As I was taught, any finite abelian group $G$ can be represented up to isomorphism by a direct product of cyclic integer groups of prime power (there is some canonical fuss there, but that's only by convention). So then, one can write:
$$G\sim \mathbb{Z}_{p_1}\times\mathbb{Z}_{p_2}\times\dots\times\mathbb{Z}_{p_n}$$
Now, on the Wikipedia page about abelian groups, I have two things confusing me.
The first is that they sometimes write $\mathbb{Z}_{p_1}\times\mathbb{Z}_{p_2}$, and other times $\mathbb{Z}_{p_1}\oplus\mathbb{Z}_{p_2}$. I don't know whether that's a matter of notation, or that it has actual meaning (e.g. having addition vs. multiplication in the end result or in the individual factors). I was taught using $\times$ only.
Secondly, it is stated that:
For another example, every abelian group of order $8$ is isomorphic to either $\mathbb {Z} _{8}$ (the integers $0$ to $7$ under addition modulo $8$), $\mathbb {Z}_{4}\oplus \mathbb {Z} _{2}$ (the odd integers $1$ to $15$ under multiplication modulo $16$), or $\mathbb {Z} _{2}\oplus \mathbb {Z}_{2}\oplus \mathbb {Z} _{2}$.
I take it that whatever is in parentheses is an example of such a group. However, how do I know which group operation is used in the factors of these decompositions, and in the end result? Neither the page nor my Abstract Algebra textbook mentions this explicitly, as far as I can tell. However, there is a big difference between $\langle\mathbb{Z}_p, +\rangle$ and $\langle\mathbb{Z}_p, \cdot\rangle$:
- For one, we don't actually consider the $0$ in $\langle\mathbb{Z}_p, \cdot\rangle$ to be part of the group, making it (or, if you will, the multiplicative group inside it) of order $p-1$ instead of $p$; there can't be a bijection between two finite groups of unequal order, so this would change the isomorphism the theorem says there is.
- For two, all $\langle\mathbb{Z}_n, +\rangle$, regardless of $n$ (generated by $1$) are cyclic. This is the case for $\langle\mathbb{Z}_p, \cdot\rangle$ if $p$ is prime, but canonically, we have structures like $\mathbb{Z}_4$ appear in the decomposition, which is not cyclic under multiplication. This also affects any isomorphism.
Which operations are used for the factors and in the result, and am I correct that they matter after all?
Best Answer
The direct product and the direct sum always coincide on finite abelian groups, so you can use either.
About your second question: the examples inside the brackets actually tell you what the operation is. For $\mathbb{Z}_8$ it says "addition modulo 8". Indeed, you can verify that $1$ is a generator (as $n=1+1+...+1$, $n$ times) and therefore that group is cyclic of order $8$.
If you meant "how do I know in the general case", when you are given an abelian group you are given elements and a group operation. Every subgroup that appears in the decomposition inherits that group operation, so there is no ambiguity.
Many cyclic groups can be realised as additive groups $(\mathbb{Z}_p, +)$ and also as multiplicative groups $(\mathbb{Z}_q^\times, \cdot)$. This is not a contradiction, as the group operation does not distinguish between $+$ and $\cdot$ when you consider the factors just up to isomorphism.