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$G$ non abelian, order $p^3$ ($p$ prime). Suppose that the center is $p^2$, prove that $\exists\ x$ outside of the center, of order p

If $|Z(G)|=p^2$, then there isn't room (in terms of order) between $Z(G)$ and the whole $G$ to accomodate the centralizers of the noncentral elements of $G$. In fact, recall that, for $x\notin Z(G)$, ...
Kan't's user avatar
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2 votes

Classify, up to isomorphism, all groups of order $5\cdot 7 \cdot 19$.

This is a particular case of the general setting: $|G|=pqr$, $p<q<r$, $p\nmid q-1$, $p\nmid r-1$ and $q\nmid r-1$. By Sylow III, $n_p\mid qr$ and $n_p\equiv 1\pmod p$. So, $n_p=1,q,r,qr$ and $...
Kan't's user avatar
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1 vote
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$G = \langle C_G(a) \mid a \in Q \setminus \{1\} \rangle.$ Proof by induction on $|G|$

Since you have proved that $H=[x, G]$ is strictly smaller than $G$, you can now use your induction hypothesis to conclude that $H= \langle C_H(a) \mid a \in Q \setminus \{ 1 \} \rangle$. That is the ...
Robert Shore's user avatar
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2 votes

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

Here is a nice fact to know: for any irreducible character $\chi$ and any $g,h \in G$ we have $$\chi(g)\chi(h) = \frac{\chi(1)}{|G|} \sum_{z \in G} \chi(gzhz^{-1}).$$ I'll leave the proof as an ...
Daniel Arreola's user avatar
0 votes

Find the number of subgroups of $(\mathbb{Z}_{7}^{*},*)$

It's well-known that $\Bbb Z_p^*=\Bbb Z_p^×$ is cyclic of order $p-1.$ So we have a cyclic group of order $6.$ Cyclic groups have a unique subgroup of each order dividing the order of the group. Thus ...
calc ll's user avatar
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2 votes
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Do all the conjugates of $H\le G$ pairwise intersect in subgroups of the same order?

For the general case, let $G = \mathbb{Z}/2\mathbb{Z} \wr \mathbb{Z}/4\mathbb{Z} = (\mathbb{Z}/2\mathbb{Z})^{\{1, 2, 3, 4\}} \rtimes \mathbb{Z}/4\mathbb{Z}$. Then $H_1 = (\mathbb{Z}/2\mathbb{Z})^{\{1, ...
David Gao's user avatar
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On the Group of order $pq$ where $p , q $ are primes .

Since $q>p$, then $n_q=1$ by a counting argument (and trivially $1\mid p$). Therefore, since the nontrivial elements of $G$ have order $p$ or $q$, there are $pq-1-(q-1)=$ $q(p-1)$ elements of order ...
Kan't's user avatar
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6 votes
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When does every element of Sylow $p$-subgroup is also a member of another Sylow $p$-subgroup?

Here's an example of order $180$. The idea is to find a group with a Sylow subgroup $P$, such that for every element $x\in P$, we have \begin{equation*} C_G(x)\not\le N_G(P) \end{equation*} This is ...
Steve D's user avatar
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2 votes

$p$-group with a cyclic subgroup

Yes this is true. Let $K\lhd G$ be the preimage of $\langle a\rangle$ in $G$, so that $[G:K]=p$. I claim that $K$ is cyclic. To see this, let's suppose it's not and reach a contradiction. Let $\langle ...
Steve D's user avatar
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3 votes
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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 $\sigma:a\mapsto b\mapsto c\...
ancient mathematician's user avatar
3 votes
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$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
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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 ...
Kan't's user avatar
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3 votes
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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
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How to find generator in ellipse curve?

Well, your algorithm for finite field multiplicative groups works for other finite groups since it doesn't use anything finite field specific and relies only on Lagrange theorem. You have the group ...
honzaik's user avatar
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A group of order $195$ has an element of order $5$ in its center

The $5$-Sylow $P_5$ is normal, and the conjugacy classes of $G$ can have sizes $1$, $3$, $5$, $13$, and their products, only. $P_5$ cannot split into $5=1+1+3$, as $|P_5\cap Z(G)|=2$ is prevented by ...
Kan't's user avatar
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0 votes

Realization of the metacyclic group of order 21

To explicitly build up a (nontrivial) semidirect product $\mathbb Z_3\stackrel{\varphi}{\ltimes}\mathbb Z_{7}$, we must send $0$ to the identity map of $\mathbb Z_{7}$, say $Id_{\mathbb Z_{7}}$, and $...
Kan't's user avatar
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0 votes

Classifying all groups of order $39$

To explicitly build up a (nontrivial) semidirect product $\mathbb Z_3\stackrel{\varphi}{\ltimes}\mathbb Z_{13}$, we must send $0$ to the identity map of $\mathbb Z_{13}$ ($Id_{\mathbb Z_{13}}$), and $...
Kan't's user avatar
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8 votes
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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
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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 ...
Kan't's user avatar
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3 votes
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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 $...
Kan't's user avatar
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2 votes
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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
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2 votes
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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
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6 votes
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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
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3 votes
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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
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0 votes

Lifting map from finite cyclic group to integers

As I was saying in my comment, the functor $P:G \mapsto G/D(G)$ from groups to abelian groups where $D(G)$ is the subgroup generated by commutators is the left adjoint of the "inclusion functor&...
julio_es_sui_glace's user avatar
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Finding all groups $H$ (up to isomorphism) such that there is a surjective homomorphism from $D_8$ to $H$

If the homomorphism were injective, we'd have an isomorphism. That leaves homomorphisms into groups of order $1,2,$ or $4;$ in particular abelian ones. Group cohomology, in particular the first ...
calc ll's user avatar
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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
  • 8,546
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\...
Kan't's user avatar
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5 votes
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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
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0 votes

Finding all groups $H$ (up to isomorphism) such that there is a surjective homomorphism from $D_8$ to $H$

If $D_8/H$ were cyclic of order $4$, then there would exist a coset $gH$ of order $4$, namely an element $g\in D_8$ such that $g,g^2,g^3\notin H$ and $g^4\in H$. So, there are two cases only to parse: ...
Kan't's user avatar
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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
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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
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3 votes
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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
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6 votes
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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
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4 votes
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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
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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
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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
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0 votes

Find all Sylow 3-subgroups of $S_3\times S_3$

The elements of order $3$ in $S_3\times S_3$ are: \begin{alignat}{1} &((123),()) \\ &((132),()) \\ &((),(123)) \\ &((),(132)) \\ &((123),(123)) \\ &((123),(132)) \\ &((132),...
Kan't's user avatar
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6 votes
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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
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