HINTS:
You'll want to decompose the groups, where appropriate.
Definition: A group $G$ is decomposable if (and only if) it is isomorphic to a direct product of two proper subgroups. Otherwise, $G$ is indecomposable.
- Note, for example, that from this definition it follows that the factor $\mathbb{Z}_{49} = \mathbb{Z}_{7^2}$ in the third and fourth groups is indecomposable.
In fact, the finite indecomposable groups are exactly the cyclic groups with order a power of a prime.
Recall that $\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn}$ if and only if $\text{ gcd}(m, n)=1$.
- For example, $\mathbb{Z}_5 \times \mathbb{Z}_9 \cong \mathbb{Z}_{45}$. Note that $\text{gcd}(5, 9) = 1.$
More generally, the group $\prod_{i=1}^n \mathbb{Z}_{m_i} \cong \mathbb{Z}_{m_1m_2\cdots m_n}$ if and only if the numbers $m_i,\;\text{for}\;i = 1, 2, \cdots n$ are pairwise relatively prime (that is, if and only if for each $m_i, m_j,\; 1 \leq i \leq n,\; 1\leq j\leq n,\; i\neq j,\;\text{ gcd}(m_i, m_j) = 1$.)
Take, as an example, the last group: $\mathbb{Z}_{175}\times\mathbb{Z}_{45} \times \mathbb{Z}_{49}$
To decompose this group, note:
- $\mathbb{Z}_{175} \cong \mathbb{Z}_{25} \times \mathbb{Z}_7 = \mathbb{Z}_{5^2} \times \mathbb{Z}_7$
- $\mathbb{Z}_{45} \cong \mathbb{Z}_{3^2} \times \mathbb{Z}_5$,
- $\mathbb{Z}_{49} \cong \mathbb{Z}_{7^2}$
Rearranging factors gives us: $\mathbb{Z}_{175}\times\mathbb{Z}_{45} \times \mathbb{Z}_{49}\cong \mathbb{Z}_{3^2} \times \mathbb{Z}_5 \times \mathbb{Z}_{5^2} \times \mathbb{Z}_7 \times \mathbb{Z}_{7^2}$
Doing this with each of the groups you list will make it clear which group(s) are isomorphic to which groups.