1
$\begingroup$

Let $G_1,G_2$ be two groups.We say that $G_1,G_2$ are similar iff for every integers $a_1,a_2,...,a_n\in \{1,-1\}$ and every function $f:\{1,...,n\}\rightarrow\{1,...,n\}$ we have the following:

$$\forall x_1,x_2,...,x_n\in G_1[x_{f(1)}^{a_1}x_{f(2)}^{a_2}...x_{f(n)}^{a_n}=e_1]$$ $$\iff \forall x_1,x_2,...,x_n\in G_2[x_{f(1)}^{a_1}x_{f(2)}^{a_2}...x_{f(n)}^{a_n}=e_2] $$

$e_1,e_2$ are the identities of $G_1,G_2$ respectively. Informally: The above definition says that for example if $G_1$ satisfies $x^2y^2zx^{-2}y^{-2}=e_1$ for every $x,y,z\in G_1$ and $G_2$ is similar to $G_1$, then $G_2$ satisfies $x^2y^2zx^{-2}y^{-2}=e_2$ for every $x,y,z\in G_2$ as well.

The additive groups $\mathbb{Z},\mathbb{Q}$ are two similar non-isomorphic groups.

Question: Are there two finite similar non-isomorphic groups ?


Robert Israel gave an answer for my above question. However, I would like to see an answer to the stronger question:

Are there two finite similar non-isomorphic groups of equal cardinality ?

In fact, this is the reason why I chose $\mathbb{Z},\mathbb{Q}$ and not $\mathbb{Q},\mathbb{R}$ in my counterexample of the infinite case.

$\endgroup$
  • $\begingroup$ If I'm reading your question right, doesn't this imply the groups would have the same presentations? $\endgroup$ – Alexander Gruber Apr 29 '14 at 15:27
  • $\begingroup$ @AlexanderGruber I don't know what the defintion of presentation is, I will check it now $\endgroup$ – Amr Apr 29 '14 at 15:29
  • $\begingroup$ "Presentation" is the wrong word. "Laws" is the right word. Different groups can have the same laws. $\endgroup$ – Jack Schmidt Apr 29 '14 at 15:35
  • $\begingroup$ A presentation gives equations that are valid for a certain set of generators, not necessarily for all members of the group. $\endgroup$ – Robert Israel Apr 29 '14 at 15:37
  • $\begingroup$ Agree. I just noted that presentation is not what I want $\endgroup$ – Amr Apr 29 '14 at 15:37
1
$\begingroup$

Consider ${\mathbb Z}_2 \times {\mathbb Z}_2$ and ${\mathbb Z}_2$. In both cases the only possible equations are where $ \{i: f(i)=j\}$ has even cardinality for all $j$.

$\endgroup$
  • $\begingroup$ Thanks. Is there an example where $G_1,G_2$ have the same cardinality? $\endgroup$ – Amr Apr 29 '14 at 15:37
  • 1
    $\begingroup$ One should be able to use Z2 x Z2 x Z4 and Z4 x Z4, I believe, though I don't have Neumann's book handy. I believe the variety generated by a finite abelian group of exponent e is the variety of abelian groups of exponent e. $\endgroup$ – Jack Schmidt Apr 29 '14 at 16:05
  • $\begingroup$ @JackSchmidt I think your sugesstion will work, I will try it. $\endgroup$ – Amr Apr 29 '14 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.