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0

Let $G = \{ 1,a,b,c,ab,ac,abc,bc\}$ be elementary abelian of order $8$. Then $$ G = \{1,a\} \cup \{c,ac\} \cup \{b,bc\} \cup \{ab,abc\}$$ which is the union of two cosets of the subgroup $\{1,a\}$ and two of the subgroup $\{1,c\}$.


1

Here's one example: in the symmetric group $S_5$ $$(1\ 2\ 3\ 4\ 5) = (1\ 2)(3\ 5)\cdot(2\ 5)(3\ 4),$$ so take $a=(1\ 2)(3\ 5)$ and $b=(2\ 5)(3\ 4).$ You can get many more examples like this using the fact that any permutation (of order $5$) can be expressed as the product of two involutions. So, for another concrete example, $$(1\ 2\ 3\ 4\ 5)(6\ 7\ 8\ 9\ ...


1

This is false. If $N=H$ you have $$G/H \times G/H \cong G/H$$ which is not possible for groups whose order is not prime ($G/H$ is not trivial and is finite). In the case that $N \neq H$ you have by minimality that $N \cap H = 1$. Now, it is quite easy to see that $$G/N \times G/H \cong G/1 =G$$ implies that $$|N||H|=|G|$$ Recall the lemma If $H ...


5

Hint: A regular pentagon can be rotated and reflected


1

Hints: Suppose the group is finite, what do you know about its order? Try writing the relations in terms of $a$ and $c=ab$. I use $1$ for the identity below.


1

If you have a bijection between two sets, then they have the same cardinality (the same "number of elements"), so if you have a bijection between $H$ and $gHg^{-1}$ and $H$ is finite, then $gHg^{-1}$ is finite too. (So that's how you use the finiteness of $H$.) For the bijection you can use $f: H \to gHg^{-1}$ defined by $f(h)=ghg^{-1}$. Surjectivity is ...


1

Conjugation by an element is an automorphism, the reciprocal of $\;x\mapsto gxg^{-1}$ being $\;x\mapsto g^{-1}xg$.


2

Lets prove it by contradiction. Let $G$ is not cyclic and finite abelian, we can find an element $g\in G$ which has maximal order among elements of $G$, say order $k$ and $k < |G|$ . Now pick an element $h\in G$ \ $<g>$ and let $|h|=m$ then obviously $m \le k$ as $k$ is maximal among orders of elements of $|G|$. Now consider $l=\text{lcm} (k,m) ...


1

For a counterexample, let $G'$ be an infinite simple group containing a nontrivial torsion element $g$, take $H=1$, $G=\langle g \rangle$. Since $G'$ is simple, it has no proper subgroups of finite index. For example, you could take $G'$ to be the alternating group on a countable set or, for a more exotic example, a Tarski Monster.


1

Assume by contradiction that $G$ is simple. Consider the action on the coset space $G/H$ by left multiplication. This induces a homomorphism $\rho:$ $G \to S_{\frac{m}{m}}$ with $\ker \rho \triangleleft H$. Since $H$ proper, and $G$ simple, then $\ker \rho=\lbrace 1 \rbrace $. So $\rho $ injective. Lagrange's theorem implies that $n$ divides $(\frac{n}{m})!$ ...


1

Assume that $G$ is simple. So the action $f$ of $G$ on the left cosets of $H$ has trivial kernel. That is, $f$ is an injective homomorphism $f:G \to S_{n/m}$ to the symmetric group of degree $n/m$. Hence $|G| = n$ divides $(n/m)!$. But then $(n/m)! < 2n$ implies $n = (n/m)!$, so $f$ must be an isomorphism, and $G \cong S_{n/m}$, which is not simple, ...


1

If $0\to\mathbb Z^n\to\mathbb Z^m$ is an exact sequence of $\mathbb Z$-modules, then $0\to S^{-1}\mathbb Z^n\to S^{-1}\mathbb Z^m$ is an exact sequence of $S^{-1}\mathbb Z$-modules, where $S=\mathbb Z-\{0\}$, hence $0\to\mathbb Q^n\to\mathbb Q^m$ is an exact sequence of $\mathbb Q$-vector spaces, so $n\le m$.


8

Let $m < n$. If $\mathbb{Z}^n$ embeds in $\mathbb{Z}^m$, then the $n$ linearly independent basis elements of $\mathbb{Z}^n$ map to $n$ linearly independent elements of $\mathbb{Z}^m$. The fact that they are linearly independent over $\mathbb{Z}$ implies they are linearly independent over $\mathbb{Q}$, in $\mathbb{Q}^m$. But as $\mathbb{Q}^m$ is a vector ...


1

It means that every chief factor of the group either has order $p$ or it has order coprime to $p$. So a (finite) group is supersoluble if and only if it is $p$-supersoluble for all primes $p$.


2

$ ( G/N )^{\prime} = G^{\prime}N/N $. Let $ K/N $ be a maximal subgroup of $ G^{\prime}N/N $, then by Dedekind modular $ K/N = K/N \cap G^{\prime}N/N = (K \cap G^{\prime}N)/N = (K \cap G^{\prime})N/N $. Hence $ K \cap G^{\prime} $ is a maximal subgroup of $ G^{\prime} $.


1

Let us study the following set of matrices $$ G=\left\{\left(\begin{array}{cc}x_1&0\\0&x_2\end{array}\right)\mid x_1,x_2\in\mu_4 \right\}\cup \left\{\left(\begin{array}{cc}0&x_1\\x_2&0\end{array}\right)\mid x_1,x_2\in\mu_4 \right\}, $$ where $\mu_4=\{\pm1,\pm i\}$ is the set of complex fourth roots of unity. We first observe that the ...


2

Here's a more structural approach. Let $G=\langle A,B\rangle$. As you already noted $A$ and $B$ have order $4$. Now note that $A^2 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$. Hence $P:=A^2B= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ is a permutation matrix. Clearly $G=\langle A,P\rangle$. If $D$ is a diagonal matrix, then ...


1

Since $A^3=A^{-1}$ and $B^3=B^{-1}$, all the elements of the group generated by $A$ and $B$ take the form $ A^{a_1}B^{b_1}A^{a_2}B^{b_2}...$ But then we notice that $BA$=$ABC$ where $C=\begin{pmatrix} -i& 0\\ 0 & i \end{pmatrix}$ We notice also that $AC=CA$ and that $CB= - BC$. Because of this, we can rewrite $ A^{a_1}B^{b_1}A^{a_2}B^{b_2}... = ...


4

Elements of $\widehat{\mathbb{Z}}$ are called profinite integers. The profinite integers have a universal property in the category of profinite groups in exactly the same way that the integers have a universal property in the category of groups: namely, $\widehat{\mathbb{Z}}$ is the free profinite group on one generator. This means precisely that elements $g ...


1

Hint: look at the order of the generators for $V_4$. Warning: spoilers ahead.


2

The quaternion group is indeed the minimal counter-example. Clearly, any group of orders $1,2,3,5$ or $7$ is cyclic, and the two (non-isomorphic) groups of order $4$ are: $\Bbb Z_4$ and $\Bbb Z_2 \times \Bbb Z_2$. There are likewise just two non-isomorphic groups of order $6$: $\Bbb Z_6$ and $S_3 \cong \Bbb Z_3 \rtimes \Bbb Z_2$ (this is the only possible ...


1

Hint: now you know there are only two possibly non-isomporphic semidirect products, namely $$\textrm{GL}(2,3)$$ and $$\mathbb{Z}_2\times \textrm{SL}(2,3).$$ Show that they are not isomorphic because their corresponding centers have different cardinality.


2

Yes, quaternion group is non abelian and not isomorphic to the direct product of $\Bbb{Z/2Z\times Z/2Z\times Z/2Z}$ or $\Bbb{Z/2Z\times Z/4Z}$, and is the one you are looking for.


1

Not sure but a good candidate should be the quaternion group. If not, the alternate group $A_5$ (much bigger), as it is a simple group.


2

In fact, it turns out that there are only finitely many finite groups with a given number of conjugacy classes. This is not hard to prove and it's not too hard to find all of them with up to four conjugacy classes, say: http://groupprops.subwiki.org/wiki/There_are_finitely_many_finite_groups_with_bounded_number_of_conjugacy_classes


1

Given $m,n\ge 1$, the set of all the $m\times n$ matrices with entries in the finite field $\Bbb F_2$ is, under addition, an abelian group which is also unipotent ($x+x=0$ for all $x$). The order of this group is $2^{m\cdot n}$.


1

Let $ M/N $ maximal subgroup of $ G/N $. Since $ M $ maximal subgroup of $ G $ and $ N $ subgroup of $ M $, Then $ P_{G/N}(M/N) = P_{G}(M)/N = G/N $. This means $ G/N $ satisfying maximal permutizer condition.


1

There are $n=2^{a−1}k$ elements of $D_{2n}$ lying outside of the cyclic subgroup $\langle r \rangle$, all of order 2. (These are usually called reflections.) Each such reflection is contained in at least one Sylow $2$-subgroup. A Sylow $2$-subgroup has order $2^a$ and contains $2^{a-1}$ reflections. So there must be at least $k$ Sylow $2$2-subgroups. But a ...


1

As an online resource, you could have a look at this website: http://www.uwyo.edu/moorhouse/pub/bol/. This is about the weaker (not necessarily associative) structure of some finite loops but, among these, you can search the isotopy classes of groups. For example http://www.uwyo.edu/moorhouse/pub/bol/htmlfiles8/8_5_2_0.html or ...


1

If you have a picture in mind of the action groupoid then the Burnside's lemma is quite intuitive. Truth is, you can see it as a 'global' version of the orbit-stabilizer theorem, in fact in any orbit the automorphisms are in bijection with the group and since the orbits form a partition of the set you have a bijection between all the automorphisms and the ...


0

Not only are modules a generalization of vector spaces over fields to vector spaces over rings, but they also arise in the representation theory of finite groups. It turns out that representations of a finite group are equivalent to left $\mathbb{C}[G]$ modules (where $\mathbb{C}[G]$ is the group algebra): Every action of a group $G$ on a complex vector ...


0

Let $x', x''\in G$ be two inverses of $x\in G$. We have the following: $$x'=ex'=(x''x)x'=x''(xx')=x''e=x''.$$ The first equality is because $e$ is neutral element. Second equality is because $x''$ is inverse of $x$. Third equality is because of associativity. The fourth equality is because $x'$ is inverse of $x$. The fifth equality is because $e$ is ...


0

a,b belongs to H,subset of G if ab^(-1) belongs to H then H is subgroup of G H1 meet H2 is nonempty as 1 belongs to both of it let a,b belongs to H1 meet H2 then ab^(-1) belongs to H1 and H2 { as H1 ,H2 are subgroups} ab^(-1) belongs to H hence h is subgroup


1

By taking the quotient, we are effectively treating any integer as the identity element. Hence an element in $\mathbb{R}/\mathbb{Z}$ has finite order if some multiple of the coset representative is an integer.But this is precisely the definition of a rational number.


2

An element of an additive group has finite order if there is a non null $n$ integer such that $n(a+ Z)= Z$. In the case of $1/2$, we have that $2(1/2 + Z)=(1+Z)=Z$. Therefore the order of $1/2$ if finite.


1

Your step (1) looks good; after that, I would look at a factor group: Assume that $a>1$ and let $N:=\langle r\rangle$ and $P\in Syl_2(D_{2n})$. Then $N$ is cyclic and therefore abelian, and $Q:=P\cap N$ has index $2$ in $P$, so $Q$ is normalized by $\langle P, N\rangle = D_{2n}$. By looking at the relations in the definition of $D_{2n}$, it's easy to see ...


7

$xyxy = yxy^2 = yx^4 = y^3 x$ by post-multiplying $xyx$ by $y$. $xyxy = x^2 yx$ by pre-multiplying $yxy$ by $x$. Therefore $y^3 x = x^2 y x$, so (by cancelling $x$ from RHS) $y^3 = x^2 y$, so (by cancelling $y$ from RHS) $y^2 = x^2$. Therefore $\underbrace{y^2 = x^3}_{\text{relation}} = x^2$, so $x = e$ and hence $y=e$.


6

It can be shown that this is a representation of the trivial group by some brute force manipulations: \begin{eqnarray*} xyxy&=&(xyx)y=(yxy)y=yxy^2=yxx^3=yx^4=y^3x,\\ xyxy&=&x(yxy)=x(xyx)=x^2yx,\\ \end{eqnarray*} so $y^3x=x^2yx$. Right-multiplying both sides by $(yx)^{-1}$ shows that $x^2=y^2$. Hence $x^3=y^2=x^2$, from which it is immediate ...


0

Ok, It seems that my doubt was correct. The from without conjugates is invariant and it is a mistake in book. Probably will be corrected in new editions. It is weird to answer your own question, but since this question doesn't need to be open anymore, I do it :),


2

The conditions imply that given any two cyclic subgroups, one is a subset of the other. This implies that the group must be cyclic because otherwise there would be two cyclic subgroups that intersect trivially. The same reason implies that the order is a prime power: if there were two relatively prime factors of the order that are both larger than $1$, then ...


4

Cauchy's theorem states that if $p$ is a prime dividing $|G|$ then $G$ has a subgroup of order $p$. If two different primes divide $|G|$ then your condition leads to a contradiction, so $|G| = p^n$ for some $n$. As for why $G$ is cyclic, consider the subgroup generated by an element in $G \setminus G_{k - 1}$.


2

In the future it will be worthwhile for you to know how to compute the order of an element of $\mathrm{Sym}(n)$, so I will detail it here. Recall that every permutation is a product of disjoint cycles, for example $\sigma=35412$ is $(134)(25)$. Disjoint cycles commute with each other, so exponentiating a permutation amounts to raising each disjoint cycle ...


2

You could do this with a constraint solver, such as Savile Row ( http://savilerow.cs.st-andrews.ac.uk/ ). While this tool does not support permutations, and permutation multiplication, directly, you can map the problem to integers fairly easily. This wouldn't use any group theory techniques at all, just tackle the problem as permutations with search, but ...


1

The quotient group $G/H$, where $G$ is a group and $H$ is a normal subgroup, is the set of equivalence classes under the equivalence relation $$ a\sim_H b \qquad\text{if and only if}\qquad ab^{-1}\in H $$ It is well known that the equivalence classes are the subsets of the form $$ aH=\{ah:h\in H\} $$ as $a\in G$, called cosets. More precisely left cosets, ...


0

There's a canonical map $\pi: G \to G/H$ which is usually defined by $\pi(g) = g+H$ and is sometimes written as $\bar g$. A system of representatives is then a set $S \subset G$ such that for every element $x\in G/H$ there is precisely one element $g\in G$ with $\pi(g) = x$. For example, take $G = \mathbb Z$ and $H = 6\mathbb Z$ then I could set $S = \{0, 1, ...


0

The Dynkin basis is the basis where we label the weights in terms of the Eigenvalues of the Cartan generators in the Chevalley basis. Or in other words The Dynkin labels are the eigenvalues of the Chevalley generators of the Cartan subalgebra The $H$-basis and the Chevalley basis are simply two different choices for the basis for the Cartan ...


3

Check out this: (resource: The Theory of Finite Groups: An Introduction. Hans Kurzweil Bernd Stellmacher pp31)


0

The function $V \otimes V^* \to \mathbb{C}[G]$ is given by $(v, \lambda) \mapsto (g \mapsto \lambda(gv))$. Let $G = SL_2$ and write the coordinate on $SL_2$ as $$ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right). $$ As shown in the post, $Span_{\mathbb{C}}\{c, d\}$ is a representation of $G$. Let $V = Span_{\mathbb{C}}\{c, d\}$ and take $e_1 ...


0

There are three things that are needed for a subset $H$ of a group $G$ to be a subgroup. They are The identity element $e \in G$ is contained in $H$. If $a, b\in H$, then $ab \in H$ as well. If $a \in H$, then $a^{-1} \in H$ as well. (It's possible to state it more compactly, for instance by saying "Whenever $a, b \in H$, we must have $a^{-1}b \in H$ as ...


2

Let $P=(x,y)$ be a point on the elliptic curve which is not the identity. We want to be determine whether $nP=O$, so for this we need to go through the arduous process of explicitly calculating what $nP$ is. This gets very messy considering you need to determine when you are adding a point to itself. Fortunately, there are these things called division ...



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