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29

Take $G=\mathbb{Z}/4\times \mathbb{Z}/4$, and $H=Q_8\times \mathbb{Z}/2$ of order $16$, where $Q_8$ denotes the quaternion group. Both groups have exactly $1$ element of order $1$, $3$ elements of order $2$ and $12$ elements of order $4$. Edit: I understood the question as follows: Is there a counterexample where two groups $G$ and $H$ have the same number ...


18

Here is an example with two groups of order $27$. Consider the group $G$ which is elementary abelian (all elements $x \in G$ satisfy $x^{3} = 1$), of order $27$. And then $H$ the non-abelian group of order $27$ and exponent $3$ (once more, $x^{3} = 1$ for all $x \in H$). Concretely, $$ \left\{\, \begin{bmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & ...


10

Let $p$ be an odd prime. Let $G$ be the non-abelian group of matrices of the form $$\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}\in\operatorname{GL}(3,\mathbb F_p).$$ Then $|G|=p^3$ and each element $g\ne 1$ has order $p$; this follows from the fact that $g-1$ is nilpotent, hence $(g-1)^p=(g-1)^3=0$ and finally ...


5

In practice, you should be able to tell that a table is a group table by matching it with a group that you already know. Assuming that you don't look at tables with more than six elements, here are the possible groups: Cyclic groups: $C_2, C_4,\dots,C_6$ The Klein four group $V_4$ The symmetric group $S_3$. So if you know what the multiplication tables ...


3

My original proof is broken. I'm trying to fix it, but in the meantime here's a proof that works for solvable groups (hence all groups of odd order). Let $G$ be solvable and let $$|G|=p_1^{a_1}p_2^{a_2}\cdots p_n^{a_n}$$ where $p_1<p_2<\cdots<p_n$ are distinct primes and $a_i\geq 1$ for all $i$. Let $N_p$ denote the number of Sylow $p$-subgroups ...


2

One easy way to check for associativity is to see if we can regard $\{a,b,c,d\}$ as functions on some set, and regard our "multiplication" as functional composition, which you should already know is associative. We already have a set at hand, $\{a,b,c,d\}$ itself (it may seems strange to use the same letters for functions and their arguments, so I will call ...


2

Like you say, in order for a multiplication table for an operation to define a group structure, its rows and columns must each contain each element exactly once. (This is a consequence of divisibility: In a group $(G, \ast)$, if $a \ast b = a \ast c$ then $b = c$.) This property of an operation $\star: X \times X \to X$, called the Latin square property, ...


2

For any finite subset $G\subset \mathbb N$ and map $\circ\colon G\times G\to \mathbb N$, we can find a polynomial $f\in\mathbb Q[X,Y]$ such that $f(a,b)=a\circ b$ for all $a,b\in G$.


2

"Transitive" means that every point in $\{1$,...,n$\}$ is mapped to every other point by the action. Equivalently, the action is transitive if the orbit of any point $x$ under the action is the whole set. In this case the orbit of any $x \in \{1,2,...n\}$ is $S_{n}(x) = \{1,...,n\}$ (consider acting by permutations $(x1),(x2),...,(xn)$), so the action is ...


2

If you can trust my sage calculations, the circulant on 15 vertices with connection set $\{\pm3,\pm5\}$ is not isomorphic to any circulant with connection set $\{\pm1,\pm i\}$ for $i=2,\ldots,7$. (I believe your assumption may be true for circulants of prime power order.)


1

For your first question, a unary relation on $G$ is just a subset of $G$. So given $g\in G$, the singleton subset $\{g\}\subseteq G$ is a relation. Formally, we can add a new relation symbol $R$ to the language and interpret it as $R^G = \{g\}$, so that $G\models R(a)$ if and only if $a = g$. Then to say that the relation is preserved by all automorphisms ...


1

No, you don't need the maps to be surjective. The category of groups is complete (all limits exist) so you don't need any requirements on the morphisms here. You can easily just construct the fibre product by taking the product of the sources $G\times H$ and then the equalizer of $f \circ \pi_1$ and $g \circ \pi_2$.


1

If we don't have the surjectivity we will loose Goursat's lemma. Since a fiber product will no longer be always a subdirect product.


1

I only know a proof without Lagrange (or Lagrange in disguise) for an abelian group. Suppose that $G=\{ a_1,\ldots ,a_n\}$, and set $g:=a_1a_2\cdots a_n$. Then for every $x\in G$ the map $a_i\mapsto xa_i$ is a permutation of $G$, so that $$ g=(xa_1)(xa_2)\cdots (xa_n)=x^na_1a_2\cdots a_n=x^ng. $$ This implies $x^{\mid G\mid}=x^n=e$.



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