# Tag Info

### 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.
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....
• 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 ...
• 2,202

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 $a^n=e.$
• 43.6k

### 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 ...
• 388k
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 ...
• 2,536

### 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 ...
• 7,124
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 ...
• 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 ...
• 14.6k

### 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 $r$ ...
• 116k

### 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 ...
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 ...
• 111
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 ...
• 87.8k
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 $h\in H,\,k\in K$. ...
• 57.9k

• 388k
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}...
• 2,925

### $a$ 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 $b$.
• 43.6k
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 ...

• 2,925
1 vote

• 109
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 ...
• 87.8k

### 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 ...
• 7,882
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, ...
• 388k

### 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 $3$, ...
• 326k
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 ...

### 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} \...
• 9,253

### 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 ...
• 666
1 vote

• 3,354
### 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,...