Let, $n_1$ and $n_2$ be a prime numbers $$ K_{n_1 n_2 *}: \cdots \xrightarrow{\text{$\partial_1$}}\mathbb{Z_{n_1n_2}} \xrightarrow{\text{$\partial_2$}} \mathbb{Z_{n_1n_2}} \xrightarrow{\text{$\partial_1$}} \mathbb{Z_{n_1n_2}} \xrightarrow{\text{$\partial_2$}} \cdots $$ be a sequence of groups, where $\partial_i : \mathbb{Z_{n_1n_2}} \to \mathbb{Z_{n_1n_2}}$ is defined as $\partial_i(\bar{x}) = n_i\bar{x} = \bar{n_ix}$ , for $i =1,2$. Observe that $\partial_2 \partial_1 $ is zero, so this is a chain complex. I'm trying to compute the homology of $K_{n_1 n_2 *}$.
My attempt
$\text{Ker}(\partial_1) =\{\bar{x} \in \mathbb{Z_{n_1n_2}}: \partial_1(\bar{x}) = n_1 \bar{x} = 0 \}$, so this is $\mathbb{Z_{n_2}}$. The same holds for $\text{Ker} \partial_2$, i.e, $\text{Ker} \partial_2= \mathbb{Z_{n_1}}$. The image of $\partial_1$ is $\mathbb{Z_{n_1}}$ and image of $\partial_2$ is $\mathbb{Z_{n_2}}$. The homology of $K_{n_1 n_2 *}$ is $$ \text{H}_n(K_{n_1 n_2 *}) = \left\{ \begin{eqnarray} \mathbb{Z_{n_2}}/\mathbb{Z_{n_1}},&\quad \text{if} \quad& i=1\\ \mathbb{Z_{n_1}}/\mathbb{Z_{n_2}} &\quad \text{if} \quad& i=2 \end{eqnarray}\right. $$
This is true? I'm new in homological algebra and its hard to me.
This is not true. $\Bbb Z_{n_2}/\Bbb Z_{n_1}$ doesn't really make sense to me.
N.B. we should assume $n_1,n_2$ are distinct primes. Or, that they are just coprime integers.
The homology is actually always trivial. Because $\ker\partial_2=\Bbb Z_{n_1}$ and $\operatorname{im}\partial_1=\Bbb Z_{n_1}$, we just get $\Bbb Z_{n_1}/\Bbb Z_{n_1}=0$, and similarly for the other case.