# Tag Info

### $G$ 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 $x\notin Z(G)$, ...
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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 ...
• 3,128
Accepted

### 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 ...
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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 $n$ ...
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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 ...
• 3,128
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). ...
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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 ...
• 298

### 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 ...
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• 3,128
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, ...
• 3,733
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?
• 9,727
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 ...
• 18.5k
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 $n_x$ ...
• 11.7k

### 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&...
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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 ...
• 8,546
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 ...
• 8,546
1 vote

• 217k