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1

To understand the explicit structure of group algebras like this, a useful idea is that of a skew group ring, which is a ring theoretic analogue of a semidirect product. Let $G$ be a group acting by ring automorphisms on a ring $A$. Then the elements of the skew group ring $A\ast G$ are formal finite sums $\sum_{g\in G}a_gg$ with $a_g\in A$, with ...


1

You should be familiar with products of nonabelian groups, say $G \times H$ is a product, with at least one of the groups not abelian, say $G$ is that group. Well $G$ is a "subgroup" of $G \times H$, and is not abelian, so $G \times H$ can not be abelian. Under the same conditions, let $G * H$ be the coproduct of $G$ and $H$, so it has the universal ...


4

By way of enrichment here is an alternate formulation using combinatorial species. The species of permutations with no transpositions is $$\mathfrak{P}(\mathfrak{C}_{=1}(\mathcal{Z}) + \mathfrak{C}_{\ge3}(\mathcal{Z})).$$ This gives the generating function $$G(z) = \exp\left(z+\frac{z^3}{3}+\frac{z^4}{4}+\cdots\right)$$ which is ...


0

Try staring hard at this picture. It should be clear that, to drive this screw into a block of wood, it has to be rotated clockwise from the top. This is a "right screw", and the orientation of this screw is different from that of a "left screw", which has to be rotated anticlockwise.


1

In "Growth of finitely generated solvable groups", by John Milnor, it is proved that if a finitely generated group $G$ is solvable group and is not polycyclic, then $G$ has exponential growth. Finitely generated nilpotent groups groups are polycyclic, or you could use Gromov's theorem to get that f.g. virtually nilpotent groups have polynomial growth, to ...


3

Hint: With the Principle of Inclusion and Exclusion, your result should be $\sum_{k=0}^{\lfloor n/2\rfloor}\,\frac{(-1)^kn!}{2^kk!}$, which is roughly the closest integer to $\frac{n!}{\sqrt{\text{e}}}$.


1

https://en.wikipedia.org/wiki/Weyl_group I am pretty sure he is discussing, without using the terms, Lie Algebras and what came to be known as Weyl Chambers. The setting for a root lattice is a (positive) quadratic form, let us call it $Q.$ The (squared) norm of a lattice vector $x$ is precisely $Q(x)$ which is an integer; for the cases of interest, this ...


2

The word automorphism can be used in a lot of contexts, as for example the sense here of a mapping of space (?) onto itself in a way that "preserves the structure". This is a reference to the Euclidean properties of space, i.e. its distances and angles. Reflection in any plane preserves the measures of distance and angles, and so is an automorphism in that ...


5

The phrase "structure of a space" has different meaning in different contexts. The quote from Weyl probably gives the structure implicitly, and I am guessing that he is talking about planes in Euclidean 3-space. In that case, an automorphism is a bijection of the points of Euclidean 3-space which preserves all the structures of Euclidean geometry: lines; ...


2

The groups $S_6$ and ${\rm PGL}(2,9)$ both have a unique composition series with normal subgroup $A_6$ and quotient group of order $2$.


1

The two sets $$A=\{ xgx^{-1} : x \in G\}, B=\{ x^{-1}gx : x \in G\}$$ are actually the same set. In fact, for all $xgx^{-1} \in A$, denote $y= x^{-1}$, so you have $$xgx^{-1} = y^{-1}gy \in B$$ Thus $A \subset B$. And you can prove similarly that $A \supset B$.


2

You should except that this is definitely not true. Generally there are a ton of groups with a given set of composition factors. The number of distinct composition series does not seem like an invariant which can carry too much information about the structure of the group, so you would expect some coincidences. One reasonable place to look for ...


3

I don't know much about this topic, but I did a quick calculation in Magma on $A = {\mathbb F}_3G$. It appears to be a direct sum of a simple algebra of dimension $18$ and a uniserial algebra of dimension $3$ with $3$ trivial composition factors. If you do this over the field of order $9$, then the $18$-dimensional algebra splits into a direct sum of two ...


2

We first note two simple facts. 1) If $\sigma=(a_1 \ \ldots\ a_n)$ is an $n$-cycle, then $\sigma(a_i \ a_j)$ is a product of two disjoint cycles, the sum of whose lengths is $n$. 2) If $\sigma=(a_1\ \ldots\ a_k)$ and $\theta=(b_1 \ \ldots\ b_l)$ are two $k$ and $l$ cycles respectively, then $\sigma\theta(a_i\ b_j)$ is a $k+l$ cycle. Now suppose we have an ...


3

If $\sigma \in S_n$ is a permutation, let $f(\sigma)$ be the number of cycles in the cycle decomposition of $\sigma$, including the 1-cycles as fixed points. Then for any transposition $\tau = (i j)$, we have $f(\sigma\tau)=f(\sigma)\pm 1$, where the sign is $+$ if $i$ and $j$ are in the same cycle in $\sigma$, and $-$ if they are in different cycles in ...


3

Bhaskar's example works (+1), though it is perhaps little unsatisfying. More generally, it is true that $\gcd(n, m) = 1 \iff \mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{nm}$; try proving this! From here, we can construct many concrete examples. For sake of demonstration, let's take $60$, which has prime factorization $2^2 \cdot 3 \cdot 5$. Notice ...


6

You don't even expect that to be true for numbers (i.e., as in $2\cdot 2 = 1\cdot 4$). A more reasonable expectation would be to have a cancelation property, i.e., if $G\times H\cong G\times H'$, does it follow that $H=H'$. This has some chance of being true, and in fact it's a rather deep theorem to classify the groups for which it does hold. What is not so ...


5

HINT- $\Bbb{Z_2}\times \Bbb{Z_3}\cong e\times \Bbb{Z_6}$


0

HINT: $$g(a) = (t(b,c) \circ t(a,b))(a) = (t(b,c))(b) = c.$$ What's $f(a)$?


1

Dietrich worded his answer in a way that I think is excessively confusing. What I think he means is that "diagonal" and "antisymmetric" are words that depend on a choice of basis of $\mathbb{R}^n$, not of the Lie algebra. You can get "diagonal" to mean something different by picking a basis that isn't just a differently-scaled version of the usual basis, and ...


5

I'll assume all of your $V_i$ are in fact the same vector space $V$, or else I don't understand what your notation means. Then $S_n$ acts on $G \times \dots \times G$ by permuting the factors (in your example $G = GL(V)$ but this argument applies more generally), and your semidirect product is a wreath product $$G \wr S_n = (G \times \dots \times G) \rtimes ...


2

The problem is that the property "diagonal matrices", or "antisymmetric matrices" depends on the basis of the underlying vector space. The standard way to represent the Lie algebra $\mathfrak{so}(n)$ faithfully by matrices, is by antisymmetric matrices. However, we may also represent $\mathfrak{so}(n)$ faithfully by matrices which are not skew-symmetric. ...


4

The problem is with #2. The rank of a Lie algebra is the dimension of any of its Cartan subalgebras (all of which are isomorphic). None of them need to contain diagonal matrices. You can see one such subalgebra computed for $SO(2n + 1, \mathbb{C})$ in this paper: Structure Theory of Semisimple Lie Groups - Knapp


4

Yes, the non-abelian group of order 63 satisfies the properties. I checked that with GAP. gap> G := SmallGroup(63, 1);; gap> A := AutomorphismGroup(G);; gap> IsNilpotent(A); false gap> Fixed := Filtered(G, g -> ForAll(A, a -> g = g^a));; #including 1 gap> List(Fixed, Order); [ 1, 3, 3 ]


0

for basics of trigonometry you can go for S.L. lony bok, and the topics what you have mentioned i will suggest two books 1) advanced engineering mathematics by erwin kreyszig 2) advanced engineering mathematics by ramana the first one is for deep understanding of the topics and second one is for practice. also if you are still looking for further basics in ...


1

Adventures in Group Theory talks somewhat extensively about the Rubik's Cube. http://www.amazon.com/dp/0801869471/


0

something like this? http://www.math.harvard.edu/~jjchen/docs/Group%20Theory%20and%20the%20Rubik's%20Cube.pdf


0

Many trig formulas are things you just look up (or memorize). They can also be sometimes be derived from each other. For example, knowing $\sin^2 x + \cos^2 x =1$, dividing by $\cos^2 x$ yields $\tan^2 x + 1 = \sec^2 x.$ A useful exercise would be to take a list of rules (maybe from your textbook) and see which ones can be derived from other ones. It ...


4

You could try Lagrange's Theorem which tells you something about the order of a subgroup. Count the elements of $H$.


0

In a f.g. nilpotent group $G$, every subgroup is closed in the profinite topology. This is proved by induction on the Hirsch length $h$ of $G$. If $h=0$, $G$ is finite, OK. Otherwise, the center of $G$ is infinite and hence contains an infinite cyclic subgroup $Z=\langle z\rangle$. Write $Z^n=\langle z^n\rangle\subset Z$. Let $H$ be a subgroup of $G$. Then ...


5

Suppose $(a,b),(c,d),(e,f)\in\Bbb Z^2$ satisfy no nontrivial relation. Then in particular, no pair of them does. Since $d(a,b)-b(c,d)=(ad-bc,0)$, if a pair of vectors in $\Bbb Z^2$ are independent then the determinant of the resulting square matrix must be nonzero. (Otherwise $b$ and $d$ would be $0$, which is impossible since $(a,0)$ and $(c,0)$ are not ...


3

Perhaps you've seen that the group of inner automorphisms of a group, $\text{Inn}(G)$ is a subgroup of $\text{Aut}(G)$ isomorphic to $G/Z(G)$? Now $Z(D_6) = \{e, r^3\}$, and it is not hard to show that $D_6/Z(D_6)$ is a non-abelian group of order $6$. So $\text{Aut}(D_6)$ has order at least (and a multiple of) $6$. Any automorphism must send only ...


0

Well, all characteristic subgroups are necessarily normal, so that suffices. As Taylor's comment points out, using that Sylow subgroups are conjugate you know that $g\left(\bigcap P_i\right)g^{-1}=\bigcap g P_i g^{-1}$, which is an intersection of Sylow subgroups. Indeed, we necessarily have $\bigcap g P_i g^{-1}= \bigcap P_i$: if $g P_i g^{-1} = g P_j ...


2

Hint: To enumerate the automorphisms, recall that an automorphism is uniquely determined by how it acts on the generators of the group. Consider the presentation, $$D_{12}=\langle r,s \mid r^6=s^2=1, srs=r^{-1}\rangle = \{ 1,r,r^2,r^3,r^4,r^5,s,sr,sr^2,sr^3,sr^4,sr^5\}.$$ The subgroup $\langle r\rangle$ is the unique subgroup of index 2, which must therefore ...


0

In general, to show that an element $g$ is a generator of a group, you need to show that every element in the group is some power of $g$. In your case here, we know that $|\mathbb{Z}_{19}^*|=18$ and the order of an element divides the order of the group, so it suffices to check that $2^2=4\neq 1$, $2^3=8\neq 1$, $2^6=13\neq 1$, and $2^9=18\neq 1$. ...


3

We know that $\mathbb{Z}_{19}^\times = 18$, and that $2$ generates a subgroup, $H$. By Lagrange's theorem, the order of $H$ must divide $18$, so we must have $|H| \in \{1,2,3,6,9,18\}$. We have \begin{align} 2^1 &= 2 \neq 1\\ 2^2 &= 4 \neq 1\\ 2^3 &= 8 \neq 1\\ 2^6 &= 64 \equiv 7 \neq 1\\ 2^9 &= 2^6 \cdot 2^3 \equiv 7 \cdot 8 = 56 ...


1

Yes, $PGL(V)$ is a linear (affine) algebra group. For more on this you can see these notes: https://www.ma.utexas.edu/users/allcock/lag/lag14.pdf See Theorem 7.1. Remember that $PGL(V)$ is defined as a quotient $GL(V) / \mathbb{G}_m$.


1

Combinatorics Solution: It might be easier to find the number of solutions where $ad - bc = 0$. There are two cases to consider: $a = 0$ and $a \neq 0$. In the second case, you can pick $b,c$ arbitrarily, and then we are forced to have $d = bc/a$ (remember nonzero elements have inverses mod $p$). Linear Algebra Solution: Think of the columns of the matrix ...


2

Yes. If $Z(G)\not\subseteq H$, then letting $x\in Z(G)$ will satisfy the problem. If $Z(G)\subseteq H$, then we take quotient by $Z(G)$ and apply induction. Suppose $x\mapsto \bar{x}$ denotes the quotient map. Then, by the induction hypothesis, there is $\bar{x}\in\bar{G}\backslash\bar{H}$ such that $\overline{xH}=\overline{Hx}$. It is easy to conclude ...


1

I define $r$ to be one rotation clockwise, and s to be reflection on the 'horizontal' line (see the figure). As an example, I want to evaluate $rsr^2$. Without lose of generality, I choose right-to-left as order of action that is first $r$ then $s$ and $r^2$ acting on id (first configuration) on the figure. Pictorially: So the final equivalence is ...


2

There are multiple levels of confusion, here (and I have deleted my previous answer because of this). Many posters here are accustomed to cycle notation, where $\sigma = (1\ 2\ 4\ 3)$ means this: $\sigma(1) = 2\\ \sigma(2) = 4\\ \sigma(3) = 1\\ \sigma(4) = 3.$ It is also possible to employ "image notation" where $\sigma = (1\ 2\ 4\ 3)$ means this: ...


2

When we write $(1243)$, we mean the permutation defined where $1 \mapsto 2$ and $2 \mapsto 4$ and $4 \mapsto 3$ and $3 \mapsto 1$. Likewise, $(1324)$ means we have $1 \mapsto 3$ and $3 \mapsto 2$ and $2 \mapsto 4$ and $4 \mapsto 1$. Finally, any number not written inside a set of parentheses with other numbers is presumed to be fixed. So, for example, if ...


2

The condition is equivalent to a Sylow $p$-subgroup being self-normalizing in $S_n$, and the answer is very simple (simple to state, anyway): $p=2$ is both necessary and sufficient for the condition to hold for all $n$. It is well-known that a Sylow $p$-subgroup $P$ of $S_n$ is a direct product of iterated wreath products $P_k = C_p \wr C_p \wr \cdots \wr ...


3

What you have written is not a presentation. In general it is not a good or efficient strategy to just write down powers of elements hoping that you will get a presentation. The best known presentation of $S_5$ is on the generators $a=(1,2)$, $b=(2,3)$, $c=(3,4)$, $d=(4,5)$: $$\langle a,b,c,d \mid a^2,b^2,c^2,d^2,(ab)^3,(bc)^3,(cd)^3,(ac)^2,(ad)^2,(cd)^2 ...


1

When you write a "number arrangement" $a,b,c,d,e,f$, you are interpreting it as, the number $a$ winds up in position $1$, $b$ winds up in position $2$, and so on. But here's a different, equally valid representation: $a,b,c,d,e,f$ is now interpreted as, vertex $1$ winds up at position $a$, vertex $2$ winds up at position $b$, and so on. With this ...


2

I assume you're using the original labeling here. I think you made some mistake in your calculation, or are using inconsistent conventions. At the top of your post, you indicate that $r^2 s$ corresponds to (15). Now (15) is what you get by doing $s$ first, then $r^2$. But then later in your post, you seem to think that $r^2 s$ means "do $r^2$ first, then ...


2

There is absolutely no reason at all. It is entirely up to the author of the textbook which way these things go, and different textbooks will use different conventions for both the symmetric group and the dihedral group. Whenever you define a group of transformations or operations, you have the choice as to whether $gh$ should mean "do $g$ first, and then ...


2

A way that doesn't use group actions (which are awesome, by the way, so try to use them when you can): We want to know the order of the centralizer of $g,\ |C(g)|$. Now $x \in C(g)$ means that $xgx^{-1} = g$. Since $x(1\ 2\ 3)x^{-1} = (x(1)\ x(2)\ x(3))$ we can immediately see two things: $x$ must map $\{1,2,3\} \to \{1,2,3\}$ and thus $\{4,5\} \to ...


0

Here is how I see it: the cosets $a^kH$ partition $G$, so we immediately have that any $g \in G$ is in some coset $a^kH$, so $g = a^kh$ for some $k \in \Bbb Z^+$ and some $h \in H$, that is: $g \in KH$ (since $a^k \in K$). Thus $G \subseteq KH$. But, obviously, $KH \subseteq G$, so the two are equal. (it might seem mystifying that I did not use normality ...



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