# Which of these numbers could be the exact number of elements of order $21$ in a group?

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).

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.

• 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? – Adam Higgins Jan 15 at 15:03
• @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 $b$. – user3482749 Jan 15 at 15:04
• @AdamHiggins $H$ might still be infinite. – Lord Shark the Unknown Jan 15 at 15:04
• Thanks all! My bad! – Adam Higgins Jan 15 at 15:05
• 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. – A. Pongrácz Jan 15 at 15:29

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.

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.)

• 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.) – A. Pongrácz Jan 15 at 15:53
• (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 :) – Shaun Jan 15 at 15:54