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Prove $S_4$ has only 1 subgroup of order 12

A subgroup of index two is normal, so it is the union of some conjugacy classes. The conjugacy classes of $S_4$ are of sizes $1,6,8,3,6$, and $1+3+8$ is the only way to sum $12$ with these numbers.
Charles Bukowski's user avatar
1 vote

What's an example of a set that's not a group?

A set and a group are different types of objects. A group $G$ is defined as a pair $G=(A,*)$, where $A$ is a set, and $*$ is an operator. So really, no set is a group–they are totally different things....
Jack's user avatar
  • 19
1 vote

How do projective representations map elements? Are they mulivalued?

I think it is the other way around. A representation is a morphism $G \rightarrow GL(V)$, and a projective representation is a morphism $G \rightarrow PGL(V)$, where $PGL(V)$ denotes the set (it is a ...
Plop's user avatar
  • 2,202
0 votes

A question on diagram of groups

Take $$\begin{array}{rcccccccl} 0&\longrightarrow &0&\longrightarrow&\mathbb{Z}&\stackrel{f}{\longrightarrow}&\mathbb{Z}\oplus\mathbb{Z}&\longrightarrow &\mathbb{Z}\\ {\...
Arturo Magidin's user avatar
3 votes

Typo? "If $G$ is a group, $H$ and $K$ are subgroups of $G$, and $K\unlhd G$. Is $H\cap K \unlhd H$ just if $K \subset H$?"

We have $H\cap K\trianglelefteq H$, with or without such inclusion. Let $h\in H$ and $x\in H\cap K$. Then $hxh^{-1}\in H$ as a product of three elements of $H$, and $hxh^{-1}\in K$ because $K\...
Mark's user avatar
  • 37.3k
4 votes

Test if a finite group is a symmetric group algorithmically

Combining the two answers gives something faster than the input size of the table. If we imagine an "oracle" type of problem, where we have functions that can invert, multiply, or sample ...
Sam Jaques's user avatar
  • 2,070
2 votes

Minimal normal, maximal and isomorphic

The issue I find with this other answer is not that it isn't elegant but that it uses the concept of composition factor as well as the Jordan-Hölder Theorem, both of which come next in the book. A ...
Leandro Caniglia's user avatar
13 votes
Accepted

Test if a finite group is a symmetric group algorithmically

The goal should be to do it in time $O(n!^2)$, which is the size of the input (the multiplication table). First determine the partition of $G$ into conjugacy classes. This can be done efficiently with ...
Sean Eberhard's user avatar
7 votes

Test if a finite group is a symmetric group algorithmically

By the Coxeter presentation of the symmetric group, a homomorphism $S_n \to G$ can be described equivalently as a sequence of elements $s_1,\dotsc,s_{n-1} \in G$ such that $$s_i^2 = 1,\, (s_i s_{i+1})^...
Martin Brandenburg's user avatar
1 vote

Order of the elements in a non-cyclic group of order 8

You're correct so far, except that you need to recognise, as said in the comments, that if $|a|\mid n$ for $a$ in a group $G$ and $n\in\Bbb N$, then $a^n=e.$
Shaun's user avatar
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0 votes

Mapping from set $A$ surjectively onto itself which is not injective.

Theorem. Let $A$ be a set. Consider the following statements: $A$ can be bijected with a proper subset of itself. There exists a function $f\colon A\to A$ that is one-to-one but not surjective. There ...
Arturo Magidin's user avatar
2 votes
Accepted

Mapping from set $A$ surjectively onto itself which is not injective.

I assume you mean, can we have a surjection $A\to A$ that is not injective. Then the answer is yes. By counting reasons, you need that $A$ is infinite, but then it is always possible. For a simple ...
SomeCallMeTim's user avatar
0 votes

Can a finite group have 2D and 3D faithful irreducible representations?

Proposition 1: Suppose that $ G $ has a faithful degree $ n $ (continuous) complex irrep whose image contains $ \zeta_n I $, where $ I $ is the identity matrix and $ \zeta_n $ is a primitive $ n $th ...
Ian Gershon Teixeira's user avatar
2 votes
Accepted

Rigid triangulations of regular $n$-gons

Yes for $n \ge 7$. Consider the triangulation that uses all the diagonals from one vertex. That is preserved only by a reflection. In the first quadrilateral on one side use the other diagonal. This ...
Ethan Bolker's user avatar
  • 91.6k
1 vote

Difference between the infinite product and infinite direct sum of cyclic groups.

$\bigoplus_{i>0}\mathbb Z_{i}\subseteq \prod_{i>0}\mathbb Z_i$ consists of tuples $(a_i)_{i>0}$ such that $a_i=0$ for $i\gg0$. In particular, the two groups are non-isomorphic because every ...
Kenta S's user avatar
  • 14.6k
4 votes

Is only one generator enough to find other generators?

Perhaps the statement you are missing is this, which is just a rewording of the statement you have quoted: For all $b \in G$, $b$ is a generator of $G$ if and only if there exists an integer $r$ ...
Lee Mosher's user avatar
  • 116k
3 votes

Is only one generator enough to find other generators?

No need to apologize, your question is perfectly suited for this forum and is an interesting topic in abstract algebra! Understanding the role of generators in a cyclic group is quite important. The ...
jack chen's user avatar
1 vote

Is only one generator enough to find other generators?

As Sean Eberhard has pointed out, the powers of a generator $a$ give you all the elements of the group. Thus the only thing left to prove is that if $(m, n) = d > 1$, then $b = a^m$ is not a ...
Eugene Kogan's user avatar
3 votes
Accepted

Are subgroup enumeration algorithms probabilistic?

Many of the most effective algorithms in group theory involve choosing random elements in the group. For example, it has long been the case that finding representatives of the conjugacy classes of ...
Derek Holt's user avatar
  • 87.8k
5 votes
Accepted

Normal subgroup of $S_9$

You've stated that you know that $H$ is normal in $HK$. Noting that $H,K$ are disjoint (i.e., $H\cap K=\{e\}$), if $K$ was normal in $HK$, it would follow that $hk=kh$ for all $h\in H,\,k\in K$. ...
quasi's user avatar
  • 57.9k
3 votes

Is this Monoid Finitely Generated?

The answer appears to be yes. $\textbf{Lemma:}$ Let $A$ be a non-empty subset of $\mathbb{N}^n$ equipped with the partial order $(a_1, \ldots ,a_n)\leq(b_1, \ldots, b_n)$ iff $a_i\leq b_i$ for all $i=...
Jonathan Hole's user avatar
4 votes

$a$ and $b$ are elements of the group $G$, then if $a \in \langle b \rangle$, then $\langle a\rangle \subseteq \langle b\rangle$

You really want to take this at a much more abstract level. If $S$ is a subset of a group $G$, then $\langle S\rangle$ has the following universal property: $\langle S\rangle$ is a subgroup; and $S\...
Arturo Magidin's user avatar
3 votes
Accepted

ord(g) = n $\cdot$ n' with gcd(n,n')=1 show that exist two elements such that g=h $\cdot$ h'=h' $\cdot$ h with ord(h)=n and ord(h')= n

You have $\operatorname{ord}(g)= nm$ with $\gcd (n,m)= 1$. Hence you have an isomorphism $$\begin{align} \phi: \langle g \rangle &\to \mathbb{Z}/nm\mathbb{Z} \\ g &\mapsto \bar{1} \end{align}$$...
julio_es_sui_glace's user avatar
0 votes

$a$ and $b$ are elements of the group $G$, then if $a \in \langle b \rangle$, then $\langle a\rangle \subseteq \langle b\rangle$

Note that $\langle x\rangle$ is all powers of $x$. If $a$ is a power of $b$, then all powers of $a$ are powers of $b$.
Shaun's user avatar
  • 43.6k
3 votes
Accepted

$a$ and $b$ are elements of the group $G$, then if $a \in \langle b \rangle$, then $\langle a\rangle \subseteq \langle b\rangle$

We have a group $(G,\cdot)$ Let's have a look at the definitions $$\langle b \rangle := \{g \in G : g = b^k, k\in \Bbb Z\}\\ \langle a \rangle := \{g \in G : g = a^h, h\in \Bbb Z\}$$ So we have that ...
Turquoise Tilt's user avatar
0 votes

Group of positive rationals under multiplication not isomorphic to group of rationals

Let $\pi_p$ be the projection onto prime $p$, i.e. if $r=\prod_{q~\mathrm{prime}} q^{\alpha_q}\in\mathbb Q$, then $\pi_p(r)=p^{\alpha_p}$. It is easily seen to be a endomorphism of $(\mathbb Q_{>0},...
Covariant's user avatar
  • 407
0 votes

Proving that the multiplicative group mod p (p is prime) is cyclic

Easy. Let $G=\mathbb{U}_p$ be this group. Each element in this group must have an order. So there must be a special element $a\in G$, such that $\text{ord}_G(a)$ is the maximum order. Let this maximum ...
MathMinded's user avatar
  • 1,314
0 votes

Set of Homomorphisms between two abelian groups is a group

Since there are already detailed explanations, I just add a short remark: Abelian groups are $\mathbb{Z}$-modules. For any $A,B\in \mathrm{obj}(_R \textbf{Mod})$, the set $\mathrm{Hom}_R(A,B)$ is an ...
Anthony's user avatar
  • 97
0 votes

Prove that the closure $\overline{H}$ of a subgroup $H \subseteq G$ of a topological group, is a subgroup.

I felt like something was missing, and as Alex Ortiz said, it was the lack of inverses. I'll post my own answer, but leave the question open just to verify my proof is correct. Part two of the proof ...
Daedalus's user avatar
0 votes

Prove that the closure $\overline{H}$ of a subgroup $H \subseteq G$ of a topological group, is a subgroup.

Don't forget that to justify $\overline H$ is a subgroup, you also need to address taking inverses of elements in $\overline H$. Here is perhaps a more natural way to write a proof using sequences. Of ...
Alex Ortiz's user avatar
  • 23.8k
0 votes

Every finite abelian group is the Galois group of some finite extension of the rationals

This is an assignment I have to do for my Galois Theory course. I reproduce the version I wrote a couple of days ago in the hope it may help other people. Let $G$ be a finite abelian group. Show that ...
Kadmos's user avatar
  • 1,266
1 vote

Maximal abelian subgroups of an extraspecial group of order $2^{2m+1}$

For the first question, this is a classical result, You know that for any $a \in U$, $\langle a \rangle \leqslant G$, so by Lagrange, you get $$|\langle a \rangle | = o(a) \mid 2^{2m+1} |G|$$ Hence, $...
julio_es_sui_glace's user avatar
1 vote

How can you show there are only 2 nonabelian groups of order 8?

Another approach: Let $G$ be a non-abelian group of order $8$. A group whith all its non-identity elements of order $2$ is abelian, so there exists some $x\in G$ of order $4$. Moreover, $|\textbf{Z}(G)...
Charles Bukowski's user avatar
0 votes
Accepted

Calculating the number of non-equivalent hands for a given suit configuration?

Lets write $[x_1, y_1][x_2, y_2]\ldots$ for number configuration with $y_1$ suits of $x_1$ cards, $y_2$ suits of $x_2$ cards and so on. We have $[x_1, y_1]\ldots = \prod_i [x_i, y_i]$ - suits with ...
mihaild's user avatar
  • 14.8k
3 votes

find the order of the element $a^{-1}b?$

The order of $a^2$ and $b^2$ are respectively $4$ and $5,$ which are coprime. The order of $(ab)^2$ is therefore $20,$ which is even. This proves that the order of $ab$ is exactly $40.$
Anne Bauval's user avatar
  • 29.3k
-1 votes
Accepted

find the order of the element $a^{-1}b?$

Hint: Note that $$\begin{align} (a^{-1}b)^{40}&=a^{-40}b^{40}\\ &=(a^8)^{-5}(b^{10})^4\\ &=e^{-5}e^4\\ &=e. \end{align}$$
Shaun's user avatar
  • 43.6k
0 votes

Looking for a simple proof that groups of order $2p$ are up to isomorphism $\Bbb{Z}_{2p}$ and $D_p$ for prime $p>2$.

By Cauchy's theorem, $G$ has an element $x$ of order $p$ and an element $y$ of order $2$. If $\text{o}(xy)=2p$ then the group is cyclic, and if $\text{o}(xy)=1,2$ then the group is dihedral because we ...
Charles Bukowski's user avatar
2 votes

Smallest $n,m$ such that $S_4\times\mathbb{Z}/5\mathbb{Z}$ and $A_5\times A_5$ is a subgroup of $S_n, S_m$

The issue is $|G|\nmid |S_8|$. The smallest group that $|G|$ divides is $S_{10}$, which then supports a natural embedding of the first $A_5$ acting on $\{1,2,3,4,5\}$ and the second $A_5$ acting on $\{...
Andrew Lys's user avatar
3 votes
Accepted

If two modules have a common Jordan-Hölder factor, is there a nonzero map between them?

There are such examples with no nonzero module homomorphisms between them. Here is an example - there may be easier ones. We let $M$ and $N$ be $KG$-modules of dimension $4$, with $K$ the field of ...
Derek Holt's user avatar
  • 87.8k
0 votes

How many cycles of length $k$ in $S_n$?

A cycle of length $k$ can be constructed by first choosing $k$ elements out of $n$ and this can be done in $C(n,k)$ ways. Further the chosen $k$ elements are to be arranged in cyclic (circular) order ...
Nitin Uniyal's user avatar
  • 7,882
2 votes
Accepted

Meaning of the terms "operation" and "invariant" in the old group theory paper

"Operation" here means "element"; groups used to be thought of as collections of "operations on sets" (even after Cayley; this is the language used in Burnside's book, ...
Arturo Magidin's user avatar
5 votes

What is wrong with my argument that every group of order $pq$ is abelian?

The subgroups $H$ and $K$ may not be normal, so you cannot take the quotients and conclude that they contain the commutator subgroup. For instance, when $G=S_3$ it has subgroups of order $2$ and $3$, ...
Eric Wofsey's user avatar
1 vote

Quotient group and classification of quotient groups $\mathbb{Z}^3/H$

I will try to use only elementary facts about groups and homomorphisms. At all stages where I assert that a map is a homomorphism, is onto, has some kernel that needs to be checked. So let's fix ...
ancient mathematician's user avatar
0 votes

Reducible representation of SO(2)

One can use the following equality true for all $\phi$: $$ \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} \begin{pmatrix} \cos \phi & - \sin \phi \\ \sin \phi & \cos \phi \end{pmatrix} \...
kabenyuk's user avatar
  • 9,253
0 votes

Reducible representation of SO(2)

Expanding on the comment above. To see that this representation is reducible over $\mathbb{C}$ you can just find a common eigenvector in $\mathbb{C}^2$ for all elements. This uses the following useful ...
Sam Ballas's user avatar
1 vote

For a normal subgroup $H$ of $G$. how to prove $XH=HX$ for any part of $G$ ? Is union distributive over multiplication?

If $X=\varnothing$, then $XH=\varnothing =HX$. Suppose $X\neq \varnothing$. Let $g\in XH$. Then $g=xh$ for some $x\in X, h\in H$. But $H\stackrel{(1)}{\unlhd}G$, so, in particular, $g\in xH\stackrel{(...
Shaun's user avatar
  • 43.6k
1 vote
Accepted

The notation $A\otimes_{\Bbb Z}G$ in Robinson's "A Course in the Theory of Groups (Second Edition)".

The tensor product is defined on p. 235 in the chapter on representations (8.4), despite being used earlier.
Steve Mitchell's user avatar
2 votes
Accepted

For a normal subgroup $H$ of $G$. how to prove $XH=HX$ for any part of $G$ ? Is union distributive over multiplication?

This is similar to your second attempt. Try proving $HX\subset XH$ and $XH\subset HX$. Hint: Let $a\in XH\implies a=xh$ for some $x\in X,~h\in H.$ Since $xH=Hx$, the element $xh\in xH\implies xh\in Hx\...
Sathvik's user avatar
  • 3,354
0 votes

Let a finite group $G$ have $n (>0)$ elements of order $p$ (a prime). If the Sylow $p$-subgroup of $G$ is normal, then does $p$ divide $n+1$?

As mentioned in the comments, a more general result holds: Let $G$ be a finite group, and $p$ a prime divisor of $|G|$. Then, there are $1\pmod p$ subgroups of order $p^k$, for every $k\in\{0,1,\dots,...
citadel's user avatar
  • 2,585
0 votes

What is the intuition for recognizing cyclic groups?

Adding to the previous answer, the intuition for this definition comes from finite cyclic groups. These are precisely the groups that consist of one cycle that can be iterated through by multiplying ...
Jack's user avatar
  • 61

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