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.

  • 1
    $\begingroup$ 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. $\endgroup$ Nov 28, 2014 at 22:24
  • 1
    $\begingroup$ Also, relevant: groupprops.subwiki.org/wiki/Krull-Remak-Schmidt_theorem $\endgroup$ Nov 28, 2014 at 22:26
  • 2
    $\begingroup$ 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, 2014 at 23:04
  • $\begingroup$ Thanks @Seirios! I'm reading counterexample, I'll answer soon. @Qiaochu, I just mean I've no counterexample. $\endgroup$ Nov 28, 2014 at 23:08
  • $\begingroup$ How do you read $\times$-compact? "product compact"? Did you invent this definition? $\endgroup$ Nov 28, 2014 at 23:39

1 Answer 1


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).

  • $\begingroup$ Thanks a lot for this comment. I'll edit definition ASAP to elude this trick. $\endgroup$ Nov 29, 2014 at 9:14
  • 1
    $\begingroup$ @Ivan: I edited my answer to prove that finitely-generated groups are $\times$-compact. $\endgroup$
    – Seirios
    Nov 29, 2014 at 10:31
  • $\begingroup$ Thanks a lot. Do you think I should leave other conjectures of just eliminate 'em? $\endgroup$ Nov 29, 2014 at 10:45
  • $\begingroup$ It seems you answered all your questions, no? $\endgroup$
    – Seirios
    Nov 29, 2014 at 14:39
  • $\begingroup$ Almost, but this comment was before counterexample to 3 and 4. $\endgroup$ Nov 29, 2014 at 15:55

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.