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I am working on the following problem.

Find the number of the elements of order $4$ in $(\mathbb Z/525\mathbb Z)^\times$.

I tried to solve it in the following way: since $525=3\cdot5^2\cdot7$, we have by the CRT $$ (\mathbb Z/525\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z)^\times \times (\mathbb Z/25\mathbb Z)^\times \times (\mathbb Z/7\mathbb Z)^\times. $$

By the fact stated by a Wikipedia article, the constituent groups on the RHS are isomorphic to the cyclic groups of order $2$, $20$, $6$, respectively. Thus, the order of an element in $(\mathbb Z/525\mathbb Z)^\times$ is the least common multiple of some subset $S\subseteq\{2, 20, 6\}$, which can never be $4$. In conclusion, the number of the elements of order $4$ in $(\mathbb Z/525\mathbb Z)^\times$ is $0$.

My question is whether my reasoning is correct (which I doubt because the result is so trivial). I would also like to ask where accessible proofs of the fact on Wikipedia (preferably on the Web) can be found.

I would appreciate your help.

user26857
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Pteromys
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1 Answers1

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Your reasoning is flawed; while it's true that all of the (non-trivial) elements in the cyclic group $(\mathbb{Z}/p\mathbb{Z})$ are of order $p$ when $p$ is prime, that same fact doesn't hold true for arbitrary groups $(\mathbb{Z}/n\mathbb{Z})$. Indeed, if the order of the element $g\in(\mathbb{Z}/n\mathbb{Z})$ is $n$, then the order of $g^k$ is $\frac{n}{\gcd(n,k)}$ (why?). In particular, there are some elements of order $4$ in $(\mathbb{Z}/25\mathbb{Z})^\times$ (how many?) and so there are elements of order $4$ in $(\mathbb{Z}/525\mathbb{Z})^\times$ (why must there be?). Your notion of using the $LCM$ is essentially solid when restricted to individual elements of the three 'constituent' groups, though, and you should be able to port that step over to determine how many elements there are.

Steven Stadnicki
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  • Thank you. Is it correct that (ℤ/25ℤ)× has two elements of order 4, and the other 'constituents' have two elements of order that divides 4 each, and thus the answer is $2^3 = 8$? – Pteromys Aug 18 '13 at 06:26
  • @Pteromys I haven't studied it very closely, but that looks correct to me at first glance. – Steven Stadnicki Aug 18 '13 at 06:56