Let $M$ be a free $\mathbb{Z}$-module. Let $a_1,a_2\in \mathbb{N}$ such that $\gcd(a_1,a_2)=1$. Then $${M}/{a_1a_2M}\cong {a_1M}/{a_1a_2M} \oplus {a_2M}/{a_1a_2M}.$$
Is the above statement true? If not, what kind of restrictions we need to consider to make it true?
Best regards.
Let me write $a=a_1$ and $b=a_2$, for simplicity and let's assume $a\ne0$ and $b\ne0$. There is an obvious homomorphism $$ f\colon\frac{aM}{abM}\oplus\frac{bM}{abM}\to\frac{M}{abM} $$ defined by $$ f(ax+abM,by+abM)=ax+by+abM $$ Let $z\in M$; can we find $x,y\in M$ such that $f(ax+abM,by+abM)=z+abM$? Since $\gcd(a,b)=1$, we can write $1=ah+bk$, so $z=ahz+bkz$.
What's the kernel of $f$? $$ f(ax+abM,by+abM)=0+abM $$ if and only if $ax+by\in abM$, that is $ax+by=abz$ for some $z\in M$, which implies $$ by=a(x-bz) $$ Write $x=\sum_{i\in I}\alpha_iv_i$, $y=\sum_{i\in I}\beta_iv_i$ and $z=\sum_{i\in I}\gamma_iv_i$, where $(v_i)_{i\in I}$ is a basis for $M$. Then $$ b\beta_i=a(\alpha_i-b\gamma_i)\qquad(i\in I). $$ Can you go on?