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Which of the following statements is/are true?

a) There are infinitely many finite groups in which every non-identity element have order $2$.

b)There exists an infinite group in which every non-identity element have order $2$.

c)There exists an infinite group in which there are elements of order $n$ for all $n\in \Bbb N$.

d)There are infinitely many infinite groups in which each non-identity element have finite order.

My attempt: Option 2 seems correct as I have an example of such a group( Power set of Natural numbers under the binary operation of symmetric difference is one such example) but I am unable to conclude anything about other options.

About option 1, it seems that it is true keeping in mind $\Bbb Z_2\times \Bbb Z_2\times\cdots$ but still I am confused. For the last two options, I don't have any conclusive idea which I can apply here. Please guide.

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    $\begingroup$ They are all true. $\endgroup$
    – Derek Holt
    Oct 31, 2020 at 9:12
  • $\begingroup$ "Infinitely many groups" should be "infinitely many isomorphism classes of groups"... $\endgroup$
    – YCor
    Nov 1, 2020 at 20:19

2 Answers 2

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(a) For $n\geq 1$, $\prod_{i=1}^n\Bbb{Z}_2$ is a finite group whose every nontrivial element has order $2$. So there are infinitely many such groups.

(b) $\prod_{i=1}^{\infty}\Bbb{Z}_2$ is the desired group.

(c) You may consider the example $\Bbb{Q}/\Bbb{Z}$. Then for every $n\in \Bbb{N}$, $\frac{1}{n}+\Bbb{Z}$ is element of order $n$ in $\Bbb{Q}/\Bbb{Z}$.

(d) For $n\geq 2$, $\prod_{i=1}^{\infty}\Bbb{Z}_n$ is an infinite group whose every nontrivial element has finite order. So there are infinitely many such groups.

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  • $\begingroup$ In (b) is there an explicit definition of the set? I mean how will the elements look like? $\endgroup$
    – Saikat
    Oct 31, 2020 at 8:36
  • $\begingroup$ I think for $(d)$ you want something like $\prod _ {i=n}^\infty\Bbb Z_i$. $\endgroup$
    – user403337
    Oct 31, 2020 at 9:08
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    $\begingroup$ I think for (c) you want a single group in which there are elements of order $n$ for all $n$, such as ${\mathbb Q}/{\mathbb Z}$ under addition. $\endgroup$
    – Derek Holt
    Oct 31, 2020 at 9:10
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    $\begingroup$ @ChrisCuster I think the given example is correct for (d). $\endgroup$
    – Derek Holt
    Oct 31, 2020 at 9:11
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    $\begingroup$ @ShashankDwivedi Yes, in d) with $n=4$, there are elements with orders $1,2$ and $4$. What's the problem with that? And yes, when $n=p$, all nontrivial elements have order $p$. By the way, in all of these examples with direct products, you could also use the direct sum (elements in which all but finitely many components are trivial). $\endgroup$
    – Derek Holt
    Oct 31, 2020 at 13:04
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If it's easier for you to conceptualize, you can take each element to be finite, even though the group as a whole is infinite.

For instance, for (2), we can take each element of $G$ to be a finite sequence consisting of $1$ and $-1$ and ending in $-1$ (the identity is the null sequence). To find $ab$, we multiply the values of $a$ and $b$ pairwise, then terminate the sequence at the last $-1$. If they are of different lengths, we append $1$ to the shorter one enough times to make them the same length, then multiply. $G$ is infinite because although each individual element has a finite length, there's an infinite number of those finite lengths.

A similar strategy works for (3) and (4). For (3), we take finite sequences in which the $n$th value is an element of $\mathbb Z_n$, unless the sequence is length 1, the last value isn't $1$. For (4), we simply take variations on (3). For instance, we can start the sequence at $\mathbb Z_k$. Or take $k$ copies of (3), etc.

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