is a subgroup of index $5$, so $|G|$ is divisible by $5$. Since $|G|$ is also divisible by $3$, it must be either $15,30$ or $60$. If $|G|=15=3\cdot 5$ then $G$ must contain an element of order $5$ by Cauchy's theorem, so must contain a $3$-cycle and a $5$-cycle, ...


1

Let me quote from the Wikipedia: Formally, let $(A(1), \ldots, A(n))$ be a sequence of $n$ distinct numbers. If $i < j$ and $A(i) > A(j)$, then the pair $(i, j)$ is called an inversion of $A$. (It seems in this case you write that $(j, i)$ is an inversion, but this is not an important difference.) So you should check for all pairs $(i, j)$, ...


1

An inversion of a permutation $f$ (of $\{1,2,\dots,n\}$) is a pair of elements $(i,j)$ whose relative order gets reversed when you apply $f$. Different authors use different conventions about how to make this precise, and your example seems to be compatible with two of these conventions. Convention 1: An inversion is a pair $(i,j)$ such that $i>j$ but ...


1

Since this is a homework problem I will give you a few things to think about to get you started: The normalizer $N_G(H)$ is the stabilizer of $H$ under the conjugation action of $G$. So you should be thinking about how $G$ acts by conjugation on the set of $p$-subgroups of $G$. The index $[G:H] = [G:N_G(H)]$ is equal to the size of the orbit of $H$ under ...


1

This isn't exactly an answer to your question, but shows how to put this example into a more general context. You can use two more general results, which prove that a group $G$ of order $p^2$ is always abelian when $p$ is a prime. The centre $Z$ of $G$ is non-trivial. Here one uses the fact that the sizes of the conjugacy classes in $G$ divide the order ...


1

There are lots of examples. Many Frobenius groups have this property (that includes the dihedral groups of twice odd order, and other examples such as nonabelian groups of order $pq$ for primes $p,q$). $A_4$, $A_5$, ${\rm SL}(2,3)$ and ${\rm SL}(2,5)$ are examples. It seems likely that $A_5$ is the only nonabelian simple group with this property, and it ...



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