This is Exercise 8.10 of Gallian's "Contemporary Abstract Algebra". Answers that use only the preceding material from the textbook are preferred.

The Question:

How many elements of order $9$ does $G=\Bbb Z_3\times\Bbb Z_9$ have? (Do not do this exercise by brute force.)

More to the point . . .

Is my calculation correct?

My Attempt:

Since $\lvert (g, h)\rvert=\operatorname{lcm}(\lvert g\rvert, \lvert h\rvert)$ for $(g, h)\in G$ by a lemma from Section 8 ibid., we have two cases to consider:

  • $\lvert g\rvert=3$ and $\lvert h\rvert=9$: There are two options for such a $g$ and $\varphi(9)=6$ options for $h$. This case thus gives $12$ elements of order $9$ in $G$.

  • $\lvert g\rvert=1$ and $\lvert h\rvert=9$: There is only one such $g$ but, as before, $6$ such $h$. This case gives $6$ elements of order $9$ in $G$.

Therefore, there are $18$ elements of order $9$ in $G$. $\square$


I just followed the examples given in the section tackling similar problems. Thus I'm not as sure as usual about the reasoning here.

At least my calculation is consistent with the following lemma:

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

Reference: Corollary of Theorem 4.4 ibid.

Please help :)

  • $\begingroup$ I could check with GAP quite easily, come to think of it. That's cheating though. $\endgroup$ – Shaun Feb 24 at 18:02

Your argumentation is correct, but you could make it shorter using Lagrange's theorem this way: the order of a $g\in \mathbf Z_3$ is anyway a divisor of $3$, hence of $9$. Therefore the solutions are $$\{(g,h)\in \mathbf Z_3\times\mathbf Z_9\mid |h|=9\}$$ and there are $3\cdot\varphi(9)$ of them.


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