# Finite/Infinite groups

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.

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

(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.

• In (b) is there an explicit definition of the set? I mean how will the elements look like? Oct 31, 2020 at 8:36
• I think for $(d)$ you want something like $\prod _ {i=n}^\infty\Bbb Z_i$.
– user403337
Oct 31, 2020 at 9:08
• 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. Oct 31, 2020 at 9:10
• @ChrisCuster I think the given example is correct for (d). Oct 31, 2020 at 9:11
• @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). Oct 31, 2020 at 13:04

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.