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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 ...
Derek Holt's user avatar
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3 votes

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 ...
Callum's user avatar
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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: ...
Long Song's user avatar
5 votes
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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 ...
Qiaochu Yuan's user avatar
4 votes

Primes in other sets?

As Randall said, the definition of prime elements applies to any structure that is a commutative ring.
Karl's user avatar
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1 vote
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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{...
Alex Ravsky's user avatar
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3 votes
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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 ...
Qiaochu Yuan's user avatar
6 votes
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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. ...
Moishe Kohan's user avatar
9 votes

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 ...
Derek Holt's user avatar
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4 votes
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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 ...
Qiaochu Yuan's user avatar
1 vote
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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 ...
Derek Holt's user avatar
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3 votes

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}{(...
ultralegend5385's user avatar
0 votes

$S_6$ 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. ...
ryan mcbean's user avatar
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4 votes
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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 ...
Qiaochu Yuan's user avatar
2 votes
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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[...
Arturo Magidin's user avatar
3 votes
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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 ...
Fançois Gatine's user avatar
2 votes
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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 ...
Nothing special's user avatar
0 votes

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 ...
jMdA's user avatar
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5 votes

$A_{11}$ 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 \...
roblich mandervach's user avatar
2 votes
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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 ...
Sammy Black's user avatar
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3 votes

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 ...
Lee Mosher's user avatar
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12 votes
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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 $...
Derek Holt's user avatar
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3 votes
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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 $|G| = 3 \cdot 19$, ...
arkeet's user avatar
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2 votes
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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 $J=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ (...
tomasz's user avatar
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4 votes
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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 ...
Qiaochu Yuan's user avatar
0 votes

"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 ...
Lee Mosher's user avatar
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2 votes

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 $...
Enguerrand Moulinier's user avatar
5 votes
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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 ...
Qiaochu Yuan's user avatar
3 votes
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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 ...
Qiaochu Yuan's user avatar
0 votes

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 ...
Kan't's user avatar
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1 vote

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 $k$. ...
testaccount's user avatar
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3 votes
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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 ...
Qiaochu Yuan's user avatar
6 votes

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 ...
Arturo Magidin's user avatar
1 vote
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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 $...
Moishe Kohan's user avatar
9 votes
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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, ...
Tim Seifert's user avatar
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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 $(n_2, k) = (1, 0), (3, 1)$. ...
Gjhdby5 Vjfhu's user avatar
2 votes
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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 ...
Geoffrey Trang's user avatar
4 votes
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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 ...
Robert Shore's user avatar
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3 votes

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 ...
TheAlertGerbil's user avatar
1 vote
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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 ...
testaccount's user avatar
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1 vote
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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 \...
Benjamin Martinez Arcos's user avatar
1 vote

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(...
Zoe Allen's user avatar
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1 vote

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: ...
4 votes
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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 $n=12$. ...
Dietrich Burde's user avatar
1 vote

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 ...
Deif's user avatar
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4 votes
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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 ...
Adayah's user avatar
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3 votes
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Let $G$ be a group of order $p^nq$ where $p$ and $q$ are distinct primes and suppose $q \nmid p^i-1$ for $1 \leq i \leq n-1$. Prove $G$ is solvable

I'm afraid almost everything is incorrect You are correct that the number of $q$-Sylow subgroups must be $1$. However, your argument via the claim "but $\mathbb{Z}_q$ is abelian, hence normal, ...
Arturo Magidin's user avatar
2 votes

Inclusions of product of groups

Let me denote the cyclic group of order $k$ by $C_k$, written multiplicatively. Claim. Let $G_2=C_n\times C_n$, with the factors generated by $x$ and $y$, and let $m$ be a positive divisor of $n$. The ...
Arturo Magidin's user avatar
2 votes

Prove the binary icosahedral group is isomorphic to ${\rm SL}(2,5)$

Your strategy looks fine to me, let's try to make it work. First, working $\bmod \sqrt{5}$ we can invert $2$ and $\tau \equiv 3 \bmod \sqrt{5}$, so we get $\zeta \equiv -1 + i + 3j \bmod \sqrt{5}$. So ...
Qiaochu Yuan's user avatar
3 votes
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Conjugacy classes of normal subgroup in group

You need every element of $N$ to have centralizer not contained in $N$. There are examples in which $C_G(N)$ is contained in $N$. A computer search shows that the smallest such example has order $64$ ...
Derek Holt's user avatar
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