Using the Fundamental Theorem of Abelian Groups to determine possible non isomorphic groups

Question

Trying to reconcile two things.

So the fundamental theorem of Abelian groups states that if G is an Abelian group with order $n$, then there are prime numbers $p_1, p_2, . . . , p_k$ , not necessarily distinct, such that $G ≈ \mathbb{Z}_{p1^{n1}} ⊕ \mathbb{Z}_{p2^{n2}} ⊕ \cdot \cdot \cdot· ⊕ \mathbb{Z}_{pk^{nk}}$ and $n = p_{1}^{n_1} \cdot p_{2}^{n_2} \cdot \cdot \cdot p_{k}^{n_k}$.

And using this fact, we can determine all non isomorphic groups of an Abelian group $G$.

Now I have two examples that seem to conflict

Firstly, for this we break the order in terms of its prime factors, and go from there ($8, 4\cdot2, 2\cdot2\cdot2$, $9, 3\cdot3$ and $7$). But why isn't $Z_{504}$ listed? Is that not a possible group G is isomorphic to?

And then here,

Why is $Z_{121}$ now listed? 121 is not prime so going off the previous examples I wouldn't list it as a group. Or am I missing some context here (because they don't specify non-isomorphic).

Would appreciate clarification!

Answer 1

It is listed. It is just written a little funny: there is a homomorphism

$$\mathbb Z \to \mathbb Z_8 \oplus \mathbb Z_9 \oplus\mathbb Z_7$$ given by $1 \mapsto (1 \mod 8, 1 \mod 9, 1 \mod 7)$ (basically send $1 \mapsto (1,1,1)$ and reduce accordingly.)

The chinese remainder theorem ensures that this is surjective, since all components are pairwise coprime.

It's kernel is precisely $504 \mathbb Z$, so the first isomorphism theorem tells that $\mathbb Z_{504} \cong \mathbb Z_8 \oplus \mathbb Z_9 \oplus\mathbb Z_7$.

The reason the same thing doesn't go through for $\mathbb Z_{121}$ and $\mathbb Z_{11} \oplus \mathbb Z_{11}$ is because the factors are not coprime. In particular, the same map clearly fails, since $1 \mapsto (1,1)$ is not surjective, and in fact its image is just $\mathbb Z_{11}$. In fact, we can see that $\mathbb Z_{121}$ is not isomorphic to $\mathbb Z_{11} \oplus \mathbb Z_{11}$, since the latter is not even cyclic.

Answer 2

The following sum of $\ 11\ $ summands $\ \mod 121\ $ is not zero:

$$ [1]_{121}\ +\ldots\ +\ [1]_{121}\,\ =\,\ [11]_{121}\,\ \ne\,\ [0]_{121} $$

however, the sum of $\ 11\ $ identical summands from $\ \mathbb Z_{11}\times\mathbb Z_{11}\ $ is zero:

$$ ([a]_{11}\ [b]_{11})\ +\ \ldots\ +\ ([a]_{11}\ [b]_{11})\,\ = \,\ ([0]_{11}\,\ [0]_{11}) $$

On the other hand, for two different primes $\ p\ne q\ $ you have group isomorphism:

$$ \mathbb Z_p \times \mathbb Z_q\,\ \longleftrightarrow \,\ \mathbb Z_{p\times q} $$

Just consult Euclid (ok, Euclid has never existed; thus consult Eratosthenes).