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I'm thinking to the famous problem of cancellation property in Grp, i.e: $$G_1 \times G_2 \cong G_1 \times G_3 \Rightarrow G_2 \cong G_3. $$ Clearly there are many counterexamples like $\prod_{i \in \omega}\mathbb{Z}_i$ or $ \oplus_{i \in \omega}\mathbb{Z}_i$ but these counterexamples can be bypassed by giving a definition.

We say that a group G is $\Pi$-compact iff $$G \cong \prod_{i\in I}G_i, \ G_i \neq \{e\} \ \Rightarrow |I| < \infty.$$

We say that a group G is $\Sigma$-compact iff $$G \cong \oplus_{i\in I}G_i, \ G_i \neq \{e\} \ \Rightarrow |I| < \infty.$$

We say that a group is $\times$-compact iff it's $\Pi$ and $\Sigma$ compact.

I've been working on many conjectures and with Seirios' help many of them have been solved. I'll tick ($\checkmark$) proved ones and refuse ($\neg$) false ones.

$$\checkmark? \ \ \ \ \ \ \ \ G_1,G_2 \times\text{-compact} \Rightarrow G_1 \times G_2 \times\text{-compact} $$ $$\checkmark \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G \text{ finitely generated} \Rightarrow G \times\text{-compact} $$ $$\neg \ \ \ \ \ \ \ \ \ \ \ \ \ \ H <G, G \times\text{-compact} \Rightarrow H \times\text{-compact} $$

$$\neg \ \ H \triangleleft G, \ \ \ \ H,G \times\text{-compact} \Rightarrow G/H \times\text{-compact} $$ $$\neg \ \ \ \ \ G \times\text{-compact} \Rightarrow \text{cancellation property holds}.$$

What's clearly true is that: finite groups are $\times$-compact (and I've seen on the web that cancellation property holds for them), simple groups are $\times$-compact, free groups are $\times$-compact. Applying $(3) \wedge (4)$ on free groups we may get (2) but $(3) \wedge (4)$ have to be false because any group is quotient of a free group. $$ \neg ( (3) \wedge (4)).$$ As Seirios has observed for countable groups it holds that $\times$-compact $\Leftrightarrow$ $\Sigma$-compact. Again Seirios noted, here is proved that cancellation is not true for finitely presented groups so if (2) is true (5) is false.

$$ (2) \Rightarrow \neg (5) $$

Seirios proved (2) here.

A sketch of proof for (1). Let's assume that $G_1 \times G_2$ is not $\times$-compact. So $G_1 \times G_2 \cong \prod_{i\in \nu}P_i$. Let's call $\pi_1$ projection on first coordinate. So $G_1\cong \pi_1 (G_1 \times G_2) \cong \pi_1(\prod P_i) \cong \prod (\pi_1 P_i)$ so finitely many $P_i$ are not in $\{0\} \times G_2$. Same argument on the other side rise to absurd $\square.$ Is this correct?

I have a counterexample for (4). Let's consider $G:=*_{i \in \omega}\mathbb{Z}_i$ Since it's free it's $\times$-compact. Its commutator [G,G] is $\times$-compact but quotient $$G/[G,G] \cong \oplus_{i \in \omega} \mathbb{Z} $$ which is not $\times$-compact. $$ \neg (4) .$$

Since this $$(2) \Rightarrow \neg (3).$$


News: None.

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    $\begingroup$ What do you mean by "claim" here? Do you mean "conjecture"? Usually when a mathematician uses the word "claim" it means they have at least the outline of a proof in mind. $\endgroup$ – Qiaochu Yuan Nov 28 '14 at 22:24
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    $\begingroup$ Also, relevant: groupprops.subwiki.org/wiki/Krull-Remak-Schmidt_theorem $\endgroup$ – Qiaochu Yuan Nov 28 '14 at 22:26
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    $\begingroup$ Non-cancellation phenomena can be found with finitely-presented groups. See for example here: artofproblemsolving.com/Forum/viewtopic.php?f=61&t=351637 $\endgroup$ – Seirios Nov 28 '14 at 23:04
  • $\begingroup$ Thanks @Seirios! I'm reading counterexample, I'll answer soon. @Qiaochu, I just mean I've no counterexample. $\endgroup$ – Ivan Di Liberti Nov 28 '14 at 23:08
  • $\begingroup$ How do you read $\times$-compact? "product compact"? Did you invent this definition? $\endgroup$ – Rudy the Reindeer Nov 28 '14 at 23:39
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Claim 1: Any countable group is $\Pi$-compact.

If $\{G_i : i \in I \}$ is an infinite collection of non trivial groups, you can find an injective map $$\{0,1\}^{\mathbb{N}} \hookrightarrow \prod\limits_{i \in I} G_i,$$ so the product $\prod\limits_{i \in I} G_i$ has cardinality at least $2^{\aleph_0}$. This proves that any countable group is $\Pi$-compact. However, clearly cancellation property does not hold for countable groups: $$\mathbb{Z} \times \bigoplus\limits_{i \geq 0} \mathbb{Z} \simeq \bigoplus\limits_{i \geq 0} \mathbb{Z} \simeq \{1\} \times \bigoplus\limits_{i \geq 0} \mathbb{Z}.$$

Claim 2: Any finitely-generated group is $\Sigma$-compact.

Let $\{G_i : i \in I \}$ be an infinite collection of non trivial groups. Suppose by contradiction that $\bigoplus\limits_{i \in I} G_i$ is finitely-generated. Let $\{s_1,\ldots,s_r\}$ be a finite generating set. Now, for each $s_k$ there exists $I_k \subset I$ such that $I \backslash I_k$ is finite and $(s_k)_i=0$ for all $i \in I_k$. Therefore, if $g \in \bigoplus\limits_{i \in I} G_i$, we can write $g$ as a product of $s_k$'s and we conclude that $g_i=0$ for all $i \in J:= \bigcap\limits_{k =1}^n I_k$: it is a contradiction since $J$ is necessarily infinite.

Conclusion: Any finitely-generated group is $\times$-compact.

However, as I mentionned before, cancellation property does not hold even for finitely-presented groups (see here).

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  • $\begingroup$ Thanks a lot for this comment. I'll edit definition ASAP to elude this trick. $\endgroup$ – Ivan Di Liberti Nov 29 '14 at 9:14
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    $\begingroup$ @Ivan: I edited my answer to prove that finitely-generated groups are $\times$-compact. $\endgroup$ – Seirios Nov 29 '14 at 10:31
  • $\begingroup$ Thanks a lot. Do you think I should leave other conjectures of just eliminate 'em? $\endgroup$ – Ivan Di Liberti Nov 29 '14 at 10:45
  • $\begingroup$ It seems you answered all your questions, no? $\endgroup$ – Seirios Nov 29 '14 at 14:39
  • $\begingroup$ Almost, but this comment was before counterexample to 3 and 4. $\endgroup$ – Ivan Di Liberti Nov 29 '14 at 15:55

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