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I'm reading "Contemporary Abstract Algebra," by Gallian.

This is Exercise 4.46.

Which of the following numbers could be the exact number of elements of order $21$ in a group: $21600, 21602, 21604$?

My Attempt:

Lemma: In a finite group, the number of elements of order $d$ is a multiple of $\varphi(d)$.

(This is a Corollary of Theorem 4.4 ibid.)

Since $\varphi(21)=12$ and $12\mid 21600$ but not the other two candidate numbers, the lemma above implies that $21600$ is the exact number of elements of order $21$ for some (finite) group.


I'm quite sure I got this right$^{\dagger}$. I intuited the result, having forgotten the Lemma (well, Corollary) above.

This question is just so I can make note of it, really, but I suppose I could use the post to ask the following:

Is the assumption that the group is finite necessary?

My guess is that it's not. I have a rough idea of using a presentation with some free generators in such a way that it has the required number of order $21$ elements all within some finite subgroup (and no other such elements in the group given by the presentation).

Please help :)


Here $\varphi$ is Euler's totient function.

$\dagger$ The Dunning-Kruger effect in action . . . It's the right number, yeah, but, as pointed out below, I didn't show that such a group actually exists.

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    $\begingroup$ I don't know if I'm being naive here, but if you had some infinite group $G$ and you were concerned about the possible precise number of elements of order $d$ that $G$ could have (given that the number of such elements is finite), could you not just consider $H$ to be the smallest subgroup of $G$ containing all elements of order $d$ (or even just the subgroup generated by the elements of order $d$) and then apply your lemma? $\endgroup$ – Adam Higgins Jan 15 at 15:03
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    $\begingroup$ @AdamHiggins $H$ is not necessarily finite: for example, if we take the group with presentation $\langle a, b | a^{21}, b^{21}\rangle$, then no finite (indeed, no proper) subgroup contains both $a$ and $b$. $\endgroup$ – user3482749 Jan 15 at 15:04
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    $\begingroup$ @AdamHiggins $H$ might still be infinite. $\endgroup$ – Lord Shark the Unknown Jan 15 at 15:04
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    $\begingroup$ Thanks all! My bad! $\endgroup$ – Adam Higgins Jan 15 at 15:05
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    $\begingroup$ You also need to show an example to a group such that the number of elements of order $21$ is exactly $21600$. The lemma you cited only shows that the other two possibilities are wrong, it does not give you that the remaining one is surely right. $\endgroup$ – A. Pongrácz Jan 15 at 15:29
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There is no need to assume that $G$ is finite. For an arbitrary group $G$, and some fixed $n >0$, let $G_n$ be the set of elements of $G$ of (finite) order $n$. Then you can show that, for $g,h \in G_n$, $g$ is a power of $h$ if and only if $h$ is a power of $g$, so "is a power of" is an equivalence relation on $G_n$, and all equivalence classes have size $\phi(n)$.

So, in particular, if $G_n$ is finite, then $|G_n|$ is a multiple of $\phi(n)$.

On the other hand, if $G_n$ is finite, then $C_G(G_n)$ has finite index in $G$, and I think that implies that $\langle G_n \rangle$ is finite.

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The thing missing from your argument: there is a group such that there are exactly $21600$ elements of order $21$, and it is not so easy to find one.

Observe that $21600= 48\cdot 450$.

In $Z_7\times Z_7$, there are $48$ elements of order $7$.

In the nontrivial semidirect product $Z_{151}\rtimes Z_3$, the possible orders of elements are $1,3$ and $151$, as $151$ is a prime, and the group is not cyclic. There is such a nontrivial semidirect product as $151\equiv 1 \pmod 3$. Elements of order $1$ and $151$ are exactly those in the normal subgroup of order $151$. Thus there are exactly $3\cdot 150= 450$ elements of order $3$ in $Z_{151}\rtimes Z_3$.

Hence, in $(Z_7\times Z_7)\times (Z_{151}\rtimes Z_3)$, there are exactly $21600$ elements of order $21$. (Think about it.)

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    $\begingroup$ It depends on which one is the normal subgroup. In my case, the thing on the left is the normal subgroup, so I had to use $\rtimes$. (The triangle points towards the normal subgroup.) $\endgroup$ – A. Pongrácz Jan 15 at 15:53
  • $\begingroup$ (Sorry: I deleted my last comment once I saw you'd edited your answer.) Yes, that's how I remember it too; I still get them mixed up every once in a while :) $\endgroup$ – Shaun Jan 15 at 15:54

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