List all abelian groups (up to isomorphism) of the given orders: a) $144$, b) $600$
a) For order $144$, I feel confident with this one so far:
- $\mathbb{Z}_4 \oplus \mathbb{Z}_{36}$ Elementary divisors: $2^2,6^2$. Invariant factors: $144$.
- $\mathbb{Z}_4 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_6$. Elementary divisors: $2^2,6,6$. Invariant factors: $6,24$.
- $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_{36}$. Elementary divisors: $2,2,6^2$. Invariant factors: $2,72$.
- $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_6$. Elementary divisors: $2,2,6,6$. Invariant factors: $12,12$.
Therefore, we have
- $\mathbb{Z}_4 \oplus \mathbb{Z}_{36} \cong \mathbb{Z}_{144}$
- $\mathbb{Z}_4 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_6 \cong \mathbb{Z}_{6} \oplus \mathbb{Z}_{25}$
- $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_{36} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_{72}$
- $\mathbb{Z}_4 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_6 \cong \mathbb{Z}_{12} \oplus \mathbb{Z}_{12}$
b) For order $600$, I am not so sure with this one:
- $\mathbb{Z}_{100} \oplus \mathbb{Z}_6$ Elementary divisors: $10^2,6$. Invariant factors: $600$.
- $\mathbb{Z}_{10} \oplus \mathbb{Z}_{10} \oplus \mathbb{Z}_6$. Elementary divisors: $10,10,6$. Invariant factors: $10,60$.
Thus,
- $\mathbb{Z}_{100} \oplus \mathbb{Z}_6 \cong \mathbb{Z}_{600}$
- $\mathbb{Z}_{10} \oplus \mathbb{Z}_{10} \oplus \mathbb{Z}_6 \cong \mathbb{Z}_{10} \oplus \mathbb{Z}_{60}$
I feel confident with part a, but with part b I feel it's incomplete.
Your last $$\mathbb Z_4\times \mathbb Z_2\times \mathbb Z_6\times \mathbb Z_6$$ has order 288. I assume that instead of $\mathbb Z_4$ you meant $\mathbb Z_2$.
Also, $$\mathbb Z_4\times \mathbb Z_{36}\not \cong \mathbb Z_{144}$$ since $\gcd(4, 36) = 4$.
I make use of the fact that $$\mathbb Z_{mn} \cong \mathbb Z_m \times \mathbb Z_n\;\;\text{if and only if}\;\;\gcd(m, n) = 1$$
$$\mathbb{Z}_4 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_6 \not \cong \mathbb{Z}_{6} \oplus \mathbb{Z}_{24}$$ because $\gcd(4, 6) = 2 \neq 1$.
$$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_6 \not\cong \mathbb{Z}_{12} \oplus \mathbb{Z}_{12}$$ because $\gcd(2, 6) = 2 \neq 1$.
Also, $3\mid 144,\;6\mid 144,\; 8\mid 144,\; 16\mid 144$...
$$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_{36} \not \cong \mathbb{Z}_2 \oplus \mathbb{Z}_{72}$$ since $\gcd(2, 36) = 2\neq 1$.
The prime factorization of $144$ is given by: $$144 = 2^43^2$$