A question about cyclic groups and their isomorphisms

Question

If I generate some cyclic groups, $C_{8k}$ with $k\in \mathbb{N}$, can these ever be isomorphic to a direct sum of other cyclic groups?

I'm generating these groups using rotation matrices corresponding to the angles $\frac{2\pi n}{8k}, n=0,1,\ldots,k-1.$

Similarly for $C_{4\left(2k+1\right)}$.

Thank you for any help or direction.

Answer 1

Cyclic groups can often be represented as direct sums of other cyclic groups. Here's the general result:

1. If $m, n$ are relatively prime then $C_n \oplus C_m \cong C_{nm}$.
2. If $m, n$ are not relatively prime then $C_n \oplus C_m$ is not cyclic.

The proof of these statements will both rely on the following fact: if $a, b \in G$ are elements of finite order and $ab=ba$ then $|ab| = lcm(|a|,|b|)$, where $|a|$ is the order of $a$. Recall furthermore that $lcm(n, m) = nm/gcd(n,m)$.

Now, let $m, n$ be relatively prime. Then $(1, 0) \in C_n \oplus C_m$ has order $n$ and $(0, 1)$ has order $m$. Then as stated above, $|(1, 1)| = lcm(n, m) = nm$. Since $|C_n \oplus C_m| = nm$, this element generates the group and it is therefore cyclic. This, by the way, is part of the Chinese Remainder Theorem.

Suppose now that $(a, b) \in C_n \oplus C_m$ and let $|a| = d$, $|b| = e$. Then of course, $d \mid n, e \mid m$. Now, we have $|(a, b)| = de/gcd(d,e)$. If $d < n$ or $e < m$ then certainly $de/gcd(d,e) < nm$. If $d=n, e=m$ then $|(a, b)| = nm/gcd(n,m) < nm$ as we assumed $gcd(n,m) > 1$. Hence, no element of $C_n \oplus C_m$ has order $nm$ so it is not cyclic.