F2 - properties

order of an element in a direct product

the order of an element in a direct product of a finite number of finite groups is the least common multiple of the orders of the components of the elements

| (g_{1}, g_{2}, \dots, g_{n}) | = lcm (| g_{1} |, | g_{2} |, \dots, | g_{n} |)

let $m, n \in Z > 0$ , divisible by $5$ , and the number of elements in $Z_{m} \oplus Z_{n}$ of order $5$ need to be determined, say $(a, b)$ such that $| (a, b) | = 5 = lcm (| a |, | b |)$
this requires $| a | = 1$ and $| b | = 5$ or vice versa, but not $(a, b) = (0, 0)$
since both $Z_{m}$ and $Z_{n}$ each have a unique subgroup of order 5, there are exactly five choices for $a$ and five for $b$ , thus $25$ choices for $(a, b)$ , including $(0, 0)$
thus, there are $24$ elements in $Z_{m} \oplus Z_{n}$ of order $5$
in general, if $m, n \in Z > 0$ and divisible by a prime $p$ , the number of elements of order $p$ in $Z_{m} \oplus Z_{n}$ is $p^{2} - 1$
to determine the number of cyclic subgroups of order $10$ in $Z_{150} \oplus Z_{50}$
so, $| (a, b) | = lcm (| a |, | b |) = 10$ , which restricts $| a |$ and $| b |$ to $1, 2, 5, 10$
$a$ must divide $10$ , in $Z_{150}$ , $10 a \equiv 0 (mod 10)$ , requiring $a \in ⟨ 15 ⟩$ , and similarly, $b \in ⟨ 5 ⟩$
thus, $(a, b) \in ⟨ 15 ⟩ \oplus ⟨ 5 ⟩ ≅ Z_{10} \oplus Z_{10}$ , which has elements of order $1, 2, 5, 10$
it has $24$ and $3$ elements of orders $5$ and $2$ , so there must be $(100 - 24 - 3 - 1 =) 72$ elements of order $10$
each subgroup of order $10$ has four order- $10$ elements, and no two cyclic subgroups can have an element in common, there must be $(72 / 4 =) 18$ cyclic subgroups of order $10$
this is analogous to counting the number of legs and dividing by $4$ to get the number of sheep
for each divisor $r$ of $m$ and $s$ of $n$ , the group $Z_{m} \oplus Z_{n}$ has a subgroup isomorphic to $Z_{r} \oplus Z_{s}$
eg: to find a subgroup of $Z_{30} \oplus Z_{12}$ isomorphic to $Z_{6} \oplus Z_{4}$ , note that $⟨ 5 ⟩$ of order $6$ and $⟨ 3 ⟩$ of order $4$ are subgroups of $Z_{30}$ and $Z_{12}$ respectively, so $⟨ 5 ⟩ \oplus ⟨ 3 ⟩$ is the desired subgroup

criterion for

G \oplus H

to be cyclic

let $G$ and $H$ be finite cyclic subgroups, then $G \oplus H$ is cyclic if and only if $| G |$ and $| H |$ are relatively prime

criterion for

\oplus_{i = 1}^{n} G_{i}

to be cyclic

$\oplus_{i = 1}^{n} G_{i}$ for finite $n$ and cyclic $G_{i}$ 's is cyclic if and only if $G_{i}$ and $G_{j}$ are relatively prime when $i \neq j$

criterion for

Z_{n_{1} n_{2} \dots n_{k}} ≅ Z_{n_{1}} \oplus Z_{n_{2}} \oplus \dots \oplus Z_{n_{k}}

let $m = n_{1} n_{2} \dots n_{k}$ , then $Z_{m}$ is isomorphic to $\oplus_{i = 1}^{k} Z_{n_{i}}$ if and only if $n_{i}$ and $n_{j}$ are relatively prime when $i \neq j$