10 votes
Accepted

If $m\mid n$ can we find a group $G$ of order $n$ with a subgroup $H$ of order $m$, i.e. can every divisibility be proved group theoretically?

More precisely, if $m\mid n$, can we always find a group $G$ of order $n$ and a subgroup $H$ of $G$ of order $m$? Yes. For each $n$, there exists a cyclic group of order $n$. In the cyclic group of ...
Shaun's user avatar
  • 45.1k
8 votes
Accepted

Center of group of order 3773

Let $P$ be a $7$-Sylow subgroup, $Q$ is an $11$-Sylow subgroup. Then $P,Q$ are normal in $G$, $P\cap Q=\{e\}$ and by comparing cardinalities, $G=PQ$. It is well known that in this case we have: $G\...
Mark's user avatar
  • 40.1k
6 votes
Accepted

Does isomorphism between group algebras imply equivalence between character tables?

Over an algebraically closed field in characteristic zero (think $\mathbb{C}$ if you like), isomorphism of group algebras depends solely on the degrees of the irreducible representations. This means ...
ahulpke's user avatar
  • 18.5k
6 votes
Accepted

Automorphisms of group ring

No, let $n = 2$ and let $C_2 = \langle g \rangle$. Then consider $\Phi: R \to R, a + bg \mapsto a - bg$. This is a ring automorphism but it is not induced by any automorphism of $C_2$.
psl2Z's user avatar
  • 2,558
6 votes
Accepted

A group isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3$

The isomorphisms are given by: $$\begin{align} \varphi: G\times (H\times K)&\to G\times H\times K,\\ (g, (h,k))&\mapsto (g,h,k);\\ \psi:G\times H\times K&\to (G\times H)\times K,\\ (g,h,k)&...
Shaun's user avatar
  • 45.1k
4 votes
Accepted

Show that a transitive subgroup of symmetric group is primitive

Here's a slightly different approach: we already know $\Gamma$ and $\Delta$ are disjoint. Since $G$ is transitive, all blocks would have the same size; assume we have blocks larger than singletons, ...
Steve D's user avatar
  • 3,630
4 votes

Show that a transitive subgroup of symmetric group is primitive

There may be easier ways of doing this; I'm just "following my nose", as I don't work much with permutation groups on a regular basis. $G_{(\Gamma)}=\{g\in G\mid \gamma^g=\gamma\text{ for ...
Arturo Magidin's user avatar
4 votes

Schur's Multiplier exercise (Problem 5A.7 Isaacs' Finite Group Theory)

Let $\Gamma$ be a Schur representation group for $G$. So $\Gamma$ has a subgroup $Z \cong M(G)$ with $Z \le Z(\Gamma) \cap \Gamma'$ and $\Gamma/Z \cong G$. Let $\bar{C}$ and $\bar{B}$ be the inverse ...
Derek Holt's user avatar
  • 90.1k
4 votes
Accepted

Need help with proof of property related to group rings.

For notational convenience I will write $$ e_i = \sum_{r=0}^{n-1} (\omega^ig)^r $$ So that "my" $\{e_0,\ldots,e_{n-1}\}$ is "your" $\{e_1,\ldots,e_n\}$. We can compute $e_ie_j$ by ...
Steve D's user avatar
  • 3,630
4 votes

Need help with proof of property related to group rings.

Firstly, I think $e_j$ should be $e_j=\dfrac{1}{n}\sum_{k=1}^n{\left( w^{j-1}g \right) ^k}$. Otherwise, $e_1=e_n$. The first part : $e_ie_i=\delta_{ij}e_i$ Let we calculate $e_ie_j.$ By the definition,...
fusheng's user avatar
  • 1,135
4 votes

Why the polynomial $x^{p+1}=y$ has exactly $p+1$ solutions in $F_{p^2}^*$ for $y \in F_p^*$?

Define $\phi: F_q^\times \to F_q^\times$ by $\phi(x)=x^{p+1}$. This is a group homomorphism. Then $\phi(x)^{p-1}=x^{(p+1)(p-1)}=x^{q-1}=1$ and so $\operatorname{im}{\phi} \subseteq \{ x \in F_q^\times ...
lhf's user avatar
  • 217k
4 votes
Accepted

Maximal subgroups of $SL(n, 2)$

You cannot hope for a general classification of maximal subgroups of this type. For that you would essentially need to find all absolutely irreducible representations over ${\mathbb F}_q$ (with $q=2$ ...
Derek Holt's user avatar
  • 90.1k
4 votes

Center of group of order 3773

If $G$ is abelian, then you are done. So, let's suppose $G$ nonabelian. From $n_{11}=1$ follows that the Sylow $11$-subgroup is normal and, as such, it is union of conjugacy classes. Now, the ...
citadel's user avatar
  • 3,035
3 votes
Accepted

GAP orthogonal groups: Specifying the invariant bilinear form

This is a feature of the forms package. The problem is that the matrix you give is not equivalent over $GF(7)$ to the matrix used internally (but its negative is). ...
ahulpke's user avatar
  • 18.5k
3 votes
Accepted

Lifting map from finite cyclic group to integers

For a free group $G= \langle X \vert\,\, \rangle$, any map (of sets) $X\to \mathbb{Z}$ determines a homomorphism $G\to \mathbb{Z}$. Thus for each $x\in X$ we just need to map $x$ to any integer $n_x$ ...
tkf's user avatar
  • 11.5k
3 votes
Accepted

Prove the index of a proper subgroup of a simple group of order 17971200 is at least 14.

Let $G$ be a nontrivial finite group and $X_G$ the set of all the proper subgroups of $G$, $X_G:=\{H\le G\mid H\ne G\}$. Lagrange's theorem already puts a limitation on the order of the elements of $...
citadel's user avatar
  • 3,035
3 votes

Is $S_5$ generated by (1 3) and (1 2 3 4 5)?

Start by recalling that, for $\sigma,\tau\in S_n$, $$\tau=(a_1,a_2,...,a_k) \implies \sigma\tau\sigma^{-1} = (\sigma(a_1),\sigma(a_2),...,\sigma(a_k))$$ For example, setting $\sigma=(1\ 2\ 3\ 4\ 5)$ ...
Tom's user avatar
  • 480
3 votes
Accepted

Why the polynomial $x^{p+1}=y$ has exactly $p+1$ solutions in $F_{p^2}^*$ for $y \in F_p^*$?

The multiplicative group of $F_{p^2}-\{0\}$ is cyclic (of order $p^2-1$, of course), which is a basic fact in field theory. Moreover, under the inclusion $\tau: F_{p}\to F_{p^2}$, the members of $\tau(...
Asigan's user avatar
  • 1,639
2 votes

Let $G$ be a finite abelian group of order n. How many distinct group homomorphisms are there from $G$ to $R/Z$?

The question does indeed turn out to be well-defined, but this is not obvious and the "order $n$" part makes it a little confusing anyway as it's much easier to answer for all $n$ at once. ...
hunter's user avatar
  • 30.1k
2 votes

Let $G$ be a finite abelian group of order n. How many distinct group homomorphisms are there from $G$ to $R/Z$?

This gets you into group cohomology. For any subgroup $A\le G$ such that $G/A$ is cyclic, it's the number of central extensions $$0\to A\to G\to G/A\to 0,$$ by the first isomorphism theorem. These ...
calc ll's user avatar
  • 8,512
2 votes
Accepted

What is the order of the following subgroup $\langle (1 \ 2 \ \cdots \ n), (a \ b)\rangle$ of $S_n$?

$\require{\mathtools}$ Here is an answer based on Serre's exercise mentioned in the comments. I'll write $c=(1,2,\ldots,n)$ and $t=(a,b)$. Note that by relabeling, we are safe to assume $a=1$. Also, ...
Steve D's user avatar
  • 3,630
2 votes

Is every element of a finite abelian group contained in a cyclic factor?

No, this is not true. Consider $A=C_8\times C_2$ and $a=(2,1)$. Then $a$ generates a cyclic subgroup of order $4$. It is not itself a direct factor, because then the other factor would be of order $4$,...
tomasz's user avatar
  • 35.5k
2 votes
Accepted

Chinese remainder theorem for finite rings.

Observe that $I^nJ^m \supseteq (I \cap J)(I^n J^m) \supseteq (IJ)I^nJ^m = I^nJ^m$. Do you see why $I^nJ^m$ is finitely generated? Can you conclude the proof using Nakayama's lemma?
Haran's user avatar
  • 9,717
2 votes
Accepted

Number of conjugacy classes in each coset of a semidirect product is the same.

I do not believe that this is true. Take $G:=\langle a,b,c \mid a^2=b^2=c^2=abc=1 \rangle$, the elementary group of order $4$. Let $\sigma$ of order $6$ act on $G$ via $\sigma:a\mapsto b\mapsto c\...
ancient mathematician's user avatar
1 vote
Accepted

$g$ lies in the centre of $G$ iff $\vert\chi(g)\vert=\vert\chi(1)\vert$ for all irreducible characters $\chi$

Let $\rho$ be the $n$-dimensional irreducible representation with character $\chi$. If $|\chi(g)|=|\chi(1)|=n$, since $\rho(g)^m=\rho(g^m)=\rho(1)=1$ for some $m$, the eigenvalues of $\rho(g)$ are $n$ ...
Just a user's user avatar
  • 15.1k
1 vote

Given a tuple of $k$ distinct integers, is there a generator list in a $\mathbb{Z}/n\mathbb{Z}$ that matches the tuple?

Basically, you want geometric sequence modulo some $n$, so $q_i^2 \equiv q_{i-1}q_{i+1}\pmod n$, i.e. $n\mid q_i^2-q_{i-1}q_{i+1}$, so to find counterexample you can take any triple $(q_1,q_2,q_3)$ ...
Ennar's user avatar
  • 23.1k
1 vote

How many homomorphisms are there from $D_5$ to $V_4$?

The commutator of $D_5$ is $\langle r\rangle \cong \Bbb Z_5.$ As a result it's in the kernel (there's two ways to see that: $V_4$ has order $4,$ and it's abelian) , and the homomorphisms have every ...
calc ll's user avatar
  • 8,512
1 vote

How many homomorphisms are there from $D_5$ to $V_4$?

Since $2\nmid 5$, all the four elements of order $5$ must be sent to $1$. As for the five elements of order $2$, $\varphi(sr^k)=$ $\varphi(s)\varphi(r^k)=$ $\varphi(s)$. Therefore, for $V_4=\{1,a,b,ab\...
citadel's user avatar
  • 3,035
1 vote

Group of order $p^{\alpha}q$ is not simple.

Since the Sylow $p$-subgroups are all pairwise conjugate (Sylow II), their intersection is the normal core in $G$ of any of them. So, once you know this intersection is not trivial, you get a ...
citadel's user avatar
  • 3,035

Only top scored, non community-wiki answers of a minimum length are eligible