New answers tagged group-theory
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"Linear independency" of Lie Brackets
Yes, indeed they always will be whatever basis you choose in the case of $\mathfrak{so}(3)$ although this is a more in depth result and not true for every Lie algebra.
Obviously the structure ...
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How to find the generator matrix for $C/C^{\perp}$?
$C^\perp$, a $[6,1,6]$ code, has $32$ cosets, $16$ of which have only even-weight vectors while the other $16$ cosets have only odd-weight vectors.
Begin Edit: To answer the numerous questions raised ...
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Alternative between 2-step solvable or perfect
Here are two observations to help you to classify such Lie groups:
If $G$ is a noncompact connected semisimple Lie group, it contains a closed subgroup locally isomorphic to $SL(2,\mathbb R)$ and, ...
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Is it possible to write $\mathbb{Z}_2^3$ using 2 generators?
Since $\Bbb Z_2$ is a field we have that $\Bbb Z_2^3$ is a vector space of dimension $3$. It therefore need $3$ generators.
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Criterion for cyclic groups in terms of its number of subgroups
I found a proof (from an old sci.math post of my own) that a finite group of order $n$ has at least $d(n)$ (the number of divisors of $n$) subgroups. This follows from the case $m=n$ in the result ...
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uniqueness of generators of Lie groups
Qiaochu has already answered this in the comments but I wanted to put together a longer answer as I see this confusion fairly regularly here.
It is quite common in Physics to use "the" a bit ...
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Let $K$ be a field. If $f = \text{irr}(c_1;K) = \text{irr}(c_2;K) (c_1 \neq c_2)$ and $\deg(f)$ is odd, then $c_1 + c_2 \not\in K$
I found my idea is correct if $f$ is separable.
Suppose $c = c_1 + c_2 \in K$, $K(c_1) = K(c_2)$. And we directly have an automorphism $\varphi$ on $K(c_1)$ which is uniquely determined by $\varphi: ...
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Structure theorem for profinite abelian groups
No. For me it's easier to think about the Pontryagin dual question: recall that Pontryagin duality $A \mapsto \hat{A} \cong \text{Hom}(A, S^1)$ restricts to a contravariant equivalence of categories ...
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Primes in other sets?
As Randall said, the definition of prime elements applies to any structure that is a commutative ring.
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Let $G$ be the special linear group $SL_2(3)$
Let $a=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ and $b=\begin{pmatrix} 1 & 2 \\ 2 & 2 \end{pmatrix}$.
Then $ab=\begin{pmatrix} 0 & 1 \\ 2 & 0 \end{pmatrix}\ne \begin{...
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Why is Aut(Griess algebra) discrete?
(All vector spaces, algebras, etc. below are real and finite-dimensional.)
So, let's discuss the following general question: suppose $A$ is a vector space equipped with a multiplication $m : A \otimes ...
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Is a transitive-on-unit-vectors subgroup of $SO(3)$ automatically all of $SO(3)$?
First of all, the group $G=SO(3)$ is simple as an abstract group: This fact was discussed several times on this site (see, say, here). Let's prove that your group $F$ is a normal subgroup of $G$, i.e. ...
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Is there a general way to find the inverse of an automorphism of the free group?
Yes. Let $X$ be the given (ordered) free generating set of the free group, and let $Y$ be the set of images of the elements of $X$ under the automorphism.
Now perform Nielsen reduction on $Y$, and ...
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A group with exactly half of the elements in one conjugacy class
Your construction in paragraph 2 gives all examples!
Now the conjugacy action $G$ on $S$ gives an injective map $i:G\to S_{|G|/2}$.
I follow you up to here but here I think you've made a small extra ...
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Derived series of a square-free order group stabilizes
Since ${\rm Aut}(G''/G'')$ is abelian, the restriction to $G'$ of the action of $G$ on $G''/G'''$ is trivial, so $G''/G''' \le Z(G'/G''')$.
But $G'/G''$ is a direct product of cyclic groups of prime ...
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How do you prove associativity for this operation?
Short answer
$$\frac{a/(1/b)}{1/c}=\frac{(a/(1/b))\color{blue}{/(1/(1/b))}}{(1/c)\color{blue}{/(1/(1/b))}}=\frac{(a\color{red}{/(1/b)})/(1\color{red}{/(1/b)})}{(1/c)/(\color{red}{1/(1/b)})}=\frac{a}{(...
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$S_6$ contains two subgroups that are isomorphic to $S_5$ but are not conjugate to each other
Transitivity of a subgroup will be preserved by conjugation.
There's $6$ obvious non-transitive $S_5$'s in $S_6.$
There's also famously a transitive $S_5,$ corresponding to an outer automorphism.
...
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Possible indices of finite index subgroups of $SL_2(\mathbb{Z})$
Actually we can give a straightforward uniform argument that $\Gamma \cong PSL_2(\mathbb{Z})$, and hence $SL_2(\mathbb{Z})$, has a subgroup of index $n$ for every positive integer $n$. The sequence of ...
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Relative divisibility of derived subgroup of free group
No. If $F$ has rank $\kappa$ (any cardinal, possibly infinite), then quotient $F/[F,F]$ is the free abelian group of rank $\kappa$. As free abelian groups are torsion free, if $w^n\in [F,F]$, then $w[...
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Show that the given representation of the group $G$ is reducible
Whenever you feel stuck on a question, a basic reflex you need to have is to write down the definitions and see if you understand them. Very often in mathematics, questions are answered effectively by ...
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Let $D_6 = \langle a, b \mid a^6 = b^2 = e, ba = a^5b \rangle$ be the dihedral group
Your solution for part (a) is correct.
Part (b): You have been asked to find all subgroups of order $4$ in $D_6$.
Your table shows that there is no element of order $4$ in $D_6$. So, we do not have ...
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Let $D_6 = \langle a, b \mid a^6 = b^2 = e, ba = a^5b \rangle$ be the dihedral group
Your answer for (a) is correct.
Your answer for (b) is not correct.
For example number 2 fails because $b \cdot ab = a^5$ which is not an element of the subgroup.
First to get a subgroup $S$ of order ...
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$A_{11}$ has no subgroups of order $\frac{11!}{14}$?
SOLUTION: Assume by way of contradiction that there exists $H \leq A_{11}$ with $|H| = 11!/14.$ Let $A_{11}/H$ be the set of left cosets of $H$ in $A_{11}$; e.g. the set of all $\sigma H$ for $\sigma \...
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Let $a$ be the reflection of the plane $\mathbb{R}^2$ over the bisector of the odd quadrants
For part (a), note that $G = \langle a, b \rangle$ indicates the subgroup generated by $a$ and $b$ (in which larger group?). So, in general, this can be rather large, since any word we write down in ...
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Understanding proof of existence of Schreier transversals
Your question is answered by a standard application of Van Kampen's Theorem, and the set $J$ is defined in the second sentence of the second paragraph of the proof. Perhaps it will be clearer if one ...
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Group of order $n$ is a subgroup of $S_{n-1}$
Yes your idea works exactly as stated. Since $n$ is not a prime power it has distinct prime divisors $p$ and $q$ and subgroups $H$ and $K$.
Let $\phi_H$ and $\phi_K$ be the homomorphisms of $G$ into $...
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Describe all non-isomorphic groups of order $57$
As noted in the comments, $|G| = pq$ a product of distinct primes doesn't imply $G$ is cyclic (for example, $S_3$ has order $2 \cdot 3$), so the attempt in the question fails.
If $|G| = 3 \cdot 19$, ...
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What is the index $ (G : C_G(B)) $?
The characteristic polynomial of $B$ is $(x-1)^2$, so it admits a Jordan decomposition over the field with $5$ elements, with Jordan normal form $J=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ (...
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The Uniqueness of the Logarithm as a Group Isomorphism between the Positive Reals and Reals
This is false without more hypotheses on $\varphi$, e.g. it suffices to require that $\varphi$ is continuous, or monotonic, or measurable. Without any hypotheses we can compose $\varphi$ with ...
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"Abstract" presentation of $SL(2,\mathbb Z)$
The group $SL(2,\mathbb Z)$ does not have a presentation of the form that you wrote. If you substitute $U=ST$ into that presentation, then you can eliminate $T$ and rewrite your presentation in the ...
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A certain inverse limit
Yes. It is a consequence of Galois theory : Let $L|K$ be a finite Galois extension, of Galois group $G$. Then there is a decreasing bijection between distinguished subgroups of $G$, and subextensions $...
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Do sets of commuting permutations with no fixed points generate Abelian groups with no fixed points?
It is difficult for permutation actions to have the property that no non-identity element has fixed points. This property is called being free, and for a group $G$ acting on a set $X$ it is equivalent ...
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subgroups of $(\mathbf Q, +)$ as direct limits
Yes, that's right. The general pattern is that the directed colimit of
$$\mathbb{Z} \xrightarrow{n_1} \mathbb{Z} \xrightarrow{n_2} \mathbb{Z} \xrightarrow{n_3} \dots $$
computes an increasing union of ...
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Show that no group of order 48 is simple
Suppose that $G$ of order $48$ is simple and $n_2=3$. Therefore, $G\hookrightarrow S_3$ (consider some $G$-action on the quotient set $G/P_2$, where $P_2$ is any Sylow $2$-subgroup): contradiction, by ...
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How to prove that all elements inside a cycle of a cyclic group are different from each other
Here is the standard way to see this.
Suppose that $n = ord(a)$, so $n$ is the smallest positive integer such that $a^n = e$.
Then note the following:
(*) $a^k = e$ if and only if $n$ divides $k$.
...
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Irreps of $SU(3)/\mathbb{Z}_3$ from irreps of $SU(3)$
(All representations are complex and finite-dimensional throughout except for at a handful of points where I talk about real representations.)
In general, if $V$ is an irreducible representation of a ...
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Surjective homomorphism $\mathbb Z * \mathbb Z \to C_2 * C_3$; can my proof be rescued?
The key is that the coproduct is generated by the canonical images of the groups it is a coproduct of.
Lemma. Let $G_1$ and $G_2$ be groups, and let $\iota_j\colon G_j\to G_1*G_2$ be the canonical ...
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Groups of homeomorphisms vs Configuration spaces
First of all, for every connected manifold $M$ (without boundary), $Homeo(M)$ acts transitively on the $n$-fold configuration space of $M$: This was discussed several times on MSE. It follows that $...
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Does there exist a group $G$ such that $\operatorname{Aut}(G)\cong D_5$, where $D_5$ denotes the dihedral group of order 10?
There is no such group.
First, as already laid out in the comments, $G$ cannot be abelian. For completeness sake, here is the argument again: If $G$ is abelian and not an $\mathbb{F}_2$-vector space, ...
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Show that no group of order 48 is simple
Analyzing the 2-Sylows:
The number $n_2$ of 2-Sylows is of the form $n_2 = 2k + 1$, for some integer $k \geq 0$, where $n_2 \mid 3$. Testing the possibilities, we see that $(n_2, k) = (1, 0), (3, 1)$.
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Let $\mathbb{R^*}$ be the multiplicative group of nonzero real numbers.Which of the following statements are true?
For statement $2$, if $x^3 \in H$ for all $x \in \mathbb{R}^*$, then $x=(\sqrt[3]{x})^3 \in \mathbb{R}^*$ for all $x \in \mathbb{R}^*$ (using the existence of cube roots).
For statement $4$, suppose ...
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Is this definition of a cycle in symmetric groups correct?"
You've correctly identified a very common (and useful) abuse of notation. When a small cycle is written in a "large" symmetric group, the implicit assumption is that all numbers not ...
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Is this definition of a cycle in symmetric groups correct?"
Yes it is correct. The cycle (2 4) tells you that something is happening to the numbers 2 and 4. Implicitly, all other numbers do not change, so the cycle bijection keeps them fixed, and it is truly a ...
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Why is the order of an element equal to the order of the group it generates?
Suppose that $ord(a) = n$ for some integer $n > 0$.
Then $a^n =e$. Thus $a \circ a^{n-1} = e$, and multiplying both sides with $a^{-1}$ gives you $$a^{-1} = a^{n-1}.$$ Now raising both powers by ...
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Show that $H$ is a normal subgroup of $G.$
Consider the action of $G$ in $G/H$ (note that it isn't necessary a group) by translation, $x.(gH)=(xg)H$. We know $\#G/H=3$, then we have a morphism from $G$ to $Biy(G/H)\cong S_3$ by $x\mapsto \...
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Is every (infinite) permutation the composition of 2 involutions in ZF?
I don't have an answer, but I want to note we know a lot about the structure of the involutions.
Let $\tau(x)=y$. Then we have
$$\sigma(y)=f(x)$$
$$\sigma(f(x))=y$$
so
$$\tau(f^{-1}(y))=f(x)$$
$$\tau(...
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How to find the order of $\text{Aut}(\text{Aut}(\mathbb{Z}_{1080}))$
This should be a comment but I need the nice formatting of an answer, so I have made it CW.
According to GAP:
...
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How to find the order of $\text{Aut}(\text{Aut}(\mathbb{Z}_{1080}))$
No, in general the group ${\rm Aut}({\rm Aut}(\Bbb Z/n))$ does not have order $\phi(\phi(n))$. The group need not even be abelian. It is useful to look at a smaller example first. Consider $n=12$. ...
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Can the sum of a nonlinear irreducible character's values on $Z(\chi)$ be zero?
This sum of values is not equal to zero if and only if $\chi$ is constant over $Z(\chi)$.
First, write the restriction of $\chi$ to $Z(\chi)$ as $\chi_{Z(\chi)}=\chi(1)\lambda$, where $\lambda$ is a ...
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If $G/Z(G)$ is isomorphic to a subgroup of $\mathbb Q$ then $G$ is abelian.
Hint: assume for contradiction that some $a, b \in G$ don't commute, and consider $H = \left< a, b \right> \leqslant G$.
PS It is also not hard to unpack the reasoning so that it does not use ...
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