Wikipedia:Reference desk/Archives/Mathematics/2012 August 27

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August 27

How do you show that two groups are not isomorphic

It's easy enough to show when two groups are isomorphic, (just find an isomorphism) but how do you show that two groups (which have the same cardinality) are not isomorphic? (Specific example: the cosets of $\{1,-1,i,-i\}$ and $\mathbb {C} ^{\times }$) Widener (talk) 02:04, 27 August 2012 (UTC)

I'm not sure what you mean by the notation $\mathbb {C} ^{\times }$, but in general, you show non-isomorphism between groups (or structures in general) by finding some property of the structures that's definable from the structure's non-logical symbols, and showing that one structure has the property and the other one doesn't. For example, for groups, as a trivial example, you might show that one group is abelian and the other is not, or one group has a finite subgroup (not counting the trivial one) and the other doesn't, or.... --Trovatore (talk) 02:27, 27 August 2012 (UTC)

(I should say, this doesn't always work. Sometimes two structures might be elementarily equivalent, but still not isomorphic. In that case, you have to think of something else — there's no general boilerplate argument as far as I know.) --Trovatore (talk) 02:40, 27 August 2012 (UTC)

There can be no boilerplate (at least for groups given in presentations) because the Group isomorphism problem is undecidable. Staecker (talk) 11:37, 28 August 2012 (UTC)

Thanks fur the help, but I still can't think of any way for my particular example. $\mathbb {C} ^{\times }$ means the group of complex numbers except zero endowed with multiplication. Widener (talk) 04:42, 27 August 2012 (UTC)

As $\mathbb {C} ^{\times }=(\mathbb {R} _{>0},\cdot )\times S^{1}$ for $S^{1}$ = unit circle, are you looking for a proof that $S^{1}$ is not isomorphic to the factor group $S^{1}/\{\pm 1,\pm i\}$? —Kusma (t·c) 08:47, 27 August 2012 (UTC)

(As these groups are isomorphic, that is going to be difficult). —Kusma (t·c) 08:56, 27 August 2012 (UTC)

They are isomorphic? What's the isomorphism? Widener (talk) 09:00, 27 August 2012 (UTC)

Think of it this way: $S^{1}$ is the unit circle. The factor group is the same circle, wrapped around four times. Work from there. --Trovatore (talk) 09:38, 27 August 2012 (UTC)

See Group cohomology. — Fly by Night (talk) 23:22, 30 August 2012 (UTC)

Isomorphic

Lets $A$ be the group of all matricies of the form ${\begin{bmatrix}a&0\\0&b\end{bmatrix}}$, $B$ the group of matrices of the form ${\begin{bmatrix}1&c\\0&1\end{bmatrix}}$ and $C$ the group of matrices of the form ${\begin{bmatrix}a&c\\0&b\end{bmatrix}}$. Show that $A\times B$ is not isomorphic to $C$. — Preceding unsigned comment added by Widener (talk • contribs) 08:59, 27 August 2012 (UTC)

I would ask you again to show how far you've gotten, and your reasoning processes so far. --Trovatore (talk) 09:03, 27 August 2012 (UTC)

These "show that X is not isomorphic to Y" questions are basically just guess and check as far as I can tell Widener (talk) 09:09, 27 August 2012 (UTC)

$\left({\begin{bmatrix}a&0\\0&b\end{bmatrix}},{\begin{bmatrix}1&c\\0&1\end{bmatrix}}\right)\left({\begin{bmatrix}d&0\\0&e\end{bmatrix}},{\begin{bmatrix}1&f\\0&1\end{bmatrix}}\right)=\left({\begin{bmatrix}ad&0\\0&be\end{bmatrix}},{\begin{bmatrix}1&c+f\\0&1\end{bmatrix}}\right)$

${\begin{bmatrix}g&h\\0&i\end{bmatrix}}{\begin{bmatrix}j&k\\0&l\end{bmatrix}}={\begin{bmatrix}gj&gk+hl\\0&il\end{bmatrix}}$

Widener (talk) 09:13, 27 August 2012 (UTC)

Exactly one of the two groups is abelian. —Kusma (t·c) 09:45, 27 August 2012 (UTC)

Cosets

If $H$ is not normal, show that there exist two left cosets of $H$ whose product is not a coset.

This is my reasoning that it cannot be a left coset: Since $H$ is not normal, there exists $b\in G$ such that $bH\neq Hc$ for any $c\in G$. Therefore, $aHbH\neq a(cH)H=acH$. But any left coset can be written in the form $acH$ so we know $aHbH$ cannot be a left coset if $H$ is not normal.

I cannot see an analogous proof to show that it cannot be a right coset though; I tried a different approach: starting with $aHbH=Hc$ and deriving that H is normal but that doesn't seem to work either. Widener (talk) 09:29, 27 August 2012 (UTC)

I don't see why your second inequality follows from the first. I think you're taking the wrong approach here; you've found the right $b$, but you haven't taken any care in choosing which $c$ to use. You should choose a particular $c$.

Remember that the definition of normality isn't just that $bH=Hc$ for some $c$, but rather that $bH=Hb$. So since $H$ isn't normal, we know that $bH\neq Hb$. While this is weaker than what you said earlier, the specificity can help guide your choice of cosets.

For showing that the product isn't itself a coset, your argument should take the following form: "If it were any coset (left or right), it would need to be ... But it can't be that, because ..."--121.73.35.181 (talk) 11:14, 27 August 2012 (UTC)