https://en.wikipedia.org/wiki/Abelian_group
I don't understand this statement from Wikipedia page: "Every abelian group of order 8 is isomorphic to either $ \Bbb Z_8\ $ (the integers 0 to 7 under addition modulo 8), $ \Bbb Z_4\ \oplus \Bbb Z_2\ $ (the odd integers 1 to 15 under multiplication modulo 16), or $ \Bbb Z_2\ \oplus \Bbb Z_2\ \oplus \Bbb Z_2\ $ ."
Isn't $ \Bbb Z_4\ $ the integers 0 to 3 under addition modulo 4 i.e. {0,1,2,3} ? What did I misunderstand?
I thought that $ \Bbb Z_4\ \oplus \Bbb Z_2\ $ should be {0,1,2,3}$\oplus${0,1} but obviously there is some misunderstanding from me, especially since it also says that The cyclic group $\Bbb {Z} _{mn}$ of order mn is isomorphic to the direct sum of ${\Bbb Z} _{m}$ and ${\Bbb Z} _{n}$ and then uses as an example $\Bbb Z_{15} \cong \{0,5,10\} \oplus \{0,3,6,9,12\}$ but {0,5,10} is not ${\Bbb Z} _{3}$, it is just order 3.
Please explain...
You are right to the extent that $\Bbb Z_4\times\Bbb Z_2$ is indeed (by definition!) the cartesian product of the additive groups of integers modulo $4$ and modulo $2$.
What the Wiki page is suggesting is that you can find a "concrete" incarnation of this group by looking at the multiplicative group $U_{16}$ of the invertible elements in $\Bbb Z_{16}$ which has clearly $8$ elements (represented by the odd natural numbers from $1$ up to $15$) and it is isomorphic to $\Bbb Z_4\times\Bbb Z_2$.
For, consider that $U_{16}$ has elements of exact order $4$ (e.g. $\bar 3$) and $3$ elements of exact order $2$, namely $\bar 7$, $\bar 9$ and $\overline{15}$. I'll leave to you the task to write down an explicit isomorphism between $U_{16}$ and $\Bbb Z_4\times\Bbb Z_2$.